144Qs

TEST

Ongoing dataset for big mc review

Revised outline

  1. Biochemistry
  2. Cells|Membranes
  3. Enzymes|Respiration
  4. DNA|RNA|Protein
  5. Mitosis|Meiosis|Genetics
  6. Evolution
  7. Animal Systems
  8. Plant Systems
  9. Ecology

Link to a genetic code table

[qwiz dataset=”144Qs” qrecord_id=”dank-144Qs”]

[q json=”true” multiple_choice=”true” unit=”Biochemistry” question_number=”1″ dataset_id=”144Qs|995ccc6720734″] In the diagram below, letter Y represents a(n)

[c] phosphate group

[f] No. The phosphate groups connect the deoxyribose sugars, and these sugars are represented as pentagons.

[c*] deoxyribose

[f] Yes. “Y” represents a deoxyribose sugar.

[c] nitrogenous base

[f] No. The nitrogenous bases form the internal “rungs” of the DNA ladder. “Y” and “X” make up the sides of the ladder. What makes up the sides?

[c] amino acid

[f] No. Amino acids are monomers of protein, and this is a molecule of DNA.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|995277725cb34″ question_number=”2″] In the diagram below, letter X represents a(n)

[c*] phosphate group

[f] Way to go. “X” represents a phosphate group.

[c] deoxyribose

[f] No. Deoxyribose is one of the sugars in DNA. “X” connects the sugars (shown at “Y”) together.

[c] nitrogenous base

[f] No. The nitrogenous bases form the internal rungs of the DNA ladder. “X” (along with “Y”) is on the outside of the ladder. What makes up the ladder?

[c] amino acid

[f] No. Amino acids are monomers of protein, and this is a molecule of DNA.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|9948227d98f34″ question_number=”3″] In the diagram below, letter Z represents a(n)

[c] phosphate group

[f] No. The phosphate groups are on the outside of DNA.

[c] deoxyribose

[f] No. Deoxyribose is the sugar on the outside of DNA.

[c*] nitrogenous base

[f] Excellent! “Z” is one of the four nitrogenous bases.

[c] amino acid

[f] No. Amino acids are monomers of protein. This is DNA.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|993da84816f34″ question_number=”4″] In the diagram below, the bonds connecting G and C (guanine and cytosine) are ________ bonds.

[c] covalent

[f] No. Covalent bonds are the strong bonds formed when atoms share electrons. They make up pretty much every bond in DNA, except the much weaker bonds between G and C. What are those weaker bonds (which play a key role in interactions between water molecules) called?

[c] peptide

[f] No. Peptide bonds are the bonds that hold amino acids together in proteins. There are no peptide bonds in DNA (because it’s not a protein, but a nucleic acid).

[c*] hydrogen

[f] Excellent! The bond between the nitrogenous bases C and G (cytosine and guanine) are hydrogen bonds.

[c] ionic

[f] No. Ionic bonds are the bonds formed when atoms trade electrons. Ionic bonds don’t play a role in DNA’s structure. The answer is a much weaker bond: the type that plays a key role in interactions between water molecules

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|99332e1294f34″ question_number=”5″] Which of the choices below correctly shows the sequence of processes that was carried out by the structure labeled “X” before it arrived at its current position?

[c] Combining with an amino acid and then binding to an anticodon.

[f] No. Structure X is carrying an amino acid, but it’s binding with a sequence of 3 letters on mRNA. What are those three letter sequences called?

[c] Binding to an anticodon and then combining with an amino acid.

[f] No. Structure X is carrying an amino acid, and it’s binding with a sequence of 3 letters on mRNA? What are those three letter sequences called?

[c] Binding to a codon and then combining with an amino acid.

[f] No, but you’re very close.The structure at X is a tRNA, and it does bind with both an amino acid and a codon. You just need to get the sequence of events right.

[c*] Combining with an amino acid and then binding to a codon.

[f] Correct! The structure at X is a tRNA. It first binds with an amino acid, and then it binds with a codon on the mRNA.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|9928b3dd12f34″ question_number=”6″] Which process is occurring below?

[c] DNA replication

[f] No. If this were DNA replication, you’d see a replication fork, nucleotides, and all types of enzymes. Here what you see is a ribosome, mRNA, and a tRNA carrying an amino acid. What process are these involved with?

[c] transcription of DNA into RNA

[f] No. If this were transcription of DNA into RNA, you’d see DNA, RNA, and enzymes like RNA polymerase.  Here what you see is a ribosome, mRNA, and a tRNA carrying an amino acid. What process are these involved with?

[c] transcription of RNA into protein

[f] No. Transcription involves changing DNA information into RNA information. Here what you see is a ribosome, mRNA, and a tRNA carrying an amino acid. What process are these involved with?

[c*] translation of RNA into protein

[f] Excellent! The image above shows a ribosome translating RNA information into protein.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|991cea60e0b34″ question_number=”7″] The structure labeled “X” below is a(n)

[c] polypeptide

[f] No. A polypeptide is a chain of amino acids, connected by peptide bonds. Above, all you see is one amino acid attached to structure X. During protein synthesis, what molecule brings amino acids to the ribosome?

[c*] tRNA

[f] Terrific! “X” is a tRNA.

[c] mRNA

[f] No. The mRNA is the single stranded nucleic acid with its 5′ and 3′ ends labeled. “X” is what brings amino acids to the mRNA at the ribosome.

[c] rRNA

[f] No. The rRNA is what makes up the ribosome. “X” is what brings amino acids to the mRNA at the ribosome.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|99108be1b5734″ question_number=”8″] To answer this question, you’ll need access to a genetic code table (clicking on the preceding link will open one up in a new tab). The third amino acid that would result from the process below would be

[c] guanine

[f] No. You can find the 3rd amino acid by identifying the third codon. It’s UAC. When you look up UAC in a genetic code table, what amino acid do you get?

[c*] tyrosine

[f] Excellent. The 3rd codon is UAC, which is translated into the amino acid tyrosine.

[c] cysteine

[f] No. You can find the 3rd amino acid by identifying the third codon. It’s UAC. When you look up UAC in agenetic code table, what amino acid do you get?

[c] proline

[f] No. You can find the 3rd amino acid by identifying the third codon. It’s UAC. When you look up UAC in agenetic code table, what amino acid do you get?

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|990611ac33734″ question_number=”9″] Transcription converts

[c] RNA into protein.

[f] No. Translation  converts information in RNA into sequences of amino acids, which make up protein. What does transcription do?

[c] protein into RNA.

[f] No. There’s no biological process that moves from protein to RNA.

[c*] DNA into RNA

[f] Excellent. Transcription converts DNA into RNA.

[c] codon sequences into amino acid sequences

[f] No. Codon sequences are translated into amino acid sequences during translation. What happens during transcription?

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|98fb9776b1734″ question_number=”10″] In eukaryotic cells, transcription moves information from ________ to _________.

[c*] the nucleus … the cytoplasm

[f] Nice job. In eukaryotic cells, transcription moves information from the nucleus to the cytoplasm.

[c] ribosomes … proteins

[f] No. The connection between ribosomes and proteins is more related to translation. In any case, ribosomes don’t have information, they translate it. Here’s a hint: transcription involves DNA, which makes up chromosomes. Where are those chromosomes found?

[c] the cytoplasm …. the nucleus

[f] No. During many cell signaling processes, information moves from the cytoplasm to the nucleus. But not in transcription. Here’s a hint: transcription involves DNA, which makes up chromosomes. Where are those chromosomes found?

[c] codon sequences … amino acid sequences

[f] No. That’s a good definition for the information transfer that happens during translation (or protein synthesis). Here’s a hint: transcription involves DNA, which makes up chromosomes. Where are those chromosomes found?

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|98f14281edb34″ question_number=”11″] In eukaryotic cells, translation moves information from ________ to _________.

[c] the nucleus … the cytoplasm

[f] No. Transcription is the process the moves information from the nucleus to the cytoplasm.

[c] ribosomes … proteins

[f] No, but you’re on the right track. Ribosomes are the translators, but they don’t provide the information that gets translated. That information is in mRNA, and its in 3 bases units. What are those units called?

[c] the cytoplasm …. the nucleus

[f] No. During many cell signaling processes, information moves from the cytoplasm to the nucleus. But not in translation. Choose another answer next time you see this question.

[c*] codon sequences … amino acid sequences

[f] Fabulous. Translation moves information from codon sequences to amino acid sequences.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|98e6c84c6bb34″ question_number=”12″] At stage I in the diagram below, the intron-free sequence of DNA would be

[c] TAC TAC GGC CTC CTG CCT GGG GGA CTA ACT

[f] No. Here’s how to do this. Write out the mRNA (shown at stage II). Then use the base pairing rules between RNA and DNA to figure out the DNA: A bonds with T, U with  A, C with G, and G with C

[c] TAC TAC GGC CTT CTG CCT CCC GGA CTA ACT

[f] No. Here’s how to do this. Write out the mRNA (shown at stage II). Then use the base pairing rules between RNA and DNA to figure out the DNA: A bonds with T, U with  A, C with G, and G with C

[c*] TAC TAC GGC CTT CTG CCT GGG GGA CTA ACT

[f] Excellent. You’ve used the the base pairing rules between RNA and DNA to figure out the DNA ( A bonds with T, U with  A, C with G, and G with C)

[c] TAC TAC GGC CTT CTG TTT GGG GGA CTA ACT

[f] No. Here’s how to do this. Write out the RNA (shown at stage II). Then use the base pairing rules between RNA and DNA to figure out the DNA: A bonds with T, U with  A, C with G, and G with C

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|98dc4e16e9b34″ question_number=”13″] In the diagram below, step E directly results in the production of

[c] messenger RNA

[f] No. Step “E” does produce RNA, but that RNA has to be processed before it’s ready to be messenger RNA and read by a ribosome. What would come before mRNA?

[c*] pre-messenger RNA

[f] Excellent. What you have after “E” is a raw RNA transcript called pre-mRNA. It still has to be processed before it can be mRNA, ready to be read by a ribosome.

[c] a polypeptide chain

[f] No. Polypeptide chains are sequences of amino acids, and that has to do with proteins. Here, your answer is going to be some type of RNA.

[c] complementary DNA (cDNA)

[f] No. Complementary DNA is used in genetic engineering to produce eukaryotic DNA that can be placed inside bacterial cells for replication, transcription, and translation. What you’re looking for here is the first result of transcription.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|98d1aea0a9734″ question_number=”14″] In the diagram below, an anti-codon is indicated by

[c*] group H

[f] Nice! Group H is an anti codon.

[c] product G

[f] No. G is mRNA. An anticodon is a short stretch of RNA on a tRNA.The anticodon complements a codon. Find a tRNA, and you’ll have your answer.

[c] structure K

[f] No. K is the ribosome. An anticodon is a short stretch of RNA on a tRNA. The anticodon complements a codon. Find a tRNA, and you’ll have your answer.

[c] bond J

[f] No. Bond “J” is a peptide bond, a bond that holds proteins together. An anticodon is a short stretch of RNA on a tRNA. The anticodon complements a codon. Find a tRNA, and you’ll have your answer.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|98c7346b27734″ question_number=”15″] The giant Pacific Octopus can lay up to 100,000 eggs (each about the size of a grain of rice). All the eggs have been fertilized with sperm from the same father. From the list below, the process that does not contribute to genetic diversity in the offspring is

[c] crossing-over during prophase 1.

[f] No. Crossing over occurs during meiosis. It involves mixing of maternal and paternal alleles, and it’s a major source of variation. What’s on the list that would decrease variation?

[c*] DNA repair during the S phase that precedes meiosis.

[f] Yes. DNA repair will take mutations and eliminate them, which reduces genetic diversity.

[c] random assortment of maternal and paternal chromosomes associated with metaphase 1 and anaphase 1

[f] No. Random assortment of maternal and paternal chromosomes (independent assortment) produces chromosomal combinations that are brand new…This is a major source of variation. What’s on the list that would decrease variation?

[c] random fertilization of gametes that follows meiosis

[f] No. Random fertilization of gametes increases variation, by combining the genes of unrelated mates. What’s on the list that would decrease variation?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|98bbdab12ff34″ question_number=”16″] Diagrams A, B and C below represent stages of cell division from the same organism. One of the three shows the first division of meiosis. Another one shows the second division of meiosis, and one shows division by mitosis.

What is the diploid number of chromosomes for this organism?

[c] 2

[f] No. Diploid means “two sets.” The diploid number can’t be two, because there are FOUR chromosomes in diagrams”A” and “C”  (which is a hint as to what the answer is).

[c*] 4

[f] Yes. Diploid means “two sets,” and you can see two sets of chromosomes in “A” and “C.”

[c] 6

[f] No. None of the cells above has six chromosomes. Remember that diploid means “two sets.” Take a good look at the diagram, find a cell with two sets of chromosomes, do some counting, and you’ll have your answer.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|98b13b3aefb34″ question_number=”17″] Diagrams A, B and C below represent stages of cell division from the same organism. One of the three shows the first division of meiosis. Another one shows the second division of meiosis, and one shows division by mitosis.

Which diagram shows a type of cell division that, in a multicellular organism, would be used for growth and repair.

[c*] A

[f] Yes! “A” looks like mitosis (metaphase, to be precise), which is used for growth and repair in eukaryotic organisms.

[c] B

[f] No. All of these cells are from the same organism. In cell “B,” there are only half of the chromosomes present (two as opposed to four). That makes B look like meiosis, which is used for reproduction, not growth and repair.

[c] C

[f] No. In cell B, it looks like homologous pairs are lining up and are about to be split apart. That’s something that happens in meiosis, which is used for reproduction, not growth and repair.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|98a69bc4af734″ question_number=”18″] The image below shows stages of cell division, but not necessarily in the correct order. Which of the following lists the correct sequence?

[c] 1-2-3-4

[f] No. You’ve chosen “1” as the first stage. But “1” shows separation of chromosomes (“anaphase.”) The first thing that happens when a cell divides is that the the nuclear membrane disintegrates. Then the chromosomes condense. Find cells that match those descriptions in the diagram, and remember these first two steps the next time you see this question.

[c] 2-1-4-3

[f] No. You’ve chosen “2” as the first stage. But “2” shows chromosomes condensing. Before that, the nuclear membrane disintegrates. Find a cell where that’s happening, and remember that first step the next time you see this question.

[c] 4-2-3-1

[f] No. You’ve chosen “4” as the first stage. But “4” shows chromosomes lined up along the cell equator, about to divide  (“metaphase.”) But before that could happen, the nuclear membrane would have to disintegrate, and then the chromosomes would condense. Find cells that match those descriptions in the diagram, and remember these first two steps the next time you see this question.

[c*] 3-2-4-1

[f] Excellent! Nice job identifying the correct sequence of events: nuclear membrane disintegration, chromosome condensation, alignment in the middle, then separation.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|989b8c8c34734″ question_number=”19″] The diagram below shows the amount of DNA per cell in a dividing cell. Which roman numeral shows the amount of DNA in an unfertilized egg cell?

[c] I

[f] No. Here’s how to think about this. You’re instructed to look for the amount of DNA in an unfertilized egg cell. That means that what you’re observing is meiosis. Meiosis starts with a diploid germ cell (with a full complement of chromosomes and DNA), and ends with a haploid gamete (a sperm, or an unfertilized egg cell, with half of the chromosomes and half the DNA). The cell that starts the process is at “I.” Look at the diagram, focus on the Y-axis (amount of DNA/cell), and figure out which roman numeral represents a cell that would have half the DNA of the cell that started the process?

[c] III

[f] No. Here’s how to think about this. You’re instructed to look for the amount of DNA in an unfertilized egg cell. That means that what you’re observing is meiosis. Meiosis starts with a diploid germ cell (with a full complement of chromosomes and DNA), and ends with a haploid gamete (a sperm, or an unfertilized egg cell, with half of the chromosomes and half the DNA). The cell that starts the process is at “I.” You chose “III,” at which point there’s twice as much DNA. Look at the diagram, focus on the Y-axis (amount of DNA/cell), and figure out which roman numeral represents a cell that would have half the DNA of the cell that started the process?

[c] IV

[f] No. Here’s how to think about this. You’re instructed to look for the amount of DNA in an unfertilized egg cell. That means that what you’re observing is meiosis. Meiosis starts with a diploid germ cell (with a full complement of chromosomes and DNA), and ends with a haploid gamete (a sperm, or an unfertilized egg cell, with half of the chromosomes and half the DNA). The cell that starts the process is at “I.” You chose “IV,” at which point there’s just as much DNA as in the cell that started the process. Look at the diagram, focus on the Y-axis (amount of DNA/cell), and figure out which roman numeral represents a cell that would have half the DNA of the cell that started the process?

[c*] VI

[f] Excellent. At point “VI,” the cell would be an unfertilized egg cell with 1/2 the DNA of the cell that started the process. It would have doubled its DNA in interphase I (“III” in the diagram above), divided chromosomes in meiosis I (resulting in “IV” above), and then divided again in meiosis II, resulting in half the amount of original DNA, as is shown in step “VI.”

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|988fc31002334″ question_number=”20″] The diagram below shows the amount of DNA per cell in a dividing cell. Which roman numeral shows the amount of DNA in a cell during prophase 1 of meiosis I?

[c] I

[f] No. You’re too early in the process. Here’s how to think about this. From the question, you know that this is meiosis. Meiosis begins by doubling of DNA (shown at “II”), then divides the DNA twice. Prophase 1 would be after the DNA doubling, but before the first division. Figure it out from there.

[c*] III

[f] Fabulous. Prophase 1 would be after the doubling of DNA in interphase I, but before the cell division of meiosis I.

[c] IV

[f] No. You’re too late in the process. Here’s how to think about this. From the question, you know that this is meiosis. Meiosis starts with doubling of DNA (shown at “II”), then divides the DNA twice. Prophase 1 would be after the DNA doubling, but before the first division. Figure it out from there.

[c] VI

[f] No. You’re too late in the process. Here’s how to think about this. From the question, you know that this is meiosis. Meiosis starts with doubling of DNA, then divides the DNA twice. Prophase 1 would be after the DNA doubling, but before the first division. Figure it out from there.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9884d91845734″ question_number=”21″] The diagram below shows the amount of DNA per cell in a dividing cell. Which roman numeral would require the involvement of enzymes like DNA polymerase, helicase, and ligase?

[c*] II

[f] Excellent! Enzymes like DNA polymerase, helicase, and ligase would be involved in DNA synthesis that happens during interphase 1, which is shown in step “II.”

[c] III

[f] No. At “III” the line is flat, indicating that the amount of DNA is staying constant over time. The enzymes mentioned above are all involved in DNA synthesis. Synthesis increases the amount of DNA. What’s the only number that shows the amount of DNA going up?

[c] IV

[f] No. At “IV” the line is flat, indicating that the amount of DNA is staying constant over time. The enzymes mentioned above are all involved in DNA synthesis. Synthesis increases the amount of DNA. What’s the only number that shows the amount of DNA going up?

[c] V

[f] No. At “V” the line is sloping downward, indicating that the amount of DNA is decreasing over time.The enzymes mentioned above are all involved in DNA synthesis. Synthesis increases the amount of DNA. What’s the only number that shows the amount of DNA going up?

[c] VI

[f] No. At “VI” the line is flat, indicating that the amount of DNA is staying constant over time. The enzymes mentioned above are all involved in DNA synthesis. Synthesis increases the amount of DNA. What’s the only number that shows the amount of DNA going up?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|98790f9c13334″ question_number=”22″] Assuming that the chromosomes shown in the germ cell on the top of the diagram represent one of the 22 homologous pairs of autosomes found in human beings, then which of the zygotes on the bottom row could develop into an individual with Down syndrome?

[c] a       [c] b         [c*] c         [c] d
[f] No. Here’s the story of cell “a.” The homologous pair of autosomes shown in the cell at the top of the diagram separated correctly, producing (in the third row) an egg with one chromosome. When that cell was fertilized by a sperm, it produced a zygote with a homologous pair. Down syndrome will not result (because Down syndrome is about having an extra chromosome). Remember that, and make another choice the next time you see this question.

[f] No. Here’s the story of cell “b.” The homologous pair of autosomes shown in the cell at the top of the diagram separated correctly, producing (in the third row) an egg with one chromosome. When that cell was fertilized by a sperm, it produced a zygote with a homologous pair. Down syndrome will not result (because Down syndrome is about having an extra chromosome). Remember that, and make another choice the next time you see this question.

[f] Yes. Here’s the story of cell “c.” The homologous pair of autosomes shown in the cell at the top of the diagram separated correctly, but there was a nondisjunction during meiosis II, producing (in the third row) an egg with an extra chromosome. When that egg cell was fertilized by a sperm, it produced a zygote with three chromosomes, instead of a homologous pair. If that extra chromosome were the 21st chromosome, then Down syndrome would result.

[f] No. Here’s the story of cell “d.” The homologous pair of autosomes shown in the cell at the top of the diagram separated correctly, but there was a nondisjunction during meiosis II, producing (in the third row) an egg with a missing chromosome. When that cell is fertilized by a sperm, it produced a zygote that has only one chromosome instead of a homologous pair. That monosomy of the 21st chromosome would not be a survivable chromosomal variation. In any case, it’s not Down syndrome, which is a trisomy (3 copies) of the 21st chromosome. Remember that, and make another choice next time you see this question.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|986ddb22d9f34″ question_number=”23″] Assuming that the chromosomes shown in the germ cell on the top of this diagram represent X chromosomes, then which of the zygotes on the bottom row could develop into an individual with Turner syndrome?

[c] a       [c] b       [c] c         [c*] d
[f] No. The zygote (fertilized egg) at “a” has two chromosomes, one that came from the sperm, and one from the egg. Since the germ cell that started the process at the top of the diagram had two chromosomes, then this zygote has the normal number of chromosomes. People with Turner syndrome are females who are missing an X chromosome. If the chromosomes shown are sex chromosomes, then which of the cells above could fit that description?

[f] No. The zygote (fertilized egg) at “b” has two chromosomes, one that came from the sperm, and one from the egg. Since the germ cell that started the process at the top of the diagram had two chromosomes, then this zygote has the normal number of chromosomes. People with Turner syndrome are females who are missing an X chromosome.If the chromosomes shown are sex chromosomes, then which of the cells above could fit that description?

[f] No. The zygote (fertilized egg) at “c” has three chromosomes, one that came from the sperm, and two from the egg. The blue arrow indicates where a nondisjunction occurred, resulting in one egg with an extra chromosome, and one with a missing chromosome. The zygote at “c” has an extra chromosome, while people with Turner syndrome are females who are missing an X chromosome. If the chromosomes shown are sex chromosomes, then which of the cells above could fit that description.

[f] Excellent. The zygote at “d” is missing a chromosome. Since people with Turner syndrome are females who are missing an X chromosome, then that cell fits the description of a zygote that could lead to a person with Turner syndrome.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9862a6a9a0b34″ question_number=”24″] Thryothorus ludovicianus, also known as the Carolina wren, has a white streak called a “supercilium” above its eye. The supercilium may extend well behind the eye, as shown below.

A family of wrens is found with a short supercilium that barely extends past the eye. To determine the type of inheritance that underlies supercilium length, a group of Carolina wrens was bred in captivity, with the following results.

In experiment 1, individuals with a long supercilium were bred together. All off spring had long supercilia.
In experiment 2, individuals with a short supercilium were bred together. All of the offspring had short supercilia.
In experiment 3, individuals with a short supercilium were bred with individuals with a long supercilium. All of the offspring had short supercilia.
In experiment 4, the short supercilia offspring from experiment three were test crossed with the true-breeding long supercilia offspring of experiment 1. Approximately 50% of the offspring had long supercilia, and 50% had short supercilia.

The data above are consistent with the theory that … If this theory is
correct, if the short offspring from experiment 3 are mated with each other the
offspring of that cross will be ….

[c] supercilium length is sex-linked…. all long.

[f] No. Sex linkage would lead to certain phenotypes being associated with one sex or the other. There’s no evidence of sex linkage here. If an trait isn’t sex-linked, then it’s most likely ______. Figure out the blank, and you’ll be on your way to the answer.

[c] supercilium length is autosomal. “Long” is controlled by a dominant allele and “short” is recessive….3/4 long and 1/4 short.

[f] No. You’re correct in that it’s autosomal, but look at the offspring of the first cross, and reassess which allele is dominant, and which is recessive.

[c*] supercilium length is autosomal.“Short” is controlled by a dominant allele and “long” is recessive …3/4 short and 1/4 long.

[f] Fabulous. Based on the data above, the allele is autosomal, with short being dominant. The F2 generation will have a ratio of three short to one long.

[c] supercilium length is autosomal.“Short” is controlled by a dominant, autosomal allele and “long” is recessive … all short.

[f] No, but you’re very close. If short is dominant (and represented by S), then what would be the result of crossing Ss with Ss?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9856b7ecb0334″ question_number=”25″] A family pedigree and a series of RFLPs for a rare genetic disease is shown below.

Which individual is a heterozygote?

[c] 4

[f] No. You can tell from the pedigree that 4 is homozygous recessive. That’s because she has a condition that she inherited from two parents, neither of whom had the condition. That means that the parents are both heterozygotes. You can check this by looking at the RFLP  patterns. 4 has one band (expected in a homozygote). Her parents, 1 and 2, both have two bands (which is what you expect in a heterozygote).

[c*] 5

[f] Fabulous. 5 doesn’t have this condition, which is caused by an autosomal recessive. 5 must be a heterozygote because his RFLP pattern shows two bands (which is what you’d expect in a heterozygote, with a dominant and a recessive allele).

[c] 8

[f] No. You can tell from the pedigree that 8 is homozygous recessive. That’s because he has a condition that he inherited from his parents, neither of whom had the condition (which is how we know that the condition is caused by a recessive allele). Additionally, if you look at 8’s RFLP pattern, you see one band, confirming the recessive genotype (recessives have two copies of the recessive allele, which will show up as one band of DNA.)

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|98425884a5334″ question_number=”26″] A family pedigree and a series of RFLPs for a rare genetic disease is shown below.

Which individual is homozygous dominant?

[c] 1

[f] No. You can tell this just by analyzing the pedigree. 1 doesn’t have the condition, nor does her partner (2), yet two of her children do (4 and 6). Therefore, you know that the condition is autosomal recessive, and that 1 has to be a heterozygote (as does her partner). You can check this by looking at the RFLPs. She has two bands, which is what you’d expect in a heterozygote (two alleles, one band for each allele).

[c] 2

[f] No. You can tell this just by analyzing the pedigree. 2 doesn’t have the condition, nor does his partner (1), yet his children do (4 and 6). Therefore, you know that the condition is autosomal recessive, and that 2 has to be a heterozygote (as does his partner). You can check this by looking at the RFLPs. He has two bands, which is what you’d expect in a heterozygote (two alleles, one band for each allele).

[c*] 3

[f] Yes. Based on the pedigree, you know that 3 (like his parents) has the dominant phenotype (because the condition is recessive, inherited from two parents who don’t have the condition). You know the 3 is a homozygote by looking at his RFLP pattern. There’s only one band, which means only one type of DNA, which is what you’d expect in a homozygote (both alleles are the same).

[c] 4

[f] No. 4 is the child of 1 and 2, neither of whom have the condition that 4 has. Therefore, the condition has to be autosomal recessive. Individual 4, having the condition, has to be homozygous recessive).

[c] 5

[f] No. 5 is showing the dominant phenotype, and the condition is clearly recessive (passed onto offspring by parents who don’t have the condition). You can tell that 5 is a heterozygote based on his RFLP pattern. The pattern has two bands, which is what you’d see in a heterozygote (two bands, one for each allele).

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|98327bf3a3f34″ question_number=”27″]  Domestic chickens have been bred for many years to increase the number of eggs laid by the females. Chick breeders need to be able to identify female chicks on the day after hatching, as only the females need to be kept for laying eggs.

Unlike mammals, which use a sex determination system with X and Y chromosomes, in chickens the sex chromosomes are known as Z and W. Male chickens have two Z chromosomes (ZZ). Female chickens have one Z chromosome and one W chromosome (ZW).

Some genes for feather color and pattern in chickens are carried on the Z chromosome but not on the W chromosome. One such example is the gene for striped feathers (barring).

The phenotypes associated with the two alleles of the barring gene are shown in the table below

AlleleAdult PhenotypeDay old chick phenotype
BBarred (black feathers striped with white bars)Black body with a white spot on the head
bNon-barred (black feathers)Black body and head

The type of inheritance associated with barring is

[c] autosomal recessive

[f] No. This question is first of all asking if you to apply your understanding of the XY system of sex linkage to this new situation. In birds, males are ZZ and females are ZW. You’re told in the problem that the allele for barring is dominant, and that it’s on the Z chromosome. Just put those two facts together, and the answer will be obvious the next time you see this question.

[c] autosomal dominant

[f] No. This question is first of all asking if you to apply your understanding of the XY system of sex linkage to this new situation. In birds, males are ZZ and females are ZW. You’re told in the problem that the allele for barring is dominant, and that it’s on the Z chromosome. Just put those two facts together, and the answer will be obvious the next time you see this question.

[c] sex linked recessive

[f] No, but you chose the second best answer. Just re-read the question carefully, looking specifically for a description of the behavior of the Barring allele. Remember what you’ve learned, and the answer will be obvious the next time you see this question.

[c*] sex linked dominant

[f] WAY TO GO! Barring is sex linked and dominant!

[c] mitochondrial

[f] No. Mitochondrial inheritance is a real thing…And since mitochondria are only passed along through the egg, there’s a particular pattern of inheritance (maternal). But it’s not at play here. This question is first of all asking if you to apply your understanding of the XY system of sex linkage to this new situation. In birds, males are ZZ and females are ZW. You’re told in the problem that the allele for barring is dominant, and that it’s on the Z chromosome. Just put those two facts together, and the answer will be obvious the next time you see this question.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|982254e126334″ question_number=”28″]  Domestic chickens have been bred for many years to increase the number of eggs laid by the females. Chick breeders need to be able to identify female chicks on the day after hatching, as only the females need to be kept for laying eggs.

Unlike mammals, which use a sex determination system with X and Y chromosomes, in chickens the sex chromosomes are known as Z and W. Male chickens have two Z chromosomes (ZZ). Female chickens have one Z chromosome and one W chromosome (ZW).

Some genes for feather color and pattern in chickens are carried on the Z chromosome but not on the W chromosome. One such example is the gene for striped feathers (barring).

The phenotypes associated with the two alleles of the barring gene are shown in the table below

AlleleAdult PhenotypeDay old chick phenotype
BBarred (black feathers striped with white bars)Black body with a white spot on the head
bNon-barred (black feathers)Black body and head

An individual with genotype ZBZb will, as an adult, have which phenotype?

[c*] Barred male

[f] Excellent. An individual with genotype ZBZb will be a barred male (ZZ = male, and B is the allele for barring)

[c] Barred female

[f] No. Just look at the problem and ask yourself two questions. Firstly, what chromosomal combinations are associated with which sex? In other words, XY makes an individual mammal male, and XX makes females. What’s the parallel in birds? Secondly, is the barring allele dominant or recessive? Put those together, hold it in memory, and you’ll get this question right next time.

[c] Non-barred female

[f] No. Just look at the problem and ask yourself two questions. Firstly, what chromosomal combinations are associated with which sex? In other words, XY makes an individual mammal male, and XX makes females. What’s the parallel in birds? Secondly, is the barring allele dominant or recessive? Put those together, hold it in memory, and you’ll get this question right next time.

[c] Non-barred male

[f] No. Just look at the problem and ask yourself two questions. Firstly, what chromosomal combinations are associated with which sex? In other words, XY makes an individual mammal male, and XX makes females. What’s the parallel in birds? Secondly, is the barring allele dominant or recessive? Put those together, hold it in memory, and you’ll get this question right next time.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|981094067bb34″ question_number=”29″]  Domestic chickens have been bred for many years to increase the number of eggs laid by the females. Chick breeders need to be able to identify female chicks on the day after hatching, as only the females need to be kept for laying eggs.

Unlike mammals, which use a sex determination system with X and Y chromosomes, in chickens the sex chromosomes are known as Z and W. Male chickens have two Z chromosomes (ZZ). Female chickens have one Z chromosome and one W chromosome (ZW).

Some genes for feather color and pattern in chickens are carried on the Z chromosome but not on the W chromosome. One such example is the gene for striped feathers (barring).

The phenotypes associated with the two alleles of the barring gene are shown in the table below

AlleleAdult PhenotypeDay old chick phenotype
BBarred (black feathers striped with white bars)Black body with a white spot on the head
bNon-barred (black feathers)Black body and head

An individual with genotype ZBW will, as an adult, have which phenotype?

[c] Barred male

[f] No. Just look at the problem and ask yourself two questions. Firstly, what chromosomal combinations are associated with which sex? In other words, XY makes an individual mammal male, and XX makes females. What’s the parallel in birds? Secondly, is the barring allele dominant or recessive? Put those together, hold it in memory, and you’ll get this question right next time.

[c*] Barred female

[f] Terrific. You’re clearly understanding how sex determination and sex linked alleles work in birds!

[c] Non-barred female

[f] No. Just look at the problem and ask yourself two questions. Firstly, what chromosomal combinations are associated with which sex? In other words, XY makes an individual mammal male, and XX makes females. What’s the parallel in birds? Secondly, is the barring allele dominant or recessive? Put those together, hold it in memory, and you’ll get this question right next time.

[c] Non-barred male

[f] No. Just look at the problem and ask yourself two questions. Firstly, what chromosomal combinations are associated with which sex? In other words, XY makes an individual mammal male, and XX makes females. What’s the parallel in birds? Secondly, is the barring allele dominant or recessive? Put those together, hold it in memory, and you’ll get this question right next time.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|97f9010e17334″ question_number=”30″]  Domestic chickens have been bred for many years to increase the number of eggs laid by the females. Chick breeders need to be able to identify female chicks on the day after hatching, as only the females need to be kept for laying eggs.

Unlike mammals, which use a sex determination system with X and Y chromosomes, in chickens the sex chromosomes are known as Z and W. Male chickens have two Z chromosomes (ZZ). Female chickens have one Z chromosome and one W chromosome (ZW).

Some genes for feather color and pattern in chickens are carried on the Z chromosome but not on the W chromosome. One such example is the gene for striped feathers (barring).

The phenotypes associated with the two alleles of the barring gene are shown in the table below

AlleleAdult PhenotypeDay old chick phenotype
BBarred (black feathers striped with white bars)Black body with a white spot on the head
bNon-barred (black feathers)Black body and head

In cross between a barred female and a non-barred male, what percentage of the offspring will be non-barred females

[c] 0%

[f] No. You have to treat this like any genetics problems that involves sex linkage. Here are the genotypes: the barred female is ZBW. The non-barred male is ZbZb. Get a pencil, set up a Punnett square like the one below, and solve the problem (determining the proportion of each phenotype).

ZBW
Zb
Zb

Hold on to the answer until the next time you see the question. 

[c] 25%

[f] No. You have to treat this like any genetics problems that involves sex linkage. Here are the genotypes: the barred female is ZBW. The non-barred male is ZbZb. Get a pencil, set up a Punnett square like the one below, and solve the problem (determining the proportion of each phenotype).

ZBW
Zb
Zb

Hold on to the answer until the next time you see the question. 

[c*] 50%

[f] You’re exhibiting marvelous mastery of (post) Mendelian genetics! 50% of the offspring are non-barred females. If you’re like me, you set up this cross. All the females are non-barred, and that’s 50% of the offspring.

ZBW
Zb
Zb

[c] 75%

[f] No. You have to treat this like any genetics problems that involves sex linkage. Here are the genotypes: the barred female is ZBW. The non-barred male is ZbZb. Get a pencil, set up a Punnett square like the one below, and solve the problem (determining the proportion of each phenotype).

ZBW
Zb
Zb

Hold on to the answer until the next time you see the question. 

[c] 100%

[f] No. You have to treat this like any genetics problems that involves sex linkage. Here are the genotypes: the barred female is ZBW. The non-barred male is ZbZb. Get a pencil, set up a Punnett square like the one below, and solve the problem (determining the proportion of each phenotype).

ZBW
Zb
Zb

Hold on to the answer until the next time you see the question. 

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|97eda7541fb34″ question_number=”31″]  In an experimental organism, the alleles for height and body color are linked on the same chromosome. Allele “T” results in a tall phenotype, while “t” results in a short phenotype. “B” results in a black body, while “b” results in brown. A test cross resulted in four types of offspring. The alleles and their position on their chromosomes for the two recombinant offspring are shown below.

Which of the following is a test cross that could have produced these results?

[c] A

[f] No. Here’s how to think about this question.  A test cross is always with an organism that is recessive for both alleles. Therefore one parent is going to be tb//tb (which should let you eliminate two of the choices above). For the other parent, choose one whose alleles are organized in such a way that crossing over will produce recombinant gametes with alleles tB/ and /Tb. Then, after the test cross, the recombinant offspring will be tB//tb and Tb//tb.

[c] B

[f] No. Here’s how to think about this question.  A test cross is always with an organism that is recessive for both alleles. Therefore one parent is going to be tb//tb (which should let you eliminate two of the choices above). For the other parent, choose one whose alleles are organized in such a way that crossing over will produce recombinant gametes with alleles tB/ and /Tb. Then, after the test cross, the recombinant offspring will be tB//tb and Tb//tb.

[c*] C

[f] Correct! A cross between TB//tb and tb//tb will result in the two recombinant offspring shown above, as well as two parental type offspring (TB//tb and tb//tb).

[c] D

[f] No, but you’re on the right track.  You correctly identified a choice where one parent is suitable to be a  test cross parent (recessive for both alleles: tb//tb). For the other parent, choose the whose alleles are organized in such a way that crossing over will produce recombinant gametes with alleles tB/ and /Tb. Then, after the test cross, the recombinant offspring will be tB//tb and Tb//tb.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|97e20318abb34″ question_number=”32″] The diagram below shows some of the steps involved in cloning. Which of the statements below is true about letters X and Y?

[c] Letter X is a sex cell removed from an adult animal. Letter Y is a nucleus removed from differentiated animal cell.

[f] No. Let’s think about how cloning works. A somatic cell is removed from the organism that you want to clone. You remove the nucleus from this differentiated cell, and implant it into an egg cell that’s had its nucleus removed. Then the resulting embryo is implanted into the uterus of a female sheep (which acts as a surrogate mother).

Study the diagram above in the light of this explanation, figure out what X and Y are, and you should be able to get this one right next time.

[c] Letter X is a sex cell removed from an adult animal. Letter Y is a nucleus removed from unfertilized egg cell.

[f] No. Let’s think about how cloning works. A somatic cell is removed from the organism that you want to clone. You remove the nucleus from this differentiated cell, and implant it into an egg cell that’s had its nucleus removed. Then the resulting embryo is implanted into the uterus of a female sheep (which acts as a surrogate mother).

Study the diagram above in the light of this explanation, figure out what X and Y are, and you should be able to this one right next time.

[c] Letter X is a differentiated cell removed from adult animal. Letter Y is a nucleus removed from differentiated animal cell.

[f] No. Let’s think about how cloning works. A somatic cell is removed from the organism that you want to clone. You remove the nucleus from this differentiated cell, and implant it into an egg cell that’s had its nucleus removed. Then the resulting embryo is implanted into the uterus of a female sheep (which acts as a surrogate mother).

Study the diagram above in the light of this explanation, figure out what X and Y are, and you should be able to get this one right next time.

[c*] Letter X is differentiated cell removed from an adult animal. Letter Y is an unfertilized egg cell about to have its nucleus removed.

[f] Nice Job. You have a good understanding of how cloning works!

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|97d6ce9f72734″ question_number=”33″]

A pair of homologous chromosomes involved in normal meiosis in an ovary carries the alleles shown below.

Look at the chromosomes below. Choose the letter that corresponds to the chromosome that could be found in any resultant eggs.

[c] I

[f] No. The problem with choice I is that two alleles are missing, making it unviable. Study the diagram, and find a chromosome that would be a viable gamete.

[c] II

[f] No. The problem with choice II is that the F allele is missing, making it unviable. Study the diagram, and find a chromosome that would be a viable gamete.

[c] III

[f]No. Choice III is a sensible choice, because it would be a viable chromosome. But take a close look at the alleles available, particularly the last one. Where did that R come from?

[c*] IV

[f] Excellent. Choice IV has all four alleles, and it’s a possible result from the chromosomes in the original germ cell.

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|97c7f6d3a4f34″ question_number=”34″] A farmer is applying a pesticide to suppress an insect called a “stem borer.” The farmer has learned that there is an allele that enables stem borers to tolerate the pesticide, and she is concerned that the stem borers will become pesticide resistant. Which of the following situations would most likely lead to the appearance of pesticide-resistant stem borers?

[c] If none of the stem borers in this area have the resistance allele.

[f] No. This is a question about natural selection, and you’re looking for the choice that would lead stem-borers that are resistant to the pesticide (meaning “not harmed by” this insect-killing chemical) to become more common in the population. If none of the stem borers in the area had an allele that would give them resistance, then they’d all be killed by the pesticide, and the farmer would have nothing to worry about. Which of the choices could lead to natural selection for the pesticide resistance allele?

[c] If stem borers that have the resistance allele also produced different pheromones, making it difficult for them to attract mates.

[f] No. If that were the case, then the resistance allele would reduce the fitness of the stem borers (because even though they had the resistance allele, they would not attract mates, and therefore wouldn’t pass the allele on to the next generation). Look at the choices again, and ask yourself which of the choices describes a situation where there would be natural selection that would increase the frequency of the pesticide resistance allele.

[c] If the resistance allele makes the stem borer adults slower and easier prey for birds.

[f] No. If that were the case, then the resistance allele would reduce the fitness of the stem borers (because even though they had the resistance allele, they would be eaten by birds, preventing the allele from being passed on to the next generation). Look at the choices again, and ask yourself which of the choices describes a situation where there would be natural selection that would increase the frequency of the pesticide resistance allele.

[c*] If neighboring farms have used this pesticide for years, and the adult stem borers can easily fly long distances.

[f] Excellent. This choice describes a situation where if the pesticide resistance allele arose in a neighboring population, it could increase in frequency within that population, and spread to neighboring farms.

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|97b2dd27e2b34″ question_number=”35″] Off of the coast of California, individuals of a species of sharks either have tall fins or short fins. Off of the coast of Australia, the same species is found, but individuals can have a third phenotype: intermediate fin height. Assuming that fin height is controlled at one locus in both populations and that both populations have only two alleles, what could explain these observations?

[c] The frequency of the short allele is higher in Australia.

[f] No. While that might be true, it wouldn’t explain the appearance of an intermediate phenotype. What kind of inheritance pattern can produce intermediate phenotypes, even when there’s only a single gene with two alleles in question?

[c] The short allele is dominant in California, but recessive in Australia.

[f] No. That wouldn’t explain the appearance of an intermediate phenotype. What kind of inheritance pattern can produce intermediate phenotypes, even when there’s only a single gene with two alleles in question?

[c] The short allele is dominant in Australia, but recessive in California.

[f] No. That wouldn’t explain the appearance of an intermediate phenotype. What kind of inheritance pattern can produce intermediate phenotypes, even when there’s only a single gene with two alleles in question?

[c*] In the Australian population, a mutation made the tall allele incompletely dominant.

[f] Nice job! Incomplete dominance, which occurs when there’s an intermediate phenotype seen in heterozygotes, could explain the appearance of intermediate fin height in the Australian sharks.

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|97a2b61564f34″ question_number=”36″] Off of the coast of California, individuals of a species of sharks either have tall fins or short fins. The protein made by the allele for short fins differs by only one amino acid from the protein that results in the tall-finned allele. Which of the following is a correct statement about the nucleotide sequences for the short and tall alleles?

[c] The nucleotide sequences could be identical.

[f] No. If there’s a different phenotype, then the allele must be coding for a different protein. In order to do that, the nucleotide sequences must be different. Note that if you’re an exceptional biological thinker you might have thought of the possibility of an epigenetic effect, but there’s a more straightforward answer among these choices.

[c] The nucleotide sequences must differ by one nucleotide.

[f] Not necessarily. The nucleotide sequences could differ by one nucleotide. But there are other ways  that the nucleotide sequences for the short and tall fin alleles could differ. See if you can think of all of the possible ways, and remember that the next time you see this question.

[c] The nucleotide sequences could differ by one, two or three nucleotides, but no more than three.

[f] No. All that’s required is that there must be a mutation that changes the amino acid sequence. Why would that be limited to one, two, or three alleles?

[c*] The nucleotide sequences could differ by more than three nucleotides.

[f] Correct. The mutation that causes this difference could come about in many ways, and of all of the ways, this is the only statement in this list of choices that is, without any qualifications, true.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|97932405e0334″ question_number=”37″] As shown below, a DNA fragment from a plasmid in E. coli has two restriction sites for the restriction endonuclease Hind III, which creates fragments X, Y, and Z.

Which of the following patterns would result from separation of these fragments by gel electrophoresis?

[c] A

[f] No. The restriction endonuclease described in the question will cut the DNA into three pieces. Just by looking at the diagram of the DNA, you can tell that Z is the longest fragment, followed by X, followed by Y. Now, connect this with the idea that electrophoresis of DNA fragments separates them based on size, with larger fragments moving less, and smaller fragments moving more. If all of these fragments start at the negative pole at the top, then which fragment will move the most, which the second most, and which the least? Figure that out, and you’ll have the answer the next time you see this question.

[c*] B

[f] Yes. Based on the diagram, Z is the longest fragment, followed by X, followed by Y. As a result, the pattern that will result from electrophoresis is B.

[c] C

[f] No. The restriction endonuclease described in the question will cut the DNA into three pieces. Just by looking at the diagram of the DNA, you can tell that Z is the longest fragment, followed by X, followed by Y. Now, connect this with the idea that electrophoresis of DNA fragments separates them based on size, with larger fragments moving less, and smaller fragments moving more. If all of these fragments start at the negative pole at the top, then which fragment will move the most, which the second most, and which the least? Figure that out, and you’ll have the answer the next time you see this question.

[c] D

[f] No. The restriction endonuclease described in the question will cut the DNA into three pieces. Just by looking at the diagram of the DNA, you can tell that Z is the longest fragment, followed by X, followed by Y. Now, connect this with the idea that electrophoresis of DNA fragments separates them based on size, with larger fragments moving less, and smaller fragments moving more. If all of these fragments start at the negative pole at the top, then which fragment will move the most, which the second most, and which the least? Figure that out, and you’ll have the answer the next time you see this question.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9787ca4be8b34″ question_number=”38″] In the diagram below, maternal chromosomes are colored white, and paternal chromosomes are colored black.

If a cell with a diploid number of six is undergoing a type of cell division, then which of the diagrams below could represent the arrangement of chromosomes in this cell at some stage of the cell division process?

[c] A  [c*] B  [c] C  [c] D

[f] No. If the diploid number is six, then the cell would never have more than six chromosomes. In this diagram, you see 12. Remember that the next time you see this question.

[f] Nice job. If the cell were going through meiosis, then you might see three doubled chromosomes after meiosis 1 (which is what you see here). Because of independent assortment, it’s completely possible that two of the chromosomes would be maternal, and one would be paternal.

[f] No. If a cell has a diploid number of six, then you could see six chromosomes lining up during metaphase 1 (the alignment phase) of mitosis. The problem is that half the chromosomes would have to be paternal, and half would be maternal. Here you have four maternal chromosomes and two paternal ones. Remember that the next time you see this question.

[f] No. While a diploid cell would line up its chromosomes in a way that’s similar to this during metaphase 1 of meiosis 1, there’s one problem: four of the chromosomes are paternal, and two are maternal. For this to be correct, half would have to be maternal, and half paternal. Remember that the next time you see this question.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|977c4b5132f34″ question_number=”39″] A plant with genotype DdEeFf was crossed with a plant with genotype ddEEFf. What is the probability of obtaining DdEeFf progeny?

Note from Mr. W: the goal is to solve this problem without making an enormous Punnett square!

[c] 1/64

[f] No. To solve this, you need to use the rule of multiplication: to calculate the probability of two or more events occurring together, we multiply their separate probabilities. In this case, the probability of Dd x dd producing Dd is 1/2. The probability of Ee x EE producing Ee is 1/2. Now figure out the probability of Ff x Ff producing Ff. Multiply that times 1/2 x 1/2, and  what do you get?

[c] 1/32

[f] No. To solve this, you need to use the rule of multiplication: to calculate the probability of two or more events occurring together, we multiply their separate probabilities. In this case, the probability of Dd x dd producing Dd is 1/2. The probability of Ee x EE producing Ee is 1/2. Now figure out the probability of Ff x Ff producing Ff. Multiply that times 1/2 x 1/2, and  what do you get?

[c*] 1/16

[f] Excellent! The probability is 1/2 x 1/2 x 1/2, or 1/16.

[c] 1/8

[f] No. To solve this, you need to use the rule of multiplication: to calculate the probability of two or more events occurring together, we multiply their separate probabilities. In this case, the probability of Dd x dd producing Dd is 1/2. The probability of Ee x EE producing Ee is 1/2. Now figure out the probability of Ff x Ff producing Ff. Multiply that times 1/2 x 1/2, and  what do you get?

[c] 1/4

[f] No. To solve this, you need to use the rule of multiplication: to calculate the probability of two or more events occurring together, we multiply their separate probabilities. In this case, the probability of Dd x dd producing Dd is 1/2. The probability of Ee x EE producing Ee is 1/2. Now figure out the probability of Ff x Ff producing Ff. Multiply that times 1/2 x 1/2, and  what do you get?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|977081d500b34″ question_number=”40″] The shapes of tabby stripes in a cat’s fur is controlled by an autosomal gene with two alleles:

T: vertical colored stripes (called mackerel tabby)

t: swirly colored stripes (called blotched tabby.}

The patterns are illustrated as follows.

A second autosomal gene (not linked to the tabby-striped gene)is the Agouti gene. This gene has the alleles

A: tabby fur pattern

a: no tabby pattern.

A cat that is aa for the Agouti locus fails to show any stripes, regardless of its genotype at the tabby stripes locus. These cats are solid black in color.

A cross was carried out between cats of the following genotypes.

TTAa X ttAa

The cats had two kittens. Which of the images below shows phenotypes that would be possible? Note that not all the kittens in a litter would have to have the phenotype shown in any one of the choices below. 

[c*] A  [c] B  [c] C  [c] D

[f] Excellent! This type of cross could only produce kittens that were black, or which were mackerel tabby (the kind with vertical stripes).

[f] No. The cross is between TTAa and ttAa

Here’s the Punnett square

tAta
TATtAATtAa
TaTtAaTtaa


The genotypes
TtAA and TtAa result in mackerel tabby. Ttaa results in black. There’s no possibility, in this cross, of a blotched tabby (which could come about through inheriting genotypes ttAa or ttAA).

[f] No. The cross is between TTAa and ttAa

Here’s the Punnett square

tAta
TATtAATtAa
TaTtAaTtaa

The genotypes TtAA and TtAa result in mackerel tabby. Ttaa results in black. There’s no possibility, in this cross, of a blotched tabby (which could come about through inheriting genotypes ttAa or ttAA).

[f] No. The cross is between TTAa and ttAa

Here’s the Punnett square

tAta
TATtAATtAa
TaTtAaTtaa

The genotypes TtAA and TtAa result in mackerel tabby. Ttaa results in black. There’s no possibility, in this cross, of a blotched tabby (which could come about through inheriting genotypes ttAa or ttAA).

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9763d8d458f34″ question_number=”41″] In a newly bred population of mice, coat color is under the influence of three alleles.

C: black coat

cR: brown coat

c: white coat

The black coat allele is dominant over both brown and white coat. The genotype cRc produces another coat color known as fawn.

Which of the following predictions is correct?

[c] CcR x CcR: all black mice.

[f] No. Here’s the Punnett square for CcR x CcR

CcR
CCCCcR
cRCcRcRcR

CC is black; Ccis black. But cRcis brown.

[c] cRcR x cRc: all brown mice.

[f] No. Here’s the Punnett square for cRcR x cRc.

cRcR
cRcRcRcRcR
ccRccRc

cRcR is brown; cRc is fawn.

[c*] cRcR x cc: all fawn mice.

[f] Fabulous. Here’s the Punnett square for cRcR x cc:

cRcR
ccRccRc
ccRccRc

All the offspring have the cR genotype, and the  fawn phenotype

[c] Cc x cc: all white mice.

[f] No. Here’s the Punnett square for Cc x cc.

Cc
cCccc
cCccc

Cc is black; cc is white. 

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9756e55234b34″ question_number=”42″] Two strains of mice that are heterozygous for four genes (AaBbCcDd)are crossed together. What proportion of the total offspring would be homozygous for all four genes?

[c] 1/256

[f] No. Think about it this way. In a monohybrid cross between Aa and Aa, the offspring can be homozygous in two ways: they can be aa or AA. If you imagine the Punnett square for Aa x Aa, you’ll see that 1/2 of the squares are filled with aa or AA. The same is true for Bb x Bb, Cc x Cc, and Dd x Dd. Now use the rule of multiplication, and you’ll have the answer.

[c] 1/32

[f] No. Think about it this way. In a monohybrid cross between Aa and Aa, the offspring can be homozygous in two ways: they can be aa or AA. If you imagine the Punnett square for Aa x Aa, you’ll see that 1/2 of the squares are filled with aa or AA. The same is true for Bb x Bb, Cc x Cc, and Dd x Dd. Now use the rule of multiplication, and you’ll have the answer.

[c*] 1/16.

[f] Excellent! In a tetrahybrid cross (AaBbCcDd x AaBbCcDd), the number of offspring that will be homozygous for all four genes is 1/2 x 1/2 x 1/2 x 1/2, or 1/16

[c] 1/8

[f] No. Think about it this way. In a monohybrid cross between Aa and Aa, the offspring can be homozygous in two ways: they can be aa or AA. If you imagine the Punnett square for Aa x Aa, you’ll see that 1/2 of the squares are filled with aa or AA. The same is true for Bb x Bb, Cc x Cc, and Dd x Dd. Now use the rule of multiplication, and you’ll have the answer.

[c] 1/4

[f] No. Think about it this way. In a monohybrid cross between Aa and Aa, the offspring can be homozygous in two ways: they can be aa or AA. If you imagine the Punnett square for Aa x Aa, you’ll see that 1/2 of the squares are filled with aa or AA. The same is true for Bb x Bb, Cc x Cc, and Dd x Dd. Now use the rule of multiplication, and you’ll have the answer.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|974b4116c0b34″ question_number=”43″] Two strains of mice that are heterozygous for four genes (AaBbCcDd)are crossed together. What proportion of the total offspring would show a recessive phenotype for all four genes?

[c*] 1/256

[f] Excellent! In a tetrahybrid cross (AaBbCcDd x AaBbCcDd) there’s a 1/256 chance of the offspring showing the recessive phenotype for all four genes (1/4 x 1/4 x 1/4 x 1/4)

[c] 1/32

[f] No. There are two steps to getting this problem right. First, note that in a monohybrid cross (Aa x Aa) there’s a 1/4 chance of getting an offspring that will show the recessive phenotype. Second, knowing that, you use the the rule of multiplication to calculate the probability of this happening for all four allele pairs. What’s 1/4 x 1/4 x 1/4 x 1/4?

[c] 1/16

[f] No. There are two steps to getting this problem right. First, note that in a monohybrid cross (Aa x Aa) there’s a 1/4 chance of getting an offspring that will show the recessive phenotype. Second, knowing that, you use the the rule of multiplication to calculate the probability of this happening for all four allele pairs. What’s 1/4 x 1/4 x 1/4 x 1/4?

[c] 1/8

[f] No. There are two steps to getting this problem right. First, note that in a monohybrid cross (Aa x Aa) there’s a 1/4 chance of getting an offspring that will show the recessive phenotype. Second, knowing that, you use the the rule of multiplication to calculate the probability of this happening for all four allele pairs. What’s 1/4 x 1/4 x 1/4 x 1/4?

[c] 1/4

[f] No. There are two steps to getting this problem right. First, note that in a monohybrid cross (Aa x Aa) there’s a 1/4 chance of getting an offspring that will show the recessive phenotype. Second, knowing that, you use the the rule of multiplication to calculate the probability of this happening for all four allele pairs. What’s 1/4 x 1/4 x 1/4 x 1/4?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|973ee29795734″ question_number=”44″] The pedigree below shows inheritance of an X-linked recessive trait.

What is the probability that the female designated as “1″ is a carrier for the trait?

[c] 0%  [c] 25%  [c*] 50%  [c] 75%  [c] 100%
[f] No. Here’s how to think about this question. Imagine that this is about any of the recessive, sex linked traits that you’ve learned about this year, such as hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  Because the female at 1 has a brother who has the condition, and parents who don’t have the conditions then we know that the mother of 1 is a carrier (XHXh), and the father was normal (XHY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XHXh
XHXHXHXHXh
YXHYXhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[f] No. Here’s how to think about this question. Imagine that this is about any of the recessive, sex linked traits that you’ve learned about this year, such as hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  Because the female at 1 has a brother who has the condition, and parents who don’t have the conditions then we know that the mother of 1 is a carrier (XHXh), and the father was normal (XHY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XHXh
XHXHXHXHXh
YXHYXhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[f] Excellent! The the probability that 1 (one of the daughters) is a carrier is 50%, as you can see by this Punnett square

XHXh
XHXHXHXHXh
YXHYXhY

[f] No. Here’s how to think about this question. Imagine that this is about any of the recessive, sex linked traits that you’ve learned about this year, such as hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  Because the female at 1 has a brother who has the condition, and parents who don’t have the conditions then we know that the mother of 1 is a carrier (XHXh), and the father was normal (XHY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XHXh
XHXHXHXHXh
YXHYXhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[f] No. Here’s how to think about this question. Imagine that this is about any of the recessive, sex linked traits that you’ve learned about this year, such as hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  Because the female at 1 has a brother who has the condition, and parents who don’t have the conditions then we know that the mother of 1 is a carrier (XHXh), and the father was normal (XHY).

If you can, your best move at this point is to make a Punnett square and figure out the answer. If you need to see the Punnett square that I made to solve this problem, read on…

Here’s the Punnett square:

XHXh
XHXHXHXHXh
YXHYXhY

Look at the two girls in the top row, and you should be able to figure out the probability that any one of the daughters will be a carrier.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9732ce99e6b34″ question_number=”45″] The pedigree below shows inheritance of an X-linked recessive trait.

What is the probability that the child indicated by ? will be a boy who is affected by the trait?

[c] 0%

[f] No. Here’s how to think about this question. Imagine that it’s a recessive, sex-linked trait, like hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. There’s a 1/2 chance that ? will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on herXh allele. Multiply all of these probabilities together, and you’ll have your answer.

[c*] 12.5%

[f] Excellent. There’s a 1/2 chance that the child will be a boy. There’s a 1/2 chance that the mom is a carrier. And there’s a 1/2 chance that this carrier mom will pass on the recessive allele. 1/2 x 1/2 x 1/2 = 1/8, or 12.5%

[c] 50%

[f] No. Here’s how to think about this question. Imagine that it’s a recessive, sex-linked trait, like hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. There’s a 1/2 chance that ? will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on herXh allele. Multiply all of these probabilities together, and you’ll have your answer.

[c] 75%

[f] No. Here’s how to think about this question. Imagine that it’s a recessive, sex-linked trait, like hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. There’s a 1/2 chance that ? will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on herXh allele. Multiply all of these probabilities together, and you’ll have your answer.

[c] 100%

[f] No. Here’s how to think about this question. Imagine that it’s a recessive, sex-linked trait, like hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. There’s a 1/2 chance that ? will be a boy (true of any child, right?). There’s a 1/2 chance that the mother is a carrier. And, if she’s a carrier, there’s a 1/2 chance that she’ll pass on herXh allele. Multiply all of these probabilities together, and you’ll have your answer.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9723870bde734″ question_number=”46″] The pedigree below shows inheritance of an X-linked recessive trait.

If the child indicated by ? is a girl, what is the probability she will be affected by the trait?

[c] 0%

[f] No. Here’s how to think about this question. Imagine that the trait under discussion is hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. It’s already established in the question that the child is a girl. There’s a 100% chance that she’ll inherit the recessive allele from her father (that’s how she wound up being a girl in the first place, right?) There’s a 1/2 chance that the mother is a carrier. And, if mom’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele, producing a daughter that’s XhXh. Multiply these probabilities out, and you’ll have your answer.

[c*] 25%

[f] Excellent! It’s already established in the question that the child is a girl. There’s a 100% chance that she’ll inherit the recessive allele from her father (that’s how she wound up being a girl in the first place, right?) There’s a 1/2 chance that the mother is a carrier. And, if mom’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele, producing a daughter that’s XhXh. 1/2 x 1/2 = 1/4.

[c] 50%

[f] No. Here’s how to think about this question. Imagine that the trait under discussion is hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. It’s already established in the question that the child is a girl. There’s a 100% chance that she’ll inherit the recessive allele from her father (that’s how she wound up being a girl in the first place, right?) There’s a 1/2 chance that the mother is a carrier. And, if mom’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele, producing a daughter that’s XhXh. Multiply these probabilities out, and you’ll have your answer.

[c] 75%

[f] No. Here’s how to think about this question. Imagine that the trait under discussion is hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. It’s already established in the question that the child is a girl. There’s a 100% chance that she’ll inherit the recessive allele from her father (that’s how she wound up being a girl in the first place, right?) There’s a 1/2 chance that the mother is a carrier. And, if mom’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele, producing a daughter that’s XhXh. Multiply these probabilities out, and you’ll have your answer.

[c] 100%

[f] No. Here’s how to think about this question. Imagine that the trait under discussion is hemophilia. For clarity, let’s have “H” stand for the normal allele, and “h” for the allele that causes the recessive condition.  The child at ? has a father who has the trait: therefore the father’s genotype is XhY. The child’s mother had a brother who was has the recessive trait. Therefore ?’s grandmother had to be a carrier, and ?’s mother has a 50% chance of being a carrier.

Now use the rule of multiplication. It’s already established in the question that the child is a girl. There’s a 100% chance that she’ll inherit the recessive allele from her father (that’s how she wound up being a girl in the first place, right?) There’s a 1/2 chance that the mother is a carrier. And, if mom’s a carrier, there’s a 1/2 chance that she’ll pass on her Xh allele, producing a daughter that’s XhXh. Multiply these probabilities out, and you’ll have your answer.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9715d94602f34″ question_number=”47″] Hemophilia A is an X-linked recessive disorder in which the ability of the blood to clot is severely reduced. Hemophilia is caused by a mutation in the gene for the clotting component Factor VIII.

Jasmine’s brother has hemophilia A, but neither Jasmine nor anyone else in her family show symptoms of the disorder.

If Jasmine has a son, what is the probability that he will have hemophilia?

[c] 0%

[f] No. Because Jasmine’s brother is a hemophiliac, we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and her father was not a hemophiliac (also known from the problem), then Jasmine has a 1/2 chance of being a carrier. And if she were a carrier, then she’d have a 1/2 chance of passing on the allele to her male child. What’s 1/2 times 1/2?

[c*] 25%

[f] Excellent! There’s a 50% probability that Jasmine is a carrier, and there’s a 50% probability that a carrier will pass on an X chromosome with the hemophilia allele on to her son. 50% of 50% is 25%.

[c] 50%

[f] No. Because Jasmine’s brother is a hemophiliac,  we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and her father was not a hemophiliac (also known from the problem), then Jasmine herself has a 1/2 chance of being a carrier. And if she were a carrier, then she’d have a 1/2 chance of passing on the allele to her male child. What’s 1/2 times 1/2?

[c] 75%

[f] No. Because Jasmine’s brother is a hemophiliac,  we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and her father was not a hemophiliac (also known from the problem), then Jasmine herself has a 1/2 chance of being a carrier. And if she were a carrier, then she’d have a 1/2 chance of passing on the allele to her male child. What’s 1/2 times 1/2?

[c] 100%

[f] No. Because Jasmine’s brother is a hemophiliac,  we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and her father was not a hemophiliac (also known from the problem), then Jasmine herself has a 1/2 chance of being a carrier. And if she were a carrier, then she’d have a 1/2 chance of passing on the allele to her male child. What’s 1/2 times 1/2?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9709c54854334″ question_number=”48″] Hemophilia A is an X-linked recessive disorder in which the ability of the blood to clot is severely reduced. Hemophilia is caused by a mutation in the gene for the clotting component Factor VIII.

Jasmine’s brother has hemophilia A, but neither Jasmine nor anyone else in her family show symptoms of the disorder.

If Jasmine’s husband had hemophilia, what would the probability be of their son being a hemophiliac?

[c] 0%

[f] No. Because Jasmine’s brother is a hemophiliac, we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and Jasmine’s father was not a hemophiliac (also known from the problem), then Jasmine herself has a 1/2 chance of being a carrier. Jasmine’s mate’s status as a hemophiliac is irrelevant, because a father passes on his Y chromosome to his sons, not his X chromosome). Now multiply the probability that Jasmine is a carrier by the probability that she would pass on her hemophilia allele, and you’ll have the answer.

[c*] 25%

[f] Excellent! There’s a 50% probability that Jasmine is a carrier, and there’s a 50% probability that a carrier will pass on an X chromosome with the hemophilia allele on to her son. As you must have understood, the father’s status as a hemophiliac is irrelevant, since he’ll only pass on his X chromosome to his daughters. 50% of 50% is 25%.

[c] 50%

[f] No. Because Jasmine’s brother is a hemophiliac, we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and Jasmine’s father was not a hemophiliac (also known from the problem), then Jasmine herself has a 1/2 chance of being a carrier. Jasmine’s mate’s status as a hemophiliac is irrelevant, because a father passes on his Y chromosome to his sons, not his X chromosome). Now multiply the probability that Jasmine is a carrier by the probability that she would pass on her hemophilia allele, and you’ll have the answer.

[c] 75%

[f] No. Because Jasmine’s brother is a hemophiliac, we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and Jasmine’s father was not a hemophiliac (also known from the problem), then Jasmine herself has a 1/2 chance of being a carrier. Jasmine’s mate’s status as a hemophiliac is irrelevant, because a father passes on his Y chromosome to his sons, not his X chromosome). Now multiply the probability that Jasmine is a carrier by the probability that she would pass on her hemophilia allele, and you’ll have the answer.

[c] 100%

[f] No. Because Jasmine’s brother is a hemophiliac, we know that his genotype is XhY. Because the hemophilia allele is on the X chromosome, then Jasmine’s brother had to have inherited the condition from his and Jasmine’s mother, who must be a carrier (it’s stated in the problem that she doesn’t have the condition, so she can only be a carrier).

If Jasmine’s mother was a carrier and Jasmine’s father was not a hemophiliac (also known from the problem), then Jasmine herself has a 1/2 chance of being a carrier. Jasmine’s mate’s status as a hemophiliac is irrelevant, because a father passes on his Y chromosome to his sons, not his X chromosome). Now multiply the probability that Jasmine is a carrier by the probability that she would pass on her hemophilia allele, and you’ll have the answer.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|96fd66c928f34″ question_number=”49″] In cats, the pattern of tabby strips can be vertical (referred to as “mackerel tabby”) or blotched (referred to as “blotched tabby”), as shown below.

When true breeding mackerel tabbies are bred with true breeding blotched tabbies, all of the F1 offspring are mackerel tabbies.

Two F1 mackerel tabbies are crossed with one another. What’s the probability that a kitten resulting from this mating will be mackerel tabby?

[c] 0%

[f] No. Here’s how to think about this problem. You know that that the mackerel tabby trait is dominant (because all of the F1 offspring are mackerel). And you also know that the F1s are heterozygotes. If you have the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x Mm, and you’d set up the Punnett square as follows:

 Mm
 M
 m


Now you draw it out, and figure out what percentage of the offspring will have genotype
MM or Mm.

[c] 25%

[f] No. Here’s how to think about this problem. You know that that the mackerel tabby trait is dominant (because all of the F1 offspring are mackerel). And you also know that the F1s are heterozygotes. If you have the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x Mm, and you’d set up the Punnett square as follows:

 Mm
 M
 m

Now you draw it out, and figure out what percentage of the offspring will have genotype MM or Mm.

[c] 50%

[f] No. Here’s how to think about this problem. You know that that the mackerel tabby trait is dominant (because all of the F1 offspring are mackerel). And you also know that the F1s are heterozygotes. If you have the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x Mm, and you’d set up

the Punnett square as follows:

 Mm
 M
 m

Now you draw it out, and figure out what percentage of the offspring will have genotype MM or Mm.

[c*] 75%

[f] Way to go! The percentage of offspring that will show the dominant trait will be 75% (25% homozygous dominant, and 50% heterozygotes).

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|96ee8efd5b734″ question_number=”50″] In cats, the pattern of tabby strips can be vertical (referred to as “mackerel tabby”) or blotched (referred to as “blotched tabby”), as shown below.

When true breeding mackerel tabbies are bred with true breeding blotched tabbies, all of the F1 offspring are mackerel tabbies.

An F1 mackerel tabby is crossed with a blotched tabby. What’s the probability that a kitten resulting from this mating will be blotched tabby?

[c] 0%

[f] No. Here’s how to think about this problem. You know that that the mackerel tabby trait is dominant (because all of the F1 offspring are mackerel). And you also know that the F1s are heterozygotes. Because the allele for blotched is recessive, a blotched tabby has to be homozygous.

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up the Punnett square as follows:

 Mm
 m
 m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[c] 25%

[f] No. Here’s how to think about this problem. You know that that the mackerel tabby trait is dominant (because all of the F1 offspring are mackerel). And you also know that the F1s are heterozygotes. Because the allele for blotched is recessive, a blotched tabby has to be homozygous.

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up the Punnett square as follows:

 Mm
 m
 m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[c*] 50%

[f] Fabulous. If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm. 50% of the offspring will be blotched tabby, with genotype mm.

[c] 75%

[f] No. Here’s how to think about this problem. You know that that the mackerel tabby trait is dominant (because all of the F1 offspring are mackerel). And you also know that the F1s are heterozygotes. Because the allele for blotched is recessive, a blotched tabby has to be homozygous.

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up

the Punnett square as follows:

 Mm
 m
 m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[c] 100%

[f] No. Here’s how to think about this problem. You know that that the mackerel tabby trait is dominant (because all of the F1 offspring are mackerel). And you also know that the F1s are heterozygotes. Because the allele for blotched is recessive, a blotched tabby has to be homozygous.

If you let the letter M represent the allele for “mackerel” and m represent the allele for “blotched, then you’re crossing Mm x mm, and you’d set up

the Punnett square as follows:

 Mm
 m
 m

Now you draw it out, and figure out what percentage of the offspring will have genotype mm (the only genotype you can have if you’re blotched tabby).

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|96dfdc724c334″ question_number=”51″] The expression of fruit pigment in a newly developed variety of apple is under genetic control. The apples can have three phenotypes: no pigment, yellow color or green color. Two genes are involved in the control of pigment production. Each of the genes has two alleles as shown.

Gene 1Gene 2
I: no color productionG: yellow fruit
i: color productiong: green fruit

These two genes are unlinked. This means that

[c*] the genes assort independently.

[f] Exactly. If the genes are not linked, then they’ll assort independently.

[c] the genes are close together on the same chromosome.

[f] No. If the genes are unlinked, then they’re not on the same chromosome. Remember that next time you see this question, and choose another answer.

[c] there is recombination through crossing over between the two genes during meiosis.

[f] No. Recombination through crossing over only occurs in genes that are linked on the same chromosome. These genes are unlinked. Remember that next time you see this question, and choose another answer.

[c] particular combinations of alleles of these genes are always inherited together.

[f] No. The key idea is that these genes (and their alleles) are unlinked. As a result, there won’t be particular combinations of alleles that are always inherited together. Instead, they’ll be inherited independently. Remember that for the next time you’ll see this question, and choose another answer.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|96d29e6eab734″ question_number=”52″] The expression of fruit pigment in a newly developed variety of apple is under genetic control. The apples can have three phenotypes: no pigment, yellow color or green color. Two genes are involved in the control of pigment production. Each of the genes has two alleles as shown.

Gene 1Gene 2
I: no color productionG: yellow fruit
i: color productiong: green fruit

The fruit color of plants with the genotype IiGg

[c] yellow.

[f] No. Here’s how to think about this question. Look carefully at the table showing the effects of the two alleles for gene 1. If a plant has the dominant allele I, what happens to fruit color? Once you have that figured out, you should have no trouble with this question the next time you see it.

[c] green.

[f] No. Here’s how to think about this question. Look carefully at the table showing the effects of the two alleles for gene 1. If a plant has the dominant allele I, what happens to fruit color? Once you have that figured out, you should have no trouble with this question the next time you see it.

[c*] no color.

[f] Excellent! if the plant has the allele I, then it won’t express any color.

[c] green and yellow.

[f] No. Here’s how to think about this question. Look carefully at the table showing the effects of the two alleles for gene 1. If a plant has the dominant allele I, what happens to fruit color? Once you have that figured out, you should have no trouble with this question the next time you see it.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|96c5f56e03b34″ question_number=”53″] The expression of fruit pigment in a newly developed variety of apple is under genetic control. The apples can have three phenotypes: no pigment, yellow color or green color. Two genes are involved in the control of pigment production. Each of the genes has two alleles as shown.

Gene 1Gene 2
I: no color productionG: yellow fruit
i: color productiong: green fruit

In a cross between two plants with genotypes IiGg and iiGg, the proportion of offspring with fruit

[c*] with no color is 1/2.

[f] Fabulous. If you just look at the I allele, then the cross is between Ii and ii. Half of the offspring will have the I allele, which means no color (and you can pretty much ignore the rest).

[c] that are green is 3/8.

[f] No. To keep things simple, just focus on the I allele. If you set up a Punnett square representing a cross between Ii and ii, you’ll have something like what’s shown below. Complete the square, and note what percentage of the offspring have the I allele.

 Ii
 i
 i


Now look at the question, and see what the consequence is of having that I allele upon the phenotype. Remember all of this for the next time you see this question.

[c] that are yellow is 1/8.

[f] No. To keep things simple, just focus on the I allele. If you set up a Punnett square representing a cross between Ii and ii, you’ll have something like what’s shown below. Complete the square, and note what percentage of the offspring have the I allele.

 Ii
 i
 i

Now look at the question, and see what the consequence is of having that I allele upon the phenotype. Remember all of this for the next time you see this question.

[c] that are either green or yellow is 1/4.

[f] No. To keep things simple, just focus on the I allele. If you set up a Punnett square representing a cross between Ii and ii, you’ll have something like what’s shown below. Complete the square, and note what percentage of the offspring have the I allele.

 Ii
 i
 i

Now look at the question, and see what the consequence is of having that I allele upon the phenotype. Remember all of this for the next time you see this question.

 

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|96b7d7e5ed734″ question_number=”54″] In the lac operon, which of the following events takes place when lactose is not available in the environment?

[c*] A regulatory protein binds at the operator, blocking transcription of the structural genes.

[f] Correct! Lactose is an inducible operon, meaning that the presence of lactose induces transcription of the structural genes for digesting lactose. When lactose is not in the environment, a regulator protein binds at the operator region, blocking transcription of the structural genes.

[c] A regulatory protein binds to an allosteric site on RNA polymerase, preventing it from binding with the promoter

[f] No. Here’s the structure of the lac operon, shown when lactose (8) is present. Note that lactose is binding with an allosteric site on a regulatory protein (3), changing its shape so that it can no longer bind with the operator region (6) allowing transcription. The regulatory protein is not binding with RNA polymerase (4). Keep that in mind the next time you see this question.

[c] A regulatory protein binds to DNA polymerase, preventing the gene for lactose production to be replicated.

[f] No. DNA polymerase isn’t involved in the function of this operon. The operon works as shown below: When lactose (8) is present,  lactose binds with an allosteric site on a regulatory protein (3), changing its shape so that it can no longer bind with the operator region (6) allowing transcription.  Keep that in mind the next time you see this question.

[c] A regulator protein binds to the start codon on mRNA, preventing the initiation of translation by the ribosome.

[f] No. The operon is working at the level of transcription, not translation. Rather, the operon works as shown below: When lactose (8) is present, lactose binds with an allosteric site on a regulatory protein (3), changing its shape so that it can no longer bind with the operator region (6) allowing transcription.  Keep that in mind the next time you see this question.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|96ac0e69bb334″ question_number=”55″] Cristae are folds in the inner mitochondrial membrane (shown at number 2 below). Electron micrographs show that mitochondria in heart muscle have much higher density of cristae than mitochondria in skin cells. That is because

[c] mutations that occur in the mitochondrial genome during development lead to distinct mitochondrial phenotypes in different parts of the body.

[f] No. This is something that occurs regularly throughout our species (and probably throughout almost all vertebrate mammals). A mutation during development wouldn’t lead to such a regular result. Here’s a hint. Think about what happens in the cristae (the inner mitochondrial membrane). Is there any way that having more cristae could be adaptive?

[c] cristae contains receptor proteins that transduce signals from the skeletal muscles, causing heart tissue to increase ATP production when oxygen is needed.

[f] No. Muscle tissue cells consume ATP that’s made in the mitochondria. Here’s a hint. Think about what happens in the cristae (the inner mitochondrial membrane). Is there any way that having more cristae could be adaptive?

[c*] cristae contain the electron transport chain and ATP synthase, which produce ATP that is needed for the contraction of heart muscle but not for skin function

[f] Bingo! That’s exactly why mitochondria in heart cells would have more cristae than those in skin cells (which don’t need very much ATP).

[c] cristae decrease inner membrane surface area, which provides for increased efficiency in heart muscle tissue, enabling heart cells to produce ATP at a higher rate.

[f] No. You’re on the right track. Keep thinking about surface area, and also think about what happens in the cristae (the mitochondrial inner membrane). If you can figure out why having more cristae would be adaptive, you’ll have your answer.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|96a01faccab34″ question_number=”56″] The reason why neurons differ from skin cells is that the neuron cells they

[c] have transposons inserted in different sites of the genome.

[f] No. This question is about genomic equivalence, a fundamental principle of  eukaryotic development. The main thing to remember is that all of the cells in a multicellular organism have the same genes. They’re different because they express different genes. My muscle cells are expressing the genes for muscle cell proteins, while the cells in the lenses of my eyes are expressing eye-lens proteins. Remember that for the next time you’ll see this question.

[c] have different kinds of genes.

[f] No. This question is about genomic equivalence, a fundamental principle of  eukaryotic development. The main thing to remember is that all of the cells in a multicellular organism have the same genes. They’re different because they express different genes. My muscle cells are expressing the genes for muscle cell proteins, while the cells in the lenses of my eyes are expressing eye-lens proteins. Remember that for the next time you’ll see this question.

[c] have different kinds of ribosomes that utilize different genetic codes.

[f] No. This question is about genomic equivalence, a fundamental principle of  eukaryotic development. The main thing to remember is that all of the cells in a multicellular organism have the same genes. They’re different because they express different genes. My muscle cells are expressing the genes for muscle cell proteins, while the cells in the lenses of my eyes are expressing eye-lens proteins. Remember that for the next time you’ll see this question.

[c*] express different genes.

[f] Fabulous! This question is about genomic equivalence, a fundamental principle of  eukaryotic development. All of the cells in a multicellular organism have the same genes. They’re different because they express different genes.

[c] have different numbers of chromosomes.

[f] No. This question is about genomic equivalence, a fundamental principle of  eukaryotic development. The main thing to remember is that all of the cells in a multicellular organism have the same genes. They’re different because they express different genes. My muscle cells are expressing the genes for muscle cell proteins, while the cells in the lenses of my eyes are expressing eye-lens proteins. Remember that for the next time you’ll see this question.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|96940baf1bf34″ question_number=”57″] In order to clone genes, bacteria are frequently used. This is because bacteria

[c] are able to randomly cut inserted DNA into fragments of varying size by using restriction enzymes.

[f] No. If bacteria randomly cut inserted DNA into fragments, that would be disastrous for cloning genes. See if there’s a choice that looks like it would be useful for making many copies of a gene.

[c*] can rapidly duplicate any DNA sequence that is successfully inserted into a bacterial cell.

[f] Nice job. Because the mechanisms for copying DNA are widely shared among all life forms, it’s possible to clone genes by inserting them into bacteria.

[c] use mitosis to exponentially increase their population.

[f] No. Mitosis is something that’s done exclusively by eukaryotes (and not by bacteria, which a prokaryotes). See if there’s a choice that looks like it would be useful for making many copies of a gene.

[c] freely allow the entry of extracellular DNA into their nuclei.

[f] No. Bacteria don’t have nuclei (a trait that is exclusive to eukaryotes like animals, plants, and fungi. See if there’s a choice that looks like it would be useful for making many identical copies of a gene.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|9685ee2705b34″ question_number=”58″] The image below represents the pGLO plasmid, which is used both in research and in education.

As indicated, the plasmid contains a restriction site and three genes:

  • ampR—confers resistance to ampicillin, and antibiotic.
  • gfp—encodes the green fluorescent protein (GFP), which glows green (fluoresces) under ultraviolet (UV) light
  • araC—encodes a protein required to promote the expression of GFP when arabinose, a disaccharide, is present

The results of a bacterial transformation experiment using the pGLO plasmid are shown in the table below.

W

(untransformed bacteria)

X

(untransformed bacteria)

Y

(transformed bacteria)

Z

(transformed bacteria)

Nutrient agarNutrient agar + ampicillinNutrient agar, ampicillin, and arabinoseNutrient agar and ampicillin

Which plate would contain bacteria that fluoresce (glow green) under UV light?

[c] plate W

[f] No. These bacteria have not been transformed (meaning that they haven’t taken up the plasmid). Without the plasmid, they lack the GFP  gene, and therefore would have no way to synthesize the protein that would enable them to glow. Next time, chose from one of the two plates with transformed bacteria.

[c] plate X

[f] No. These bacteria have not been transformed (meaning that they haven’t taken up the plasmid). On top of that, they’ve been placed upon the antibiotic ampicillin, which is preventing any bacterial growth. Next time, chose from one of the two plates with transformed bacteria.

[c*] plate Y

[f] Fabulous. These bacteria have taken up the plasmid, which provides them with the gene to synthesize GFP. On top of that, they have been provided with arabinose, which induces the cell to synthesize GFP.

[c] plate Z

[f] No. While these bacteria have been transformed (as demonstrated by the fact that they’re growing in the presence of the antibiotic ampicillin), they have not been fed arabinose, which is required to induce the cell to synthesize GFP. Remember that next time you see this question, and make another choice.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|96796a671c334″ question_number=”59″] The image below represents the pGLO plasmid, which is used both in research and in education.

As indicated, the plasmid contains a restriction site and three genes:

  • ampR—confers resistance to ampicillin, and antibiotic.
  • gfp—encodes the green fluorescent protein (GFP), which glows green (fluoresces) under ultraviolet (UV) light
  • araC—encodes a protein required to promote the expression of GFP when arabinose, a disaccharide, is present

The results of a bacterial transformation experiment using the pGLO plasmid are shown in the table below.

W

(untransformed bacteria)

X

(untransformed bacteria)

Y

(transformed bacteria)

Z

(transformed bacteria)

Nutrient agarNutrient agar + ampicillinNutrient agar, ampicillin, and arabinoseNutrient agar and ampicillin

What was the purpose of plates W or X?

[c] Plate W showed successful destruction of the plasmid by bacterial restriction enzymes.

[f] No. Plate W has untransformed bacteria that never took up the plasmid in the first place. Next time you see this question, think about the idea experimental controls, and make a different choice.

[c] Plate W showed that there was extremely high transformation efficiency.

[f] No. The growth on plate W is not about transformation efficiency, because the bacteria on that plate were not transformed.  Next time you see this question, think about the idea experimental controls, and make a different choice.

[c] Plate X showed how nutrient agar can enhance the growth of viable bacteria.

[f] No. There’s no growth on plate X, and that’s due to the presence of the antibiotic ampicillin. Next time you see this question, think about the idea experimental controls, and make a different choice.

[c*] Plate X showed that ampicillin was effective in inhibiting the growth of bacteria that had not taken up the plasmid.

[f] That’s correct! Plate X shows that there won’t be any growth in untransformed bacteria, which haven’t taken up the plasmid that provides them with antibiotic resistance (in addition to the gene for GFP).

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|966c0722bd334″ question_number=”60″] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.

In a very schematic form, the above process represents

[c] DNA replication

[f] No. The “U” nucleotide indicates uracil, found in RNA but not DNA. So, you can rule out DNA replication. What process converts RNA nucleotides into amino acids.

[c] transcription

[f] No. Transcription changes information in DNA into RNA. Here, information in RNA is being changed into protein. It’s similar to what happens when you change the words in one language into equivalent words in another language. What process is that?

[c] transformation

[f] No. The term transformation refers to a when a bacterial cell takes up DNA from a plasmid (or from the environment), thereby acquiring new genes. Here, information in RNA is being changed into protein. It’s similar to what happens when you change the words in one language into equivalent words in another language. What process is that?

[c*] translation

[f] Fabuloso! The process at work above is translation, which translates RNA sequences into amino acid sequences.

[c] transduction

[f] No. Transduction refers to several processes in biology, such as when a virus brings DNA from one bacterial cell to another during a viral replication cycle. Here, information in RNA is being changed into protein. It’s similar to what happens when you change the words in one language into equivalent words in another language. What process is that?

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|965f8362d3b34″ question_number=”61″] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.

The processes that would change the nucleic acid information in the top row into the amino acid information in the bottom row occur

[c*] at a ribosome

[f] Exactly! The process is translation, at it occurs at a ribosome.

[c] in enzymes embedded in the cell’s membrane.

[f] No. The process that converts nucleic acid information into amino acid information is translation. Where does that occur?

[c] in the nucleus.

[f] No. The process that converts nucleic acid information into amino acid information is translation. Where does that occur?

[c] in the smooth Endoplasmic reticulum

[f] No. The process that converts nucleic acid information into amino acid information is translation. Where does that occur?

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|964f11ced9734″ question_number=”62″] The following image shows a sequence of nucleic acid bases coding for a sequence of amino acids. The third amino acid in the sequence (lysine) has been filled in.

The sequence of nucleic acid bases on the top line of the image above make up

[c] DNA

[f] No. The presence of the nucleotide uracil (U) tells you that this can’t be DNA. What nucleic acid contains uracil, and gets translated into amino acids?

[c] tRNA

[f] No. You’re right about the RNA part, but you’ve chosen the wrong type of RNA. What type of RNA gets translated into amino acids?

[c] rRNA

[f] No. You’re right about the RNA part, but you’ve chosen the wrong type of RNA. What type of RNA gets translated into amino acids?

[c*] mRNA

[f] Fantastic. mRNA is the nucleic acid that gets translated into protein.

[c] RNAi

[f] No. RNAi is amazing stuff, responsible for all type of epigenetic effects…but it’s not the RNA you’re looking for. What type of RNA gets translated into amino acids?

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|9641ae8a7a734″ question_number=”63″] Certain restriction enzymes cut DNA in a way that leaves the fragments with “sticky ends.” These “sticky ends” are useful in genetic engineering because they allow for

[c] screening of plasmids with antibiotic resistance genes.

[f] No. Take a look at this picture of DNA that’s been cut in a way that leaves it with sticky ends. Now, look again at the choices and see if you can find a better answer.

[c] identification of bacteria that have been successfully transformed by plasmids.

[f] No. Take a look at this picture of DNA that’s been cut in a way that leaves it with sticky ends. Now, look again at the choices and see if you can find a better answer.

[c] replication of messenger RNA within the bacterial cell

[f] No. Take a look at this picture of DNA that’s been cut in a way that leaves it with sticky ends. Now, look again at the choices and see if you can find a better answer.

[c*] DNA from different sources to be joined together

[f] Excellent. DNA that’s been cut in a way that leaves it with sticky ends. Now, look again at the choices and see if you can find a better answer.

 

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9635754c0d734″ question_number=”64″] A primatologist studying lowland gorillas has developed a scheme for classifying gorilla noses as long, medium, or short in length, and either wide or narrow. She observes the results of several naturally-occurring matings, which are recorded below.

  1. Long, wide X long, wide | All long, wide
  2. Short, narrow X long, narrow | all medium, narrow
  3. Medium, wide X medium, narrow | 1/4 long wide, 1/2 medium wide, 1/4 short wide
  4. Short, wide X medium, narrow | 1/2 medium, wide; 1/2 short, wide
  5. Short, wide X short, narrow | all short, wide

Which of the following best describes the underlying genetics for inheritance of nose length.

[c] One locus governs nose length. That locus can hold three possible alleles. One codes for a short nose, the second for a medium nose, and the third for a long nose.

[f] No. You don’t need three alleles to explain the inheritance described in this problem. Rather, the way to approach this question is to look at each cross, think of a mode of inheritance that could explain the outcome, and then compare that to each choice. Crosses number 2, 3, and 4 tell us that nose length can be explained by two alleles. However, you need to account for the fact that some of the offspring had an intermediate phenotype. What kind of inheritance allows for that?

[c] Two loci govern nose length. If there are dominant alleles at both loci, the nose will be Long. If one locus has a dominant allele, the nose will be short. If there are no dominant alleles at either locus, the nose will be medium.

[f] No. If you’re just considering nose length, you can explain it through two alleles at one locus. However, you need to account for the fact that some of the offspring have an intermediate phenotype. What kind of inheritance allows for that?

[c*] One locus governs nose length. The locus has two alleles. A gorilla can be homozygous long or homozygous short. Heterozygotes have medium nose length.

[f] Excellent. This model works perfectly for describing inheritance of nose length.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|9628f18c23f34″ question_number=”65″] A research team is studying the genetics of beak length in house sparrows, where beak length has two phenotypes, long and short. Pedigree studies have shown that this phenotype is under the control of a single autosomal gene with two alleles. The DNA for each allele is isolated and sequenced, and it’s determined that a single nucleotide difference accounts for the short mutation.

When the DNA of both alleles is subjected to restriction fragment analysis, the pattern below is produced.

Based on the diagram, the most reasonable explanation for the different RFLPs associated with the long and short beaked alleles is that a mutation in the long allele DNA

[c] eliminated a restriction site.

[f] No. The key thing to notice is that the long allele DNA has four fragments, and the short allele DNA has five. A mutation that eliminated a restriction site would produce fewer fragments, not more. Remember that, and make another choice next time you see this question.

[c] activated an inactive gene, creating new DNA fragments.

[f] No. It’s not impossible for a mutation to activate a previously inactive gene. But activating a gene doesn’t create new restriction fragments. Next time you see this question, think about this: what kind of mutation in DNA would increase the number of restriction fragments? This what you have to explain (because there are four fragments in the long DNA’s allele, and five fragments in the short DNA).

[c*] created a new restriction site.

[f] Correct. RFLP analysis produces four fragments in the DNA for the long allele, and five in the DNA for the short allele. If a mutation created a new restriction site in the second biggest DNA fragment in the long DNA, then it would produce two additional fragments, each of intermediate length. That would create precisely the patten that’s seen in the short allele.

[c] disabled an active gene.

[f] No. Disabling an active gene would change gene expression, but it wouldn’t necessarily change the RFLP pattern. Next time you see this question, think about this: what kind of mutation in DNA would increase the number of restriction fragments? This what you have to explain (because there are four fragments in the long DNA’s allele, and five fragments in the short DNA).

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|961a6441d2f34″ question_number=”66″] In Drosophila, allele W codes for normal wings, while allele w codes for vestigial wings. Allele G codes for normal body color, while allele g codes for ebony. A heterozygous normal winged, ebony male is crossed with a vestigial winged, heterozygous normal body color female.

What is the genotype of the male fly?

[c] WwGg

[f] No. The male is described as heterozygous normal winged, ebony. WwGg would be described as heterozygous normal winged, heterozygous normal body color. What genotype would a fly have to have to be ebony?

[c] WWGG

[f] No. The male is described as heterozygous normal winged, ebony.WWGG would be described as homozygous normal winged, homozygous normal body color. What genotype would a fly have to be heterozygous for normal wings, and ebony bodied.

[c] wwgg

[f] No. The male is described as heterozygous normal winged, ebony.The genotype wwgg would be described as homozygous vestigial winged, homozygous ebony body color (so you got the second part right) What genotype would a fly have to be heterozygous for normal wings?

[c] wwGG

[f] No. The male is described as heterozygous normal winged, ebony.The genotype wwGG would be described as homozygous vestigial winged, homozygous normal body color. What genotype would a fly have to have to be heterozygous for normal wings, and ebony bodied (and remember that ebony bodied is a recessive trait).

[c*] Wwgg

[f] Fabulous. A genotype of Wwgg means that the fly is heterozygous for normal wings, and ebony colored.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|960d263e32334″ question_number=”67″] In Drosophila, allele W codes for normal wings, while allele w codes for vestigial wings. Allele G codes for normal body color, while allele g codes for ebony coloring. A heterozygous normal winged, ebony male is crossed with a vestigial winged, heterozygous normal body color female. What ratio of of phenotypes is expected in the offspring?

[c*] 1 normal wings, normal body : 1  normal wings, ebony body : 1 vestigial wings, normal body : 1  vestigal wings, ebony body

[f] Excellent. A cross between Wwgg and wwGg will give you a 1:1:1:1 ratio of phenotypes in the offspring.

[c] 3 normal wings, normal body : 3 normal wings, ebony body : 3 vestigial wings, normal body : 3 vestigal wings, ebony body

[f] No. Start by reading the description of the parents, and converting them into genotypes. The male is heterozygous normal winged and ebony-bodied, so his genotype is Wwgg. The female is vestigial winged, heterozygous normal body color, so her genotype is wwGg. In the Punnett square below, I put the males gametes on the left, and the mother’s gametes on top, and I even filled in the first square for you. You do the rest, and then convert the genotypes back into phenotypes. That should give you the answer when you next see this question.

wGwg
Wg WgGg
wg

[c] 3 normal wings, normal body : 1 normal wings, ebony body : 3 vestigial wings, normal body : 1 vestigial wings, ebony body

[f] No. Start by reading the description of the parents, and converting them into genotypes. The male is heterozygous normal winged and ebony-bodied, so his genotype is Wwgg. The female is vestigial winged, heterozygous normal body color, so her genotype is wwGg. In the Punnett square below, I put the males gametes on the left, and the mother’s gametes on top, and I even filled in the first square for you. You do the rest, and then convert the genotypes back into phenotypes. That should give you the answer when you next see this question.

wGwg
Wg WgGg
wg

[c] 9 normal wings, normal body : 3 normal wings, ebony body : 3 vestigial wings, normal body : 1 vestigal wings, ebony body

[f] No. Start by reading the description of the parents, and converting them into genotypes. The male is heterozygous normal winged and ebony-bodied, so his genotype is Wwgg. The female is vestigial winged, heterozygous normal body color, so her genotype is wwGg. In the Punnett square below, I put the males gametes on the left, and the mother’s gametes on top, and I even filled in the first square for you. You do the rest, and then convert the genotypes back into phenotypes. That should give you the answer when you next see this question.

wGwg
Wg WgGg
wg

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|960057fccc334″ question_number=”68″] Imagine that an error has during DNA replication in a cell. The error substitutes a G in place of a T in one of the cell’s genes. Which of the following is the most likely effect that this will have on the cell?

[c] Each of the cell’s proteins will contain an incorrect amino acid.

[f] No. The main problem with this choice is that the mutation will affect only a single protein (not each of the cell’s proteins). When you next see this question, think of how information in DNA gets translated into protein, and consider the likely effect of a single base pair substitution.

[c] The amino acid sequence of one of the cell’s proteins will be completely altered.

[f] No. All that this mutation does is substitute one nucleotide for another. The effect described here (completely altering the amino acid sequence) is more commonly associated with base pair deletions or insertions that change the reading frame. When you next see this question, think of how information in DNA gets translated into protein, and consider the likely effect of a single base pair substitution.

[c] One amino acid will be missing from one of the cell’s proteins,

[f] No. All that this mutation does is substitute one nucleotide for another. That might result in an incorrect amino acid, but it wouldn’t result in a missing amino acid.

[c*] One of the cell’s proteins might contain an incorrect amino acid.

[f] Correct. A likely effect of a single nucleotide change is that one of the cell’s protein might contain an incorrect amino acid.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|95f07b6bcaf34″ question_number=”69″] A team of biologists is studying gene expression in rat liver tissue. They find that two different proteins with different structures were translated from two different mRNAs. When they traced the source of the mRNA, however, they found the both mRNAs were transcribed from the same gene in the cell’s nucleus. Which of the following explanation best accounts for what the team has uncovered?

[c] The gene might have been altered by a mutation.

[f] No. The question states that both of these mRNAs were transcribed from the same gene. If the gene were altered by a mutation, then the RNA would be changed, and the resulting protein would be changed. But it wouldn’t produce the two RNAs that are described above. The next time you see this question, think about gene expression in eukaryotes, and see if you can remember a mechanism by which two RNAs could result from one gene.

[c] The different functions of each protein require them to have different underlying RNAs which code for them.

[f] No. Remember the central dogma of molecular biology: DNA makes RNA makes protein. This choice is proposing a process that violates this key idea. The next time you see this question, think about gene expression in eukaryotes, and see if you can remember a mechanism by which two RNAs could result from one gene.

[c] Different systems of histone acetylation could result in two different mRNAs.

[f] No. Different patterns of histone acetylation can result in different patterns of gene expression, but it wouldn’t produce two mRNAs from one gene. The next time you see this question, think about gene expression in eukaryotes, and see if you can remember a mechanism by which two RNAs could result from one gene.

[c*] During pre-mRNA processing, exons could be spliced together in various ways to make different mRNAs.

[f] Way to go! If one gene is resulting in two mRNAs, then what’s happening is alternative splicing of RNA. These different RNAs would then go on to be translated as distinct proteins.

[c] One set of ribosomes read the mRNA in the 5′ to 3′ direction. Another set reads them in the 3′ to 5′ direction.

[f] No. All ribosomes (whether in an E. coli bacterium or an orchid, read mRNA in the 5′ to 3′ direction. The next time you see this question, think about gene expression in eukaryotes, and see if you can remember a mechanism by which two RNAs could result from one gene.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|95e362a8e8734″ question_number=”70″] In shorthorn cattle, coat color can be solid red, solid white, or red-roan (a mixture of white and pigmented hairs). When a true breeding solid red male is crossed with a white female, the offspring are roan. This demonstrates that

[c] coat color is controlled by at least three alleles.

[f] No. You might be thinking of systems like inheritance of blood type, where the A allele and the B allele produce blood type AB. But note that you can achieve the intermediate phenotype with just two alleles (as opposed to three). Keep that in mind when you next see this question.

[c*] red and white show incomplete dominance.

[f] Yes. If r is the allele for Red and w is the allele for white, then the cross above could be represented as rr x ww, with all the offspring having the genotype rw, and the roan phenotype.

[c] the red allele is dominant over the white allele.

[f] No. If R represented the dominant allele for red, and r represented the recessive allele for white, then a cross between true breeding red and white would be RR x rr. All the offspring would be Rr, and we’d expect them to be red, not roan. What kind of inheritance system would result in pure bred red crossed with purebred white resulting in an intermediate phenotype?

[c] Crossing over and recombination occurring during meiosis.

[f] No. Crossing over and recombination almost certainly did occur during meiosis, but it wouldn’t achieve the result described above. What kind of inheritance system would result in pure bred red crossed with purebred white resulting in an intermediate phenotype?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|95d5da23cb334″ question_number=”71″] Blood type inheritance involves three alleles. While there are several notation systems for representing these alleles, a common one has alleles IA and IB as codominant to one another, with each of these dominant over i. If a child has type O blood (genotype ii), which of the following blood types is impossible in either parents?

[c] Type O

[f] No. The best way to think about this is to draw a Punnett square, and put the genotype of the type O child in the lower right box, as shown below.

 ii

If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.

 ___i
 ___
 i ii

Each parent is __i. It’s completely possible that the child could have a parent that’s type O (genotype ii). But what genotype is impossible for either of the parents to have?

[c] Type B

[f] No. The best way to think about this is to draw a Punnett square, and put the genotype of the type O child in the lower right box, as shown below.

 ii

If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.

 ___i
 ___
 i ii

Each parent is __i. It’s completely possible that the child could have a parent that’s type B (genotype IBi). But what genotype is impossible for either of the parents to have?

[c] Type A

[f] No. The best way to think about this is to draw a Punnett square, and put the genotype of the type O child in the lower right box, as shown below.

 ii

If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.

 ___i
 ___
 i ii

Each parent is __i. It’s completely possible that the child could have a parent that’s type A (genotype IAi). But what genotype is impossible for either of the parents to have?

[c*] Type AB

[f] Way to go. Gregor Mendel would be proud. A child with type O blood needs each of their parents to give them a i allele. Given that, type AB is impossible (because a type AB parent could give their child either IA  or IBbut not a i.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|95c8519eadf34″ question_number=”72″] Blood type inheritance involves three alleles. While there are several notation systems for representing these alleles, a common one has alleles IA and IB as codominant to one another, with each of these dominant over i. If a man with type B blood has children with a woman with type O blood, which of the following blood types would be impossible in their children.

[c] Type O

[f] No. The man with type B blood could have one of two genotypes: IBi (if he were heterozygous), or IBIB, (if he were homozygous type B). The woman is type O, so her genotype has to be ii. If you can set up two Punnett squares on your own, then do so, and you’ll see the two genotypes that are impossible. If not, then continue reading below.

Here’s the Punnett square that shows the father as a heterozygote:

 IB i
 i
 i ii

Here’s the Punnett square that shows the father as a homozygote:

 IB IB
 i
 i ii

Complete both Punnett squares by bringing the alleles down and over. As you can see by the first Punnett square, a type O child is possible. But which blood type is impossible? 

[c] Type B

[f] No. The man with type B blood could have one of two genotypes: IBi (if he were heterozygous), or IBIB, (if he were homozygous type B). The woman is type O, so here genotype has to be ii. If you can set up two Punnett squares on your own, then do so, and you’ll see the two genotypes that are impossible. If not, then continue reading below.

Here’s the Punnett square that shows the father as a heterozygote:

 IB i
 i
 i ii

Here’s the Punnett square that shows the father as a homozygote:

 IB IB
 i
 i ii

Complete both Punnett squares by bringing the alleles down and over. As you can see, in either Punnett square, a type B child is possible. But which blood type is impossible? 

[c*] Type AB

[f] Correct! The man with type B blood could have one of two genotypes: IBi (if he were heterozygous), or IBIB, (if he were homozygous type B). The woman is type O, so her genotype has to be ii. The offspring will all be type B or type O. Type AB is impossible (as is type A).

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|95bb38dbcb734″ question_number=”73″] Down syndrome and and Klinefelter’s syndrome have in which of the following in common?

[c] Both involve the X or Y chromosomes

[f] No. While Klinefelter’s syndrome does involve the X chromosome, Down syndrome has no connection to the X or Y chromosome.  Down syndrome occurs when a person receives three copies of chromosome 21, rather than two. Klinefelter’s syndrome occurs when a chromosomally male person has an extra X chromosome. Next time, choose another answer (and note that I’ve just given you a huge hint as to what the answer is).

[c] Both are the result of a point mutations.

[f] No. Neither Down syndrome nor Klinefelter’s syndrome involve point mutations. Down syndrome occurs when a person receives three copies of chromosome 21, rather than two. Klinefelter’s syndrome occurs when a chromosomally male person has an extra X chromosome. Next time, choose another answer (and note that I’ve just given you a huge hint as to what the answer is).

[c*] Both are associated with of changes in chromosome number

[f] Yes. Both Down syndrome and Klinefelter’s syndrome involve changes in chromosome number. Down Syndrome occurs when a person receives three copies of chromosome 21, rather than two. Klinefelter’s syndrome occurs when a chromosomally male person has an extra X chromosome.

[c] Both conditions are caused by autosomal dominant alleles

[f] No. Down and Klinefelter syndromes are not caused by autosomal dominant alleles. Down syndrome occurs when a person receives three copies of chromosome 21, rather than two. Klinefelter’s syndrome occurs when a chromosomally male person has an extra X chromosome. Next time, choose another answer (and note that I’ve just given you a huge hint as to what the answer is).

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|95ab818b88734″ question_number=”74″] Conditions like hemophilia, Red-green colorblindness, and Duchenne’s muscular dystrophy are inherited a sex-linked recessive disorders. The term “sex-linked” means:

[c] The allele is carried on an autosome

[f] No. The reason these conditions are “sex linked” is because they’re not carried on autosomes. If they were carried on autosomes, their inheritance would not be linked to sex. What chromosomes would a condition have to be linked to in order to be linked to sex?

[c] The allele is carried on the Y chromosome

[f] No, but you’re thinking about this the right way. Hemophilia, red-green colorblindness, and Duchenne’s muscular dystrophy show up more often in males than females, but not because they are on the Y chromosome. What chromosome would a condition have to be linked to in order to be linked to sex?

[c*] The allele is carried on the X~chromosome

[f] Fabulous! The alleles for hemophilia, red-green colorblindness, and Duchenne’s muscular dystrophy are all found on the X chromosome, making them sex linked.

[c] The allele is carried in the mitochondrial DNA

[f] No. Alleles carried in mitochondrial DNA have their own pattern of inheritance, logically called “mitochondrial inheritance.” The pattern is not the same as what’s seen in sex linked conditions (such as hemophilia, red-green colorblindness, and Duchenne’s muscular dystrophy). All of these conditions show up more often in males than females, and they’re all linked to a specific chromosome. What chromosome would a condition have to be linked to in order to be linked to sex?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|959e68c8a5f34″ question_number=”75″] A man and a woman, both with normal vision, have a son who is red-green colorblind. Which of the following is the most likely explanation.

[c] Both maternal grandparents are colorblind

[f] No. The key thing to remember here is that red-green colorblindness is a recessive, sex-linked condition, with the allele for the condition found on the X chromosome. The genotype of a colorblind male is represented as XcY, whereas a male with normal vision would have the genotype XCY. Now, think about what you know about inheritance, and ask yourself who that colorblind male would have inherited his X chromosome from. Remember that the next time you see this question.

[c*] The mother carries a recessive allele

[f] Excellent! Because red-green colorblindness is a recessive sex linked condition, it’s always transmitted by the mother, who passes a copy of the recessive allele to her sons through an X chromosome in her egg.

[c] The father passed his allele for colorblindness to his son.

[f] No. The key thing to remember here is that red-green colorblindness is a recessive, sex-linked condition, with the allele for the condition found on the X chromosome. The genotype of a colorblind male is represented as XcY, whereas a male with normal vision would have the genotype XCY. Now, think about what you know about inheritance, and ask yourself who that colorblind male would have inherited his X chromosome from. Remember that the next time you see this question.

[c] Both the mother and the father were heterozygous for the color-blind gene.

[f] No. The key thing to remember here is that red-green colorblindness is a recessive, sex-linked condition, with the allele for the condition found on the X chromosome. The genotype of a colorblind male is represented as XcY, whereas a male with normal vision would have the genotype XCY. While a female can be heterozygous normal, with the genotype  XCXc, a male can’t be heterozygous. He either inherits the normal allele, or the recessive one (with no countervailing allele on his Y chromosome)

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|958ed6b921334″ question_number=”76″] The karyotype shown below is an example of a

[c*] trisomy.

[f] Nice job! As you observed, there are three copies of chromosome 15, creating a trisomy.

[c] monosomy.

[f] No. A monosomy would have just one chromosome, instead of a pair. You might have been confused by the single X and Y chromosomes, but that’s the pair that creates a chromosomally male mammal. Take a closer look at the chromosomes, and see if you can find another type of chromosomal difference in this karyotype.

[c] polyploidy.

[f] No. Polyploidy involves duplication of entire chromosome sets. This is a single set. Take a closer look at the chromosomes, and see if you can find some type of chromosomal difference that corresponds to one of the answers

[c] chromosomal inversion.

[f] No. A chromosomal inversion wouldn’t be possible to discern in this type of black and white karyotype. An inversion results when chromosomal segments are flipped, as is shown in the diagram below (from nobelprize.org)

Take a closer look at the chromosomes, and see if you can find some type of chromosomal difference that corresponds to one of the other answers.

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|958128f345b34″ question_number=”77″] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype.

What is the frequency of the recessive allele for this trait?

[c] 0.36

[f] No. This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.

[c] 0.74

[f] No. This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.

[c*] 0.6

[f] Fabulous! Hardy and Weinberg (who developed the math for this) would be proud. If the frequency of the recessive phenotype is 0.36, then (assuming the population is in Hardy-Weinberg equilibrium), the frequency of the recessive allele will be 0.6.

[c] 0.4

[f] No. This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.

[c] 0.48

[f] No. This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|957355ecabf34″ question_number=”78″] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype. What is the frequency of the dominant allele for this trait?

[c] 0.36

[f] No. This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?

[c] 0.74

[f] No. This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?

[c] 0.6

[f] No. 0.6 is the frequency of the recessive allele. I’m not sure where you went wrong, so I’ll walk you through the whole thing.

This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?

[c*] 0.4

[f] Fabulous!  If the frequency of the recessive phenotype is 0.36, then (assuming the population is in Hardy-Weinberg equilibrium), then the frequency of the recessive allele will be 0.6. Since p + q = 1, then you simply subtract 1 – 0.6 to get the answer, 0.4.

[c] 0.48

[f] No. 0.48 is the frequency of heterozygotes in this population. I’m not sure where you went wrong, so I’ll walk you through the whole thing.

This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|95651323d7734″ question_number=”79″] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype. What is the frequency of the heterozygotes within this population?

[c] 0.36

[f] No. 0.36 is the frequency of homozygous dominants in this population. I’m not sure where you went wrong, so I’ll walk you through the whole thing.

This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.

If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.

There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.

[c] 0.74

[f] No. I’m not sure where you went wrong, so I’ll walk you through the whole thing.

This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.

If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.

There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.

[c] 0.6

[f] No. 0.6 is the frequency of the recessive allele. That’s a good start, but since I’m not sure where you went wrong, I’ll walk you through the whole thing.

This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.

If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.

There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.

[c] 0.4

[f] No. 0.4 is the frequency of the dominat allele. That’s a good start, but since I’m not sure where you went wrong, I’ll walk you through the whole thing.

This is a population genetics problem, and the easiest way to solve this problem is to use a cross multiplication table (a kind of Punnett square). You start from the idea that p + q = 1, where p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.

If you plugged these values into a cross multiplication table, you’d get:

 pq
p p2pq
 q pqq2

You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.

 pq
p p2pq
 q pq.36

To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.

There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.

[c*] 0.48

[f] Excellent! The percentage of heterozygotes in the population is 0.48.

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|9557655dfbf34″ question_number=”80″] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype. What percentage of individuals will have the dominant phenotype?

[c] 0.36

[f] No. 0.36 is the frequency of the individuals with the recessive phenotype. You’re looking for the percentage of individuals with the dominant phenotype.

[c] 0.74

[f] No. You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. If 36% of the population has the recessive phenotype, what percentage has the dominant phenotype?

[c*] 0.64

[f] Excellent. If 36% of the population has the recessive phenotype, then everybody else in the population has to have the dominant phenotype. 1 – 0.36 = 0.64.

[c] 0.40

[f] No. You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. If 36% of the population has the recessive phenotype, what percentage has the dominant phenotype?

[c] 0.48

[f] No. 0.48 is the frequency of heterozygotes. You looking for the number of individuals with the dominant phenotype. You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. If 36% of the population has the recessive phenotype, what percentage has the dominant phenotype?

 

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|954a4c9b19734″ question_number=”81″] The image below is a scatter graph that shows the growth rate and egg productivity in a flock of chickens.

A breeder has divided this flock into four groups, A, B, C, and D. If the breeder wants to increase the flocks’ growth rate and egg-laying productivity, which group should she select from?

[c] A

[f] No. This is a question about artificial selection, and the basic idea is that you want to use animals that have desired traits as breeding stock. The breeder wants both high growth rate and egg laying productivity. Take a look at the labels for each axis on this graph. The chickens in group A have high egg laying productivity (that’s why they’re up on the Y axis, which is for egg laying productivity). They are, however, low in terms of growth rate (which is why they’re on the left of the X axis, which is growth rate). Which group is high for both egg laying and growth rate?

[c*] B

[f] Way to go! The best breeding stock would be those chickens with high growth rate and high egg laying, and that’s going to be group B.

[c] C

[f] No. This is a question about artificial selection, and the basic idea is that you want to use animals that have desired traits as breeding stock. The breeder wants both high growth rate and egg laying productivity. Take a look at the labels for each axis on this graph. The chickens in group C have low egg laying productivity (that’s why they’re down on the Y axis, which is for egg laying productivity). But they’re high in terms of growth rate (which is why they’re on the right of the X axis, which is growth rate). Which group is high for both egg laying and growth rate?

[c] D

[f] No. This is a question about artificial selection, and the basic idea is that you want to use animals that have desired traits as breeding stock. The breeder wants both high growth rate and egg laying productivity. Take a look at the labels for each axis on this graph. The chickens in group D have low egg laying productivity (that’s why they’re down on the Y axis, which is for egg laying productivity). They’re also low in terms of growth rate (which is why they’re on the left of the X axis, which is growth rate). Which group is high for both egg laying and growth rate?

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|953ce956ba734″ question_number=”82″] The image below is a scatter graph that shows the growth rate and egg productivity in a flock of chickens.

A breeder has divided this flock into four groups, A, B, C, and D. If the breeder is selecting for maximum growth rate (for meat production), and has less interest in egg production, which group should she select from?

[c] A

[f] No. This is a question about artificial selection, and the basic idea is that you want to use animals that have desired traits as breeding stock. The breeder wants a high growth rate, but has less interest in egg laying productivity. Take a look at the labels for each axis on this graph. The chickens in group A have high egg laying productivity (that’s why they’re up on the Y axis, which is for egg laying productivity). They are, however, low in terms of growth rate (which is why they’re on the left of the X axis, which is growth rate). Which group is high for growth rate, but low for egg laying?

[c] B

[f] No. This is a question about artificial selection, and the basic idea is that you want to use animals that have desired traits as breeding stock. The breeder wants a high growth rate, but isn’t interested in high egg production. Take a look at the labels for each axis on this graph. The chickens in group B have high egg laying productivity (that’s why they’re up on the Y axis, which is for egg laying productivity). They are also high in terms of growth rate (which is why they’re on the left of the X axis, which is growth rate). Which group will provide the breeder with breeding stock that will result in high growth rate, but lower egg laying capability?

[c*] C

[f] Terrific! Group C is the group to choose from if you want breeding stock for chickens with high growth rate, but low egg-laying capability.

[c] D

[f] No. This is a question about artificial selection, and the basic idea is that you want to use animals that have desired traits as breeding stock. The breeder wants a high growth rate, but isn’t interested in high egg production. Take a look at the labels for each axis on this graph. The chickens in group D have low egg laying productivity (that’s why they’re down on the Y axis, which is for egg laying productivity). They are also low in terms of growth rate (which is why they’re on the right of the X axis, which is growth rate). Which group will provide the breeder with breeding stock that will result in high growth rate, but lower egg laying capability?

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|952fab5319b34″ question_number=”83″] A team of biologists has started a project to clone the rare blue lobster through somatic cell nuclear transfer. The blue lobster is a variety of the more common purple lobster. The difference in phenotype has been traced to a genetic mutation that results in a blue exoskeleton.

For somatic cell nuclear transfer to be successful, the best choice for a donor cell will be a(n)

[c] haploid cell.

[f] No. Cloning by somatic cell nuclear transfer involves taking the nucleus from a diploid cell, inserting that nucleus into an egg cell that’s had its genetic material removed, and then inducing that egg cell (now with a complete diploid genome) to divide and develop. Haploid cells, with only a single chromosome set, can’t be used for this purpose. Remember that the next time you see this question.

[c] an undifferentiated sperm cell.

[f] No. Cloning by somatic cell nuclear transfer involves taking the nucleus from a diploid cell, inserting that nucleus into an egg cell that’s had its genetic material removed, and then inducing that egg cell (now with a complete diploid genome) to divide and develop. An undifferentiated sperm cell is haploid, and haploid cells, with only a single chromosome set, can’t be used for this purpose. Remember that the next time you see this question.

[c*] intestinal cell.

[f] Excellent. Intestinal cells are diploid, and could be used as a source of genetic material for cloning.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|9520f8c80a734″ question_number=”84″] If you wanted to clone a dog, the least promising tissue to serve as donor cells would be the

[c*] testes

[f] Correct. Cells in the testes will be haploid, and you’ll need a diploid cell to provide the genetic material for a newly cloned organism.

[c] retina

[f] No. While retinal cells are highly differentiated (which could pose some technical problems), they have all the genetic material need to create a newly cloned organism. One of the tissues listed only has half the genetic material needed. Think about meiosis, think about what that process does to chromosome number, think about where it occurs, and remember that the next time you see this question.

[c] stomach

[f] No. Cells in the stomach they have all the genetic material need to create a newly cloned organism, and could therefore be a source of donor cells. However, one of the tissues listed only has half the genetic material needed. Think about meiosis, think about what that process does to chromosome number, think about where it occurs, and remember that the next time you see this question.

[c] brain

[f] No. While brain cells are highly differentiated (which could pose some technical problems), they have all the genetic material need to create a newly cloned organism. However, one of the tissues listed only has half the genetic material needed. Think about meiosis, think about what that process does to chromosome number, think about where it occurs, and remember that the next time you see this question.

[c] intestines

[f] No. Cells in the intestines have all the genetic material need to create a newly cloned organism, and could therefore be a source of donor cells. However, one of the tissues listed only has half the genetic material needed. Think about meiosis, think about what that process does to chromosome number, think about where it occurs, and remember that the next time you see this question.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|95139583ab734″ question_number=”85″] A team of biologists has started a project to clone the rare blue lobster through somatic cell nuclear transfer. The blue lobster is a variety of the more common purple lobster. The difference in phenotype has been traced to a genetic mutation that results in a blue exoskeleton.

The team has discovered that rate of production of viable clones is highest when intestinal cells are used. The least likely reason for this would be the difference in

[c] histone acetylation

[f] No. Histone acetylation is one way that DNA expression is controlled in eukaryotic organisms. Levels of histone acetylation in donor cells could affect the viability of cloned organisms created from these cells, as certain genes might not express themselves at the right level for proper embryonic development. Next time you see this question, see if you can find something that doesn’t seem to be tightly connected with DNA structure.

[c*] Ca++ level

[f] Terrific. Every other choice is connected either with DNA packaging or with control or gene expression, all of which could affect how genes are expressed in a developing embryo, which could in turn affect the viability of those embryos. Ca++ levels seem like the factor that would have the least impact.

[c] DNA methylation

[f] No. DNA methylation is one way that DNA expression is controlled in eukaryotic organisms, with methyl groups added to cytosine bases in DNA sequences that, in particular tissues, are not expressed.  Levels of DNA methylation in donor cells could affect the viability of cloned organisms created from these cells, as certain genes might not express themselves at the right level for proper embryonic development. Next time you see this question, see if you can find something that doesn’t seem to be tightly connected with DNA structure.

[c] chromatin structure

[f] No. Control of chromatin structure is one way that DNA expression is controlled in eukaryotic organisms. Basically, if chromatin is tightly wound up (a form known as heterochromatin) the DNA in that chromatin is not expressed. In more loosely bound chromatin (a form known as euchromatin), the DNA is available for expression. Consequently, chromatin structure in donor cells could affect the viability of cloned organisms created from these cells, as certain genes might not express themselves at the right level for proper embryonic development. Next time you see this question, see if you can find something that doesn’t seem to be tightly connected with DNA structure.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|9505e7bdcff34″ question_number=”86″] A team of biologists has started a project to clone the rare blue lobster through somatic cell nuclear transfer. The blue lobster is a variety of the more common purple lobster. The difference in phenotype has been traced to a genetic mutation that results in a blue exoskeleton.

In the new population of blue lobster clones that the biologists have created, many have behavioral defects. From the list below, the most likely reason for this would be

[c] complete reprogramming of the DNA in the newly formed zygote to an embryonic state after nuclear transfer.

[f] No. Complete reprogramming of the newly formed zygote to an embryonic state after nuclear transfer is exactly what these biologists would want to happen, because it means that any epigenetic modification of the DNA (through histone acetylation or DNA methylation) would be removed, and that this zygote could develop with normal gene expression. Look at the list again, and see if you can find something that could lead to genetic problems during development.

[c] mitochondrial incompatibility within the egg

[f] No. In somatic cell nuclear transfer, just the nucleus of the donor cell is implanted into an egg cell that has had its nucleus removed. In other words, all the mitochondria should be coming from the egg (which is what happens during the normal fertilization process). As a result, there’s no chance of mitochondrial incompatibility. Look at the list again, and see if you can find something that could lead to genetic problems during development.

[c*] donor cell telomeres that were too short

[f] Great job. As a multicellular eukaryotic organism ages, the telomeres in its cell shorten. At a certain point, the telomeres are gone, and further cell division results in loss of genetic DNA. This is a known problem in cloned organisms, and could account for the behavioral problems described above.

[c] actin polymerization levels in the egg that were too high

[f] No. While actin polymerization levels that were too high could cause problems (or be a symptom of a problem), there’s a more obvious choice that could lead to genetic problems during development.

 

 

 

 

 

 

 

 

 

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|94f59b6a93f34″ question_number=”87″] The diagram below represents a sequence from two DNA molecules from the same species of bacteria. In the experiment’s protocol, the bacteria were first grown for several generations in a medium containing non-radioactive nitrogen. Subsequently, a fraction of the bacteria were transferred to a medium containing radioactive nitrogen.

Which of the diagrams below best represents the DNA molecules produced after the bacteria were transferred to radioactive nitrogen and then allowed to proceed through one replication cycle.

[c] A

[f] No. Model A shows the two daughter strands that result from DNA replication as consisting of one double stranded daughter consisting entirely of DNA made from 14N (the lighter one on the left), and the other one entirely made of 15N (the darker one on the right). But that’s not how DNA replication works. When DNA replicates itself, the DNA “unzips” and each strand subsequently acts as a template for formation of a new complementary strand. The resulting daughter strands will each be half old and half new. See if you can figure out which diagram represents this correct model.

[c] B

[f] No. Model B shows the two daughter strands that result from DNA replication as consisting entirely of DNA made from 15N. But that’s not how DNA replication works. When DNA replicates itself, the DNA “unzips” and each strand subsequently acts as a template for formation of a new complementary strand. The resulting daughter strands will each be half old and half new. See if you can figure out which diagram represents this correct model.

[c] C

[f] No. Model C has two key features: 1) It shows the two daughter strands that result from DNA replication as consisting of a mixture of DNA made from 15N and 14N, and it shows each individual strand consisting of both 15N and 14N. But that’s not how DNA replication works. When DNA replicates itself, the DNA “unzips” and each strand subsequently acts as a template for formation of a new complementary strand. The resulting daughter strands will each be half old and half new. See if you can figure out which diagram represents this correct model.

[c*] D

[f] Fabulous. Model D correctly represents what happens during DNA replication, which each daughter strand consisting of one of the original parent strands and one new complementary strand.

 

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|94e812e576b34″ question_number=”88″] The diagram below represents a sequence from two DNA molecules from the same species of bacteria. In the experiment’s protocol, the bacteria were first grown for several generations in a medium containing non-radioactive nitrogen. Subsequently, a fraction of the bacteria were transferred to a medium containing radioactive nitrogen.

One of the diagrams below represents the DNA molecules that would produced after the bacteria were transferred to radioactive nitrogen and then allowed to proceed through one replication cycle.

This model is called

[c] conservative

[f] No. The conservative model of DNA replication is represented by model B. This model is incorrect, because when DNA replicates itself, the DNA “unzips” and each strand subsequently acts as a template for formation of a new complementary strand. The resulting daughter strands will each be half old and half new. What’s a logical name for this model of replication?

[c] dispersive

[f] No. The dispersive model of DNA replication is represented by model C. This model is incorrect, because when DNA replicates itself, the DNA “unzips” and each strand subsequently acts as a template for formation of a new complementary strand. The resulting daughter strands will each be half old and half new (with one strand in each molecule entirely old, and one strand entirely new). What’s a logical name for this model of replication?

[c*] semi-conservative

[f] Nice job! DNA replication is best described as semi-conservative, and it’s shown by model D.

[c] interduplicative

[f] No. When DNA replicates itself, the DNA “unzips” and each strand subsequently acts as a template for formation of a new complementary strand. The resulting daughter strands will each be half old and half new (with one strand in each molecule entirely old, and one strand entirely new). What’s a logical name for this model of replication?

[q json=”true” multiple_choice=”true” unit=”Evolution” dataset_id=”144Qs|94da8a6059734″ question_number=”89″] The diagram below shows the results of an experiment in artificial selection with the fruit fly
Drosophila melanogaster. During the first 25 generations the smallest flies were selected to produce the next generation. After generation 25, the selection was reversed: from generations 25 through 35 only the largest flies were chosen to breed the next generation.

In relationship to alleles coding for body size, the experimental results suggest that

[c*] after 25 generations there was little genetic variation left in the population

[f] Excellent. From the choices given, it seems most reasonable that 25 generations of selection for smaller size had fixed the alleles for body size in the population, making it impossible to effectively select for larger size after generation 25.

[c] for the first 25 generations there was no genetic variation within the population.

[f] No. The drop in body size between generations 0 and 25 indicate that there was genetic variation, and that it could be selected for. Now think about why the body size remains constant after generation 25.

[c] selection between generations 25 and 35 had a significant effect on average body size.

[f] No. Take a good look at the graph. The line is flat after generation 25, indicating that selection for large body size had no effect on selection.

[c] if selection for small body size continued after generation 25, average body size would continue to fall.

[f] No. That seems unlikely, if only because selection for small body size seems to have little or no effect between generation 20 and 25.

[q json=”true” multiple_choice=”true” unit=”Mitosis|Meiosis|Genetics” dataset_id=”144Qs|94ccdc9a7df34″ question_number=”90″] In a group of organisms, individuals are genetically identical at a particular single gene locus. However, these individuals show a variety of phenotypes for the trait. The most likely explanation for the variation in phenotypes for this trait is

[c] codominant alleles.

[f] No. Alleles are codominant when they both have an effect on the phenotype of the offspring. For example, in blood type, the A and B alleles, if inherited together, both express themselves by coding for A and B glycoproteins on red blood cells. In the case described above, the individuals are genetically identical at a particular genetic locus, so codominance can’t be at play. What could produce phenotypic variation in genetically identical organisms?

[c] polygenic inheritance for the trait under investigation.

[f] No (but this is a smart choice, that could be true under some circumstances). Polygenic inheritance involves multiple genes having an effect on a phenotype. Height in humans (and many other organisms) is a polygenic trait. In the problem above, individuals are described as being genetically identical at a particular gene locus, so polygenic inheritance doesn’t seem to apply (unless this were a trick question, and I assure you that it’s not). Next time you see this, ask yourself this question: What could produce phenotypic variation in genetically identical organisms?

[c*] environmental influences creating phenotypic differences.

[f] Nice! If individuals are genetically identical, then the most straightforward explanation for the difference would be environmental influences. Same seed, different soil.

[c] multiple alleles at the gene locus within the gene pool.

[f] No. In this question, individuals are described as being identical a a particular gene locus. Even if there were multiple alleles for this gene, it wouldn’t explain their phenotypic variation (because they all have the same allele). Next time you see this, ask yourself this question: What could produce phenotypic variation in genetically identical organisms?

 

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|94bee45325f34″ question_number=”91″] A lysogenic virus infects a particular cell type. and integrates its genome into a site that contains a proto-oncogene. This transforms the cell and increases the level of a protein X. Protein X, in turn, increases cellular proliferation.

A compound P is known to increase the level of tumor suppressor proteins in that cell type. A second compound, Q helps in stimulating protein Z. Protein Z has been shown to be capable of binding to X rendering it inactive. Which one of the following graphs correctly represents the mode of action of P and Q?

[c] A

[f] No. This is the kind of question that requires some biology knowledge, and some good text and graph reading skills. In terms of the biology, you have to know that proto-oncogenes can be mutated into oncogenes, which promote uncontrolled cell proliferation (promoting cancer). Tumor suppressor genes (and proteins), by contrast, inhibit cell proliferation (preventing cancer). From careful reading of the question, we know that P increases tumor suppressor proteins (inhibiting cell proliferation). Q, by a very different mechanism (stimulating Z, which binds to and inactivates X) also inhibits cell proliferation. In other words, both P and Q inhibit cell proliferation.

Knowing that, take a look at graph A. The plus sign means the substance is present, and the minus sign is where the substance is absent. This graph is saying that when P alone is present (second bar), it has no effect (because it’s the same height as when P is absent, as in the first bar). This graph is also saying that Q has no effect (third bar) and that both P and Q together have no effect (fourth bar). This graph doesn’t match the biology (which is why this answer is incorrect).

Now look over the four graphs again. Find a graph where the bar for P alone is lower, and the bar for Q alone is lower, and the bar for P and Q together is the lowest. Remember that, and choose that graph the next time you see this question.

[c] B

[f] No. This is the kind of question that requires some biology knowledge, and some good text and graph reading skills. In terms of the biology, you have to know that proto-oncogenes can be mutated into oncogenes, which promote uncontrolled cell proliferation (promoting cancer). Tumor suppressor genes (and proteins), by contrast, inhibit cell proliferation (preventing cancer). From careful reading of the question, we know that P increases tumor suppressor proteins (inhibiting cell proliferation). Q, by a very different mechanism (stimulating Z, which binds to and inactivates X) also inhibits cell proliferation. In other words, both P and Q inhibit cell proliferation.

Knowing that, take a look at graph B. The plus sign means the substance is present, and the minus sign is where the substance is absent. This graph is saying that when P alone is present (second bar), it has an effect (because it’s lower than when P is absent, as in the first bar). This graph is also saying that Q has an effect (the third bar is lower than the 1st bar). However, you’d expect that both P and Q together would be even more effective. The fourth bar says the opposite (it’s the same height as the first bar).

Now look over the four graphs again. Find a graph where the bar for P alone is lower, and the bar for Q alone is lower, and the bar for P and Q together is the lowest. Remember that, and choose that graph the next time you see this question.

[c*] C

[f] Fantastic! Graph C matches the biology: P alone is lower, Q alone is lower, and the bar for P and Q together is the lowest of all.

[c] D

[f] No. This is the kind of question that requires some biology knowledge, and some good text and graph reading skills. In terms of the biology, you have to know that proto-oncogenes can be mutated into oncogenes, which promote uncontrolled cell proliferation (promoting cancer). Tumor suppressor genes (and proteins), by contrast, inhibit cell proliferation (preventing cancer). From careful reading of the question, we know that P increases tumor suppressor proteins (inhibiting cell proliferation). Q, by a very different mechanism (stimulating Z, which binds to and inactivates X) also inhibits cell proliferation. In other words, both P and Q inhibit cell proliferation.

Knowing that, take a look at graph D. The plus sign means the substance is present, and the minus sign is where the substance is absent. This graph is saying that when P alone is present (second bar), it has no effect (because it’s the same height as  when P is absent, as in the first bar). This graph is also saying that Q has an effect (the third bar is lower than the 1st bar). However, you’d expect that both P and Q together would be even more effective. The fourth bar says the opposite (it’s the same height as the first bar).

Now look over the four graphs again. Find a graph where the bar for P alone is lower, and the bar for Q alone is lower, and the bar for P and Q together is the lowest. Remember that, and choose that graph the next time you see this question.

[q json=”true” multiple_choice=”true” unit=”DNA|RNA|Protein” dataset_id=”144Qs|94af77845f734″ question_number=”92″] Two cloning vectors used in genetic engineering are bacterial viruses and plasmids. Which of
the following statements about viruses and plasmids is false?

[c] Plasmids use the translation machinery of the cell.

[f] No. Remember that you’re looking for a false statement, and this one is true. Once you introduce a plasmid into a host cell, it can be replicated, transcribed, and translated using the host cell’s molecular machinery. Remember that, and next time choose another answer.

[c] Viruses have a protein capsid.

[f] No.Remember that you’re looking for a false statement, and this one is true. A typical virus has a nucleic acid core (composed of DNA or RNA) surrounded by a protein coat (or capsid). Remember that, and next time choose another answer.

[c*] Viruses can replicate in the absence of a cellular host.

[f] Excellent. You’ve obviously remembered that viruses are obligate intracellular parasites, that depend on their host in order to replicate.

[c] Plasmids carry genes.

[f] No.Remember that you’re looking for a false statement, and this one is true. The reason why plasmids are used as vectors is that they carry genes. Through genetic engineering, these genes can be moved from organisms like humans to organisms like bacteria, which will express these genes. This is how we make genetically engineered insulin or clotting factor (to treat hemophilia). Remember that, and next time choose another answer.

[c] Plasmids have restriction sites.

[f] No. Remember that you’re looking for a false statement, and this one is true. Plasmids do have restriction sites (sites where they can be cut open by enzymes). That’s how plasmids can be engineered to combine with other DNA, making recombinant DNA technology possible. Remember that next time, and choose another answer.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|94a00ab598f34″ question_number=”93″] Researchers are investigating two steroid hormone receptors that they designate X and Y. Both receptors contain a ligand binding domain and a DNA binding domain. The researchers genetically engineer a third, hybrid receptor. This receptor, designated “H,” contains the ligand binding
domain of X and DNA binding domain of Y.

The researchers next isolate three groups of cells, each of which over-expresses one of the three receptors: X, Y or H. Each group of cells was then treated separately either with hormone X or with hormone Y.

Assuming that there is no cross-reactivity, which one of the following graphs best represent the receptor-ligand binding in each case?

[c] A

[f] No. This question assumes that you know that a ligand is something that binds with a receptor. Hormone X, for example, is the ligand that binds with receptor X.

Now, let’s look at the three bars on the left side of graph A. What the height of these bars are saying is that when cells over-expressing the receptor for X (white bar) are treated with hormone X, there’s a high percentage of binding. That makes sense. But it’s also saying that cells over-expressing the receptor for hormone Y are treated with X, they’ll respond more than cells over-expressing the receptor for H. This doesn’t make sense, because receptor H has the same ligand binding receptor as receptor X.

If you’ve got the idea, take a look at the graphs and figure out which one makes sense. If not, read below for another hint…

Here’s the hint: you’re looking for a graph where cells X and H will have about the same vigorous response to hormone X, and where cell Y will be the only cell vigorously responding to hormone Y.

[c] B

[f] No. This question assumes that you know that a ligand is something that binds with a receptor. Hormone X, for example, is the ligand that binds with receptor X.

Now, let’s look at graph B, and focus on the left side (response hormone X). This graph is saying that cells over-expressing the receptor to H will respond the most to X, followed by cells overexpressing the receptor to Y, followed by cells over-expressing the receptor to X. That doesn’t make sense. The cells with the receptor to X should respond the most to X, followed by the cells that overexpress the receptor for H (because it also binds with X). The cells with the receptor for Y should barely respond to X.

If you’ve got the idea, take a look at the graphs and figure out which one makes sense. If not, read below for another hint…

Here’s the hint: you’re looking for a graph where cells X and H will have about the same vigorous response to hormone X, and where cell Y will be the only cell vigorously responding to hormone Y.

[c*] C

[f] Nice job! You get high bars for reading bar graphs! Graph C is showing cells with the receptor for X and H responding to hormone X, and cells with the receptor for Y responding to hormone Y.

[c] D

[f] No. This question assumes that you know that a ligand is something that binds with a receptor. Hormone X, for example, is the ligand that binds with receptor X.

Now, let’s look at graph D, and focus on the left side (response hormone X). This graph is saying that cells overexpressing the receptor to Y will respond more to hormone X than cells overexpressing the receptor to X or H…and that doesn’t make any sense. Cells overexpressing the receptor to X should respond the most to hormone X, and because receptor H binds with hormone X, its response should be about the same as Xs.

If you’ve got the idea, take a look at the graphs and figure out which one makes sense. If not, read below for another hint…

Here’s the hint: you’re looking for a graph where cells X and H will have about the same vigorous response to hormone X, and where cell Y will be the only cell vigorously responding to hormone Y.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|94917d6b47f34″ question_number=”94″] The image below shows a sensory neuron and its associated receptor cells.

The axon is at

[c] A

[f] No. You’ve chosen a dendrite, which is responsible for input. The axon’s function is output. Take a good look at the diagram, and see if you can find something analogous to a wire sending messages away from the cell body (which is at C.)

[c] B

[f] No. B is bringing information to the cell body (shown at C). The axon’s function is output. Take a good look at the diagram, and see if you can find something analogous to a wire sending messages away from the cell body.

[c] C

[f] No. C is the cell body, which is responsible for integrating incoming messages, and “deciding” on the cell’s output. That output will flow through the axon. Take a good look at the diagram, and see if you can find something analogous to a wire sending messages away from the cell body.

[c*] D

[f] Correct! D is the axon.

[c] E

[f] No, but you were close. E is a Schwann cell making up the myelin sheath. This myelin sheath is a kind of insulation that makes nerve impulse transmission much faster. The axon is often wrapped by the myelin sheath, and it’s responsible for output. So take a good look at the diagram, and find the only other part that could be the axon.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|9483cfa56c734″ question_number=”95″] The image below shows a sensory neuron and its associated receptor cells.

The myelin sheath is at

[c] A

[f] No. You’ve chosen a dendrite, which is responsible for input. The myelin sheath is a kind of insulation that wraps around the neuron’s axon, the long process that’s responsible for output of information away from the cell body.

[c] C

[f] No.C is the cell body, which is responsible for integrating incoming messages, and “deciding” on the cell’s output. The myelin sheath is a kind of insulation that wraps around the neuron’s axon, the long process that’s responsible for output of information away from the cell body.

[c] D

[f] No. D is the axon, the long process responsible for output. The myelin sheath is a kind of insulation that wraps around the neuron’s axon.

[c*] E

[f] Excellent! You’ve correctly identified the myelin sheath.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|947647204f334″ question_number=”96″] The image below shows a sensory neuron and its associated receptor cells.

The dendrites are at

[c*] A

[f] Nice Job. You’ve correctly identified the dendrites.

[c] C

[f] No. C is the cell body, which is responsible for integrating incoming messages, and “deciding” on the cell’s output. The dendrites are responsible for input: bringing information to the cell body (which is at C).

[c] D

[f] No/ D is the axon, which is responsible for output. The dendrites are responsible for input: bringing information to the cell body (which is at C).

[c] E

[f] No. E is a Schwann cell making up the myelin sheath. This myelin sheath is a kind of insulation that makes nerve impulse transmission much faster. The dendrites are responsible for input: bringing information to the cell body (which is at C).

 

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|9468995a73b34″ question_number=”97″] In the image below, the threshold potential is indicated by

[c*] W

[f] Excellent. The threshold potential is the depolarization that is sufficient to initiate an action potential. In most cells, that’s between -50 and -55 mV.

[c] X

[f] No. You’ve identified the rising phase of an action potential, during which the membrane completely depolarizes. The threshold potential is the depolarization that is sufficient to initiate an action potential.

[c] Y

[f] No. You’ve identified the falling phase of an action potential, during which the membrane repolarizes. The threshold potential is the depolarization that is sufficient to initiate an action potential.

[c] Z

[f] No. You’ve identified the undershoot, the phase that follows the action potential. The threshold potential is the depolarization that is sufficient to initiate an action potential.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|945aeb9498334″ question_number=”98″] In the image below, the “undershoot,” or “hyperpolarization,” is indicated by letter

[c] W

[f] No. W is the threshold potential, which is the depolarization needed to initiate an action potential. The undershoot occurs immediately after the membrane repolarizes, and it’s a moment when the membrane is even more negatively charged than it was during the resting potential.

[c] X

[f] No. You’ve identified the rising phase of an action potential, during which the membrane completely depolarizes. The undershoot occurs immediately after the membrane repolarizes, and it’s a moment when the membrane is even more negatively charged than it was during the resting potential.

[c] Y

[f] No. You’ve identified the falling phase of an action potential, during which the membrane repolarizes.The undershoot occurs immediately after the membrane repolarizes, and it’s a moment when the membrane is even more negatively charged than it was during the resting potential.

[c*] Z

[f] Nicely done! Z is the undershoot, a hyperpolarization that immediately follows the action potential.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|944d3dcebcb34″ question_number=”99″] In the image below, the action potential is indicated by

[c] A

[f] No. “A” represents a graded potential. Graded potentials either hyperpolarize or depolarize the membrane, based on the strength of the stimulus. The action potential is a massive, “all or none” depolarization.

[c*] B

[f] Correct. “B” represents an action potential.

[c] C

[f] No. “C” represents the undershoot,  a hyperpolarization that immediately follows the action potential. The action potential is a massive, “all or none” depolarization.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|943f6ac822f34″ question_number=”100″] In the image below, repolarization is indicated by

[c] 1

[f] No. 1 is the resting potential. Repolarization is when the membrane, depolarized by the rising phase of the action potential, restores its negative charge.

[c] 2

[f] No. 2 is the threshold potential. Repolarization is when the membrane, depolarized by the rising phase of the action potential, restores its negative charge.

[c] 3

[f] No. 3 is the rising phase of the action potential, which is a massive depolarization that switches the charge across the membrane from negative to positive. Repolarization is when the membrane restores its negative charge.

[c*] 4

[f] Excellent. 4 is when the membrane is repolarizing.

[c] 5

[f] No. 5 is a hyperpolarization, also known as the undershoot. At this point, the membrane is even more negative than it is during its resting potential.  Repolarization is when the membrane, depolarized by the rising phase of the action potential, restores its negative charge.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|9431bd0247734″ question_number=”101″] The diagram below shows synaptic transmission. What is shown by letters X and Y?

[c] X shows neurotransmitter entering the synaptic knob. Y shows Ca2+ ions.

[f] No. Here’s a hint. The little spheres that diffuse from the axonal bulb (on top) to the postsynaptic membrane (below) are neurotransmitter. What has to diffuse into the membrane in order to induce the vesicles containing neurotransmitter to move to the tip of the axon and release their contents.

[c] X shows K+ ions diffusing into the synaptic knob. Y shows neurotransmitter.

[f] No. K+ ions are involved in the action potential, but that’s not what’s entering the cell at X . Here’s a hint. The little spheres that diffuse from the axonal bulb (on top) to the postsynaptic membrane (below) are neurotransmitter. What has to diffuse into the membrane in order to induce the vesicles containing neurotransmitter to move to the tip of the axon and release their contents.

[c] X shows Na+ ions diffusing into the synaptic knob. Y shows  Ca2+ ions.

[f] No. Na+ ions are involved in action potentials, and Ca2+ ions are involved in release of neurotransmitter, but not in the way you’ve identified. Here’s a hint. The little spheres that diffuse from the axonal bulb (on top) to the postsynaptic membrane (below) are neurotransmitter. What has to diffuse into the membrane in order to induce the vesicles containing neurotransmitter to move to the tip of the axon and release their contents.

[c*] X shows Ca2+ ions diffusing into the synaptic knob. Y shows neurotransmitter.

[f] Fabulous! After Ca2+ ions diffusing into the synaptic knob (at X) they’ll induce vesicles containing neurotransmitter (at Y) to release their contents into the synapse.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|9423e9fbadb34″ question_number=”102″] In a neuron at rest, which of the following conditions is true?

[c] Chlorine ions can freely diffuse from the intracellular fluid out of the cell.

[f] No. The membrane of neurons is relatively impermeable to negative ions.

[c*] Potassium ions are in higher concentration inside the cell than outside the cell.

[f] Yes. Due to the activity of the sodium potassium pump, a neuron at rest has a higher concentration of potassium ions inside the cell than outside.

[c] Large negatively charged ions are concentrated in the extracellular fluid.

[f] No. It’s actually the opposite: large negatively charged ions are concentrated in the intracellular fluid.

[c] Sodium ions are constantly entering the cell because of the sodium potassium pump.

[f] No. The sodium potassium pump pumps sodium ions out of the cell, while pumping potassium ions into the cell.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|94163c35d2334″ question_number=”103″] In the image below, which represents a signal transduction pathway, which letter represents a receptor?

[c] A

[f] No. A is the ligand. You’re looking for the receptor, which is the membrane protein that binds with the ligand.

[c*] B

[f] Correct! “B” is the receptor.

[c] C

[f] No. If this diagram were a representation of a G-coupled receptor, the letter “C” would represent a G protein. The G protein’s function is to activate a second messenger that would send a message through various relay molecules in the cytoplasm to activate a cellular response. You’re looking for the receptor, which is the membrane protein that binds with the ligand.

[c] D

[f] No.If this diagram were a representation of a G-coupled receptor, the letter “D” would represent the activity of a second messenger, or an entire phosphorylation cascade. You’re looking for the receptor, which is the membrane protein that binds with the ligand.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|9408692f38734″ question_number=”104″] The image below represents a signal transduction pathway. Which letter represents a ligand?

[c*] A

 

 

[f*]Exactly. A is the ligand.

[f] Good!

[c] B

[f] No. “B” is the receptor. The ligand is the molecule that binds with the receptor.

[c] C

[f] No. If this diagram were a representation of a G-coupled receptor, the letter “C” would represent a G protein. The G protein’s function is to activate a second messenger that would send a message through various relay molecules in the cytoplasm to activate a cellular response. The ligand is the molecule that binds with the receptor.

[c] D

[f] No.If this diagram were a representation of a G-coupled receptor, the letter “D” would represent a phosphorylation cascade. The ligand is the molecule that binds with the receptor.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|93f8fc6071f34″ question_number=”105″] The image below represents a signal transduction pathway. Which letter represents second messengers?

[c] A

 

 

[f*]No. A is the ligand. The second messengers are molecules that relay the message into the cytoplasm to bring about a response.

[f] No.

[c] B

[f] No. “B” is the receptor. The second messengers are molecules that relay the message into the cytoplasm to bring about a response.

[c] C

[f] No. If this diagram were a representation of a G-coupled receptor, the letter “C” would represent a G protein. The G protein’s function is to activate a second messenger that would send a message through various relay molecules in the cytoplasm to activate a cellular response…and that’s a pretty big hint as to which part of the diagram represents the second messengers.

[c*] D

[f] Fabulous! If this diagram were a representation of a G-coupled receptor, the letter “D” would represent a second messenger (or an entire phosphorylation cascade)

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|93e7d088c0734″ question_number=”106″] The image below represents a signal transduction pathway. Which letter could represent a hormone?

[c*] A

 

 

[f*]Exactly. “A” would represent a hormone.

[f] Excellent!

[c] B

[f] No. “B” is the receptor. The hormone is the molecule that binds with the receptor.

[c] C

[f] No. If this diagram were a representation of a G-coupled receptor, the letter “C” would represent a G protein. The G protein’s function is to activate a second messenger that would send a message through various relay molecules in the cytoplasm to activate a cellular response. The hormone is the molecule that binds with the receptor.

[c] D

[f] No.If this diagram were a representation of a G-coupled receptor, the letter “D” would represent a second messenger, or an entire phosphorylation cascade. The hormone is the molecule that binds with the receptor.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|93d9fd8226b34″ question_number=”107″] The image below represents a signal transduction pathway. Which letter represents a protein kinase?

[c] A

[f] No. “A” represents a ligand. The protein kinase is found within the cytoplasm, relaying the message from the membrane to some effector within the cell.

[c] B

[f] No. “B” represents a receptor. The protein kinase is found within the cytoplasm, relaying the message from the membrane to some effector within the cell.

[c] C

[f] No. If this diagram were a representation of a G-coupled receptor, the letter “C” would represent a G protein. The G protein’s function is to activate a second messenger that would send a message through various relay molecules in the cytoplasm to activate a cellular response. The protein kinase is found within the cytoplasm, relaying the message from the membrane to some effector within the cell.

[c*] D

[f] Yes! “D” represents protein kinases, participating in a phosphorylation cascade that is amplifying the cellular response.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|93cc4fbc4b334″ question_number=”108″] The image below represents a signal transduction pathway. Which roman numeral represents reception?

[c*] I

[f] Yes. “I” represents reception.

[c] II

[f] No. “II” represents signal transduction (conversion of the initial signal, the hormone, into one that can be carried into the cytoplasm to bring about a cellular response).

[c] III

[f] No. “III” represents the cellular response.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|93be7cb5b1734″ question_number=”109″] The image below represents a signal transduction pathway. Which roman numeral represents transduction?

[c] I

[f] No. “I” represents reception.

[c*] II

[f] Yes. “II” represents signal transduction (conversion of the initial signal, the hormone, into one that can be carried into the cytoplasm to bring about a cellular response).

[c] III

[f] No. “III” represents the cellular response.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|93b0a9af17b34″ question_number=”110″] In the diagram below, which stage represents signal transduction?

[c] I

[f] No. Stage I is reception. Transduction is taking the initial signal (from the ligand, at number 1) and then passing that signal on, but in another form. An analogy would be this: Joe rings your doorbell (that’s the signal). You answer the door, and then shout to your sister, “Joe’s here” (that’s transduction). Where do you see transduction in the diagram above?

[c*] II

[f] Correct. Stage II is transduction.

[c] III

[f] No. Stage III is response. Transduction is taking the initial signal (from the ligand, at number 1) and then passing that signal on, but in another form. An analogy would be this: Joe rings your doorbell (that’s the signal). You answer the door, and then shout to your sister, “Joe’s here” (that’s transduction). Where do you see transduction in the diagram above?

[c] IV

[f] No. In stage IV, something (5) is heading toward the Golgi apparatus, making 5 a newly synthesized protein.Transduction is taking the initial signal (from the ligand, at number 1) and then passing that signal on, but in another form. An analogy would be this: Joe rings your doorbell (that’s the signal). You answer the door, and then shout to your sister, “Joe’s here” (that’s transduction). Where do you see transduction in the diagram above?

[c] V

[f] No. In stage V, something is being exported from the cell (which is part of the cellular response). Transduction is taking the initial signal (from the ligand, at number 1) and then passing that signal on, but in another form. An analogy would be this: Joe rings your doorbell (that’s the signal). You answer the door, and then shout to your sister, “Joe’s here” (that’s transduction). Where do you see transduction in the diagram above?

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|93a2b167bfb34″ question_number=”111″] In the diagram below, which stage represents reception?

[c*] I

[f] Nice!. Stage I is reception.

[c] II

[f] No. Stage II is transduction. Reception is when the ligand (usually a hormone) binds with a receptor (usually on the membrane).

[c] III

[f] No. Stage III is response. Reception is when the ligand (usually a hormone) binds with a receptor (usually on the membrane).

[c] IV

[f] No. In stage IV, something that left the nucleus (5) is headed toward the Golgi apparatus. Reception is when the ligand (usually a hormone) binds with a receptor (usually on the membrane).

[c] V

[f] No. In stage V, something is being exported from the cell (which is part of the cellular response). Reception is when the ligand (usually a hormone) binds with a receptor (usually on the membrane).

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|9394b92067b34″ question_number=”112″] In the diagram below, which stage represents exocytosis?

[c] I

[f] No. Stage I is reception. Exocytosis means “export from the cell.”

[c] II

[f] No. Stage II is transduction.Exocytosis means “export from the cell.”

[c] III

[f] No. Stage III is response. Exocytosis means “export from the cell.”

[c] IV

[f] No. In stage IV, a protein is headed toward the Golgi apparatus.Exocytosis means “export from the cell.”

[c*] V

[f] Excellent. Stage V is exocytosis, the export of something from the cell.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|9386765793334″ question_number=”113″] If 5 represents a newly synthesized protein, then 6 would most likely be

[c] a ribosome

[f] No. Ribosomes are where proteins are synthesized. If 5 represents a protein, then 6 would represent a part of the cell where proteins go after synthesis. What part of a eukaryotic cell is responsible for modifying and packaging proteins?

[c] a mitochondrion

[f] No. Mitochondria are organelles involved in ATP synthesis, and aren’t represented anywhere above. Here’s a hint: If 5 represents a protein, then 6 would represent a part of the cell where proteins go after synthesis. What part of a eukaryotic cell is responsible for modifying and packaging proteins?

[c*] the golgi complex

[f] Fabulous! If 5 represents a protein, then 6 would be the Golgi complex.

[c] the rough endoplasmic reticulum

[f] No. The rough endoplasmic reticulum is where proteins destined for export (or for a lysosome) are synthesized. The question tells you that 5 is a protein that has already been synthesized. Where would proteins go after being synthesized for modification and packaging?

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|9374fffe65334″ question_number=”114″] In the diagram below, which part represents a ligand?

[c*] 1

[f] Nice! 1 is the ligand.

[c] 2

[f] No. 2 is a receptor. The ligand is what binds with a receptor.

[c] 3

[f] No. 3 is taking a message from the membrane, and bringing it to the nucleus, making it a second messenger. The ligand is what binds with a receptor.

[c] 4

[f] No. 4 is the nucleus. The ligand is what binds with a receptor at the membrane.

[c] 5

[f] No. 5 is something that’s heading toward the Golgi apparatus, making 5 a newly synthesized protein. The ligand is what binds with a receptor.

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|9365ddb11b334″ question_number=”115″] In the diagram below, which part represents a second messenger?

[c] 1

[f] No. 1 represents a ligand. The second messenger takes the initial signal (which occurs when the ligand binds with the receptor) and then passes that signal on, but in another form. An analogy would be this: Joe rings your doorbell. You answer the door, and then shout to your sister, “Joe’s here.” Joe is the first messenger, and you’re the second messenger. What could be the second messenger in the diagram above?

[c] 2

[f] No. 2 is the receptor. The second messenger takes the initial signal (which occurs when the ligand binds with the receptor) and then passes that signal on, but in another form. An analogy would be this: Joe rings your doorbell. You answer the door, and then shout to your sister, “Joe’s here.” Joe is the first messenger, and you’re the second messenger. What could be the second messenger in the diagram above?

[c*] 3

[f] Way to go! 3 is the second messenger.

[c] 4

[f] No. 4 represents the nucleus. The second messenger takes the initial signal (which occurs when the ligand binds with the receptor) and then passes that signal on, but in another form. An analogy would be this: Joe rings your doorbell. You answer the door, and then shout to your sister, “Joe’s here.” Joe is the first messenger, and you’re the second messenger. What could be the second messenger in the diagram above?

[c] 5

[f] No. 5 is part of the cellular response (a protein made in response to binding of the ligand at the receptor). The second messenger takes the initial signal (which occurs when the ligand binds with the receptor) and then passes that signal on, but in another form. An analogy would be this: Joe rings your doorbell. You answer the door, and then shout to your sister, “Joe’s here.” Joe is the first messenger, and you’re the second messenger. What could be the second messenger in the diagram above?

[q json=”true” multiple_choice=”true” unit=”Cells|Membranes” dataset_id=”144Qs|9357e569c3334″ question_number=”116″] In the diagram below, transcription factors would be active at

[c] 1

[f] No. 1 is a ligand. Transcription factors are substances that enter into the nucleus and induce transcription of genes. Take another look at the diagram, and identify a part that could fit that description.

[c] 2

[f] No. 2 is a receptor. Transcription factors are substances that enter into the nucleus and induce transcription of genes. Take another look at the diagram, and identify a part that could fit that description.

[c] 3

[f] No, but you’re on the right track. 3 represents second messengers that are heading toward the nucleus. Based on what’s depicted in this diagram, these second messengers are undoubtedly transcription factors. But they won’t be active until they enter the nucleus and bind with DNA in a way that initiates transcription of specific genes. What number represents the location where these transcription factors will be active?

[c*] 4

[f] Excellent. 4 represents the nucleus, which is where the second messengers (which are transcription factors) are going to activate transcription of specific genes.

[c] 5

[f] No. 5 represents proteins that have been produced because of the activity of transcription factors. Transcription factors are substances that enter into the nucleus and induce transcription of genes. Take another look at the diagram, and identify a part that could fit that description.

 

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|9349c7e1acf34″ question_number=”117″] In a physiology laboratory exercise, students place electrodes at a point on a nerve cell of a living organism. The nerve cell is then stimulated, resulting in the graph below.

From the following list, the most reasonable claim is

[c] J, represents the resting potential of the nerve.

[f] No. J represents the hyperpolarization, or undershoot.

[c*] during phase F, more sodium ions are entering the cell than exiting the cell.

[f] Nice. F is the rising phase of the action potential. At this point, the membrane is depolarizing as voltage sensitive sodium gates open, allowing sodium ions to enter the cell.

[c] during phase K, the cell will no longer respond to additional stimulation.

[f] No. This question is referring to the refractory period, which is when the membrane can’t respond to stimulation. That happens during phase J.  By phase K, the resting potential has been restored, and further stimulation is possible.

[c] the stimulus will only have an effect when the membrane potential is positive.

[f] No. Nerve cells can only respond to stimuli when they’re at resting potential. During resting potential, the membrane is polarized at about – 70 mV.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|933af015df734″ question_number=”118″] In a physiology laboratory exercise, students place electrodes at a point on a nerve cell of a living organism. The nerve cell is then stimulated, resulting in the graph below.

Which of the following is closest to the threshold potential?

[c] -70 mV

[f] No. -70 mV is the resting potential. What’s tricky about this question is that the threshold potential isn’t directly shown. All I’ll give you as a hint is that it’s less negative than the resting potential.

[c*] -55 mV

[f] Correct. The threshold potential for a neuron is about -55mV.

[c] -30 mV

[f] No. -30mV is above the threshold potential. What’s tricky about this question is that the threshold potential isn’t directly shown. All I’ll give you as a hint is that it’s less negative than the resting potential.

[c] 0 mV

[f] No. 0mV is way above the threshold potential. What’s tricky about this question is that the threshold potential isn’t directly shown. All I’ll give you as a hint is that it’s less negative than the resting potential.

[c] +30 mV

[f] No. +30 mV is way above the threshold potential. What’s tricky about this question is that the threshold potential isn’t directly shown. All I’ll give you as a hint is that it’s less negative than the resting potential.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|932cd28dc9334″ question_number=”119″] In a physiology laboratory exercise, students place electrodes at a point on a nerve cell of a living organism. The nerve cell is then stimulated, resulting in the graph below.

Membrane depolarization is occurring during

[c*] F

[f] Way to go! F is showing membrane depolarization.

[c] H

[f] No. At H, the membrane is repolarizing (becoming negative again). Depolarization is when the membrane is shifting from its negative resting potential to positive. What letter fits that description?

[c] J

[f] No. At J, the membrane is at a phase called a hyperpolarization, or undershoot. Depolarization is when the membrane is shifting from its negative resting potential to positive. What letter fits that description?

[c] K

[f] No. At K, the membrane potential has been restored. Depolarization is when the membrane is shifting from its negative resting potential to positive. What letter fits that description?

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|931eda4671334″ question_number=”120″] In a physiology laboratory exercise, students place electrodes at a point on a nerve cell of a living organism. The nerve cell is then stimulated, resulting in the graph below.

The refractory period, during which the membrane can’t respond to further stimulation, is represented by point

[c] F

[f] No. F represents the membrane when it’s depolarizing. Here’s a hint. The refractory period is when the membrane is hyperpolarized, and even more negative than it is during the resting potential.

[c] G

[f] No. G represents the height of the action potential, when the membrane is at its maximum depolarization. Here’s a hint. The refractory period is when the membrane is hyperpolarized, and even more negative than it is during the resting potential.

[c] H

[f] No. H shows when the membrane is repolarizing (becoming negative again). Here’s a hint. The refractory period is when the membrane is hyperpolarized, and even more negative than it is during the resting potential.

[c*] J

[f] Good job. J shows the refractory period.

[c] K

[f] No. K shows the membrane after it has restored itself to its resting potential. Here’s a hint. The refractory period is when the membrane is hyperpolarized, and even more negative than it is during the resting potential.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|9310e1ff19334″ question_number=”121″] The diagram shows a ball approaching the eye, and a representation of part of the pathway of an automatic blinking response.

Which number below represents a sensory neuron?

[c] 1

[f] No. 1 shows the muscle that’s going to contract so that the eyelid will close (the final part of this reflex). The sensory neuron is taking information in, and bringing it to the motor neuron. To find the sensory neuron, find the “wire” that’s bringing information from the eye to the brain.

[c] 2

[f] No. 2 shows a motor neuron, bringing a message to the muscle that’s attached to the eyelid. The sensory neuron is taking information in, and bringing it to the motor neuron. To find the sensory neuron, find the “wire” that’s bringing information from the eye to the brain.

[c] 3

[f] No. 3 is the synapse between the sensory neuron and the motor neuron. The sensory neuron is taking information in, and bringing it to the motor neuron. To find the sensory neuron, find the “wire” that’s bringing information from the eye to the brain.

[c*] 4

[f] Correct! 4 is the sensory neuron.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|9302c47702f34″ question_number=”122″] The diagram shows a ball approaching the eye, and a representation of part of the pathway of an automatic blinking response.

Which number below represents a motor neuron?

[c] 1

[f] No. 1 shows the muscle that’s going to contract so that the eyelid will close (the final part of this reflex). The motor neuron brings the impulse to contract to this muscle.

[c*] 2

[f] Nice job. Number 2 is the motor neuron.

[c] 3

[f] No. 3 is the synapse between the sensory neuron and the motor neuron. The motor neuron brings the impulse to contract to the muscle that closes the eyelid (which is shown at 1)

[c] 4

[f] No. 4 is the sensory neuron. The motor neuron brings the impulse to contract to the muscle that closes the eyelid (which is shown at 1)

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|92f4a6eeecb34″ question_number=”123″] The diagram shows a ball approaching the eye, and a representation of part of the pathway of an automatic blinking response.

Which number below represents a synapse?

[c] 1

[f] No. 1 shows the muscle that’s going to contract so that the eyelid will close (the final part of this reflex). The synapse is the connection between the sensory neuron (which brings information in) and the motor neuron (which sends the impulse out).

[c] 2

[f] No. 2 is a motor neuron. The synapse is the connection between the sensory neuron (which brings information in) and the motor neuron (which sends the impulse out).

[c*] 3

[f] Yes. 3 is the synapse between the sensory neuron and the motor neuron.

[c] 4

[f] No. 4 is the sensory neuron. The synapse is the connection between the sensory neuron (which brings information in) and the motor neuron (which sends the impulse out).

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|92e68966d6734″ question_number=”124″] The diagram shows a ball approaching the eye, and a representation of part of the pathway of an automatic blinking response.

Which number below represents an effector (such as a muscle)?

[c*] 1

[f] Yes.1 shows the muscle that’s going to contract so that the eyelid will close (the final part of this reflex).

[c] 2

[f] No. 2 is the motor neuron. To find the effector, find the muscle cell at the end of the motor neuron that’s going to contract, closing the eyelid (the final move in this reflex arc).

[c] 3

[f] No. 3 is the synapse between the sensory and motor neurons. To find the effector, find the muscle cell at the end of the motor neuron that’s going to contract, closing the eyelid (the final move in this reflex arc).

[c] 4

[f] No. 4 is the sensory neuron. To find the effector, find the muscle cell at the end of the motor neuron that’s going to contract, closing the eyelid (the final move in this reflex arc).

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|92d78c5a4ab34″ question_number=”125″] In the diagram below, the most likely site for an action potential to be initiated is

[c] 1

[f] No. 1 shows calcium ions diffusing into the pre-synaptic neuron (the neuron before the synapse). The action potential is going to be initiated in the post-synaptic neuron.

[c] 3

[f] No. 3 is a vesicle in the pre-synaptic neuron. The action potential is going to be initiated in the post-synaptic neuron.

[c] 5

[f] No. 5 is a receptor in the post-synaptic neuron. The action potential is going to be initiated very close by, and as a result of that receptor binding with neurotransmitter. Find another spot on the presynaptic neuron.

[c*] 9

[f] Yes. If the neurotransmitter’s effect is to cause a depolarization sufficient to exceed the threshold potential, then an action potential will be intitiated along the membrane of the post-synaptic neuron.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|92c96ed234734″ question_number=”126″] In the diagram below, a receptor is shown at

[c] 1

[f] No. 1 shows calcium ions diffusing into the pre-synaptic neuron (the neuron before the synapse). The receptor is found in the post-synaptic neuron, where it binds with the neurotransmitter (shown at 4).

[c] 3

[f] No. 3 is a vesicle containing neurotransmitter. The receptor is found in the post-synaptic neuron, where it binds with the neurotransmitter (shown at 4).

[c*] 5

[f] Excellent. 5 is the receptor.

[c] 9

[f] No. 9 is the membrane of the post-synaptic neuron. The receptor is also found in the post-synaptic neuron, where it binds with the neurotransmitter (shown at 4).

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|92bb514a1e334″ question_number=”127″] In the diagram below, a neurotransmitter is shown at

[c] 1

[f] No. 1 shows calcium ions diffusing into the pre-synaptic neuron (the neuron before the synapse). To find the neurotransmitter, find the molecule that moves from the pre-synaptic neuron, through the synapse, and then binds with a receptor in the post-synaptic neuron.

[c*] 4

[f] Excellent! 4 is the neurotransmitter.

[c] 5

[f] No. 5 is the receptor in the post-synaptic neuron. To find the neurotransmitter, find the molecule that moves from the pre-synaptic neuron, through the synapse, and then binds with a receptor in the post-synaptic neuron.

[c] 6

[f] No, but you’re very close. 6 is an enzyme that’s degrading the neurotransmitter. To find the neurotransmitter, identify the molecule that moves from the pre-synaptic neuron, through the synapse, and then binds with a receptor in the post-synaptic neuron.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|92ad0e8149b34″ question_number=”128″] In the diagram below, an enzyme degrading a neurotransmitter is shown at

[c] 1

[f] No. 1 shows calcium ions diffusing into the pre-synaptic neuron (the neuron before the synapse). To find the enzyme that degrades the neurotransmitter, you first have to locate the neurotransmitter itself. To do that, find the molecule that binds with the receptor in the post-synaptic neuron (shown at 5)

[c] 4

[f] No, but your very close. 4 is the neurotransmitter. Take another look at the diagram, and find the enzyme that’s breaking this neurotransmitter into two fragments.

[c] 5

[f] No. 5 is a receptor in the post-synaptic neuron. To find the enzyme that degrades the neurotransmitter, you first have to locate the neurotransmitter itself. To do that, find the molecule that binds with this receptor.

[c*] 6

[f] Marvelous! 6 is the enzyme that breaks down the neurotransmitter.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|929ef0f933734″ question_number=”129″] When an action potential arrives at a presynaptic cell, something causes the presynaptic cell to release neurotransmitter. In the diagram below, that something is represented by

[c*] 1

[f] Nice job. When the action potential arrives at the axon’s tip, voltage gated calcium channels open up, allowing calcium to flow into the cell. These calcium ions, in turn, induce vesicles filled with neurotransmitter to dump their contents into the synapse.

[c] 2

[f] No. 2 represents the axon of the pre-synaptic neuron. You’re looking for something in the pre-synaptic neuron that will induce vesicles filled with neurotransmitter to dump that neurotransmitter into the synapse.

[c] 3

[f] No. 3 represents a vesicle filled with neurotransmitter. You’re looking for something in the pre-synaptic neuron that will induce that vesicle to dump that neurotransmitter into the synapse.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|92903e6e24334″ question_number=”130″] Actin filaments (which make up a key part of the cytoskeleton, are capable of growing at either end. The “plus” end (also called the “barbed end”) grows faster, while the “minus” end (also called the “pointed end”) grows more slowly.

The figure below shows the growth rates at the plus and minus ends of actin filaments as a function of actin monomer concentration.

Actin filaments of defined length are added to a solution of actin monomers at the concentration indicated as A, B, C, D, or E. Which concentration ofthe solution will let you grow actin filaments at the plus end, but not the minus end?

[c] Concentration at A or B

[f] No. This is one of those questions that’s all about graph reading. Look at what’s happening at A. At that concentration of actin monomers (which you read on the X axis), the rate of filament growth (which you read on the Y axis) on both the plus end and the minus end is negative (below zero), which means that the size of the filament is decreasing on both ends. Your job is to find the values where the filament is increasing on the plus end, but not on the minus end.

[c] Concentration at B or C

[f] No. This is one of those questions that’s all about graph reading. You were right about C, but look at what’s happening at B. At that concentration of actin monomers (which you read on the X axis), the rate of filament growth (which you read on the Y axis) is zero on the plus end and below zero on the minus end. Your job is to find the values where the filament is increasing on the plus end, but not on the minus end. C is one such value. What’s the other?

[c*] Concentration at C or D

[f] Excellent. At these concentrations, the filament will increase in size on the plus end, but either decrease in size (point C) or stay the same in size (point D) on the minus end.

[c] Concentration at D or E

[f] No. This is one of those questions that’s all about graph reading. Look at what’s happening at E. At that concentration of actin monomers (which you read on the X axis), the rate of filament growth (which you read on the Y axis) is positive on both the plus end and on the minus end. Your job is to find the values where the filament is increasing on the plus end, but not on the minus end.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|9281d66491734″ question_number=”131″] As applied to proteins, which of the following best describes “denaturation?”

[c] cleaving peptide bonds

[f] No. Cleaving peptide bonds is best described as protein digestion or degradation. Denaturation is much less severe and permanent.

[c] covalently modifying a protein

[f] No. Covalent modification would constitute a chemical change to the protein, and is much more drastic than what happens during denaturation.

[c*] a changing a protein’s tertiary structure

[f] Yes. Denaturation is a change in protein shape brought about by a change in a protein’s environment. Often, these changes involve disruption of tertiary level interactions between amino acids.

[c] a changing a protein’s the primary structure

[f] No. A change in a protein’s primary structure is a fundamental shift in a proteins identify, chemistry, and structure. Denaturation is often temporary, and involve a much less drastic level of change.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|9273491a40734″ question_number=”132″] People with chronic high blood pressure have elevated blood levels of ACE (Angiotensin-converting enzyme). ACE acts on the polypeptide Angiotensin I to produce Angiotensin II, which as the effect of raising blood-pressure.

The diagram below represents the active site of ACE. The “+” sign represents positively charged regions in the active site.

A pharmaceutical company is trying to develop a drug to lower blood pressure. As part of this effort, a range of drugs was designed and manufactured. A sample of the molecular shape of each is shown below. Notice the “+” and “-” signs, which represent positively and negatively charged regions in the molecules making up the candidate drugs.

Which drug is likely to be the most effective in preventingexcessive high blood pressure?

[c] Drug A

[f] No. What you’re looking for is a drug with a shape that will enable it to act as a competitive inhibitor of ACE. To do that, you’ll need a drug candidate with a shape and charge that’s most complementary to ACE. Take a look at the four drug candidates, and find the one that best fits that description.

[c*] Drug B

[f] Excellent: Drug B has a charge and shape that best complements ACE>

[c] Drug C

[f] No. What you’re looking for is a drug with a shape that will enable it to act as a competitive inhibitor of ACE. To do that, you’ll need a drug candidate with a shape and charge that’s most complementary to ACE. Take a look at the four drug candidates, and find the one that best fits that description.

[c] Drug D

[f] No, but you’re close. What you’re looking for is a drug with a shape that will enable it to act as a competitive inhibitor of ACE. To do that, you’ll need a drug candidate with a shape and charge that’s most complementary to ACE. You’ve chosen a good shape, but you need to be more careful about the charge.

[q json=”true” multiple_choice=”true” unit=”Plant_Systems” dataset_id=”144Qs|9264968f31334″ question_number=”133″] The graph on the right below shows the amount of sunlight received by various levels of a forest on a sunny summer’s day. For example, at noon, the tallest trees receive 100% of full sunlight. By contrast, the bushes receive between about 12% and 50% of full sunlight (depending on their height and position). Note that the light curve experienced by the layers is a triangle, although a more flattened triangle for the lower layers.

The graph on the left shows the plants’ response to light, as shown by their rate of photosynthesis.

Using the average light intensity for their level between about 10 am and 2pm, what percent of maximum photosynthesis would the tallest trees have? What percent would the bushes have?

[c] 75% and 9%

[f] No. This is one of those questions that’s all about interpreting graphs. Start by looking at the triangle graph on the right. Between 10am and 2pm, the tallest trees are receiving at least 75% of full sunlight. Now look at the graph on the left. At 75% percent of full sunlight (X axis), what percentage of photosynthesis can a plant perform? For the bushes, the % of full sunlight might be as low as 12%. But look at how much photosynthesis a plant can do with 12% of full sunlight.

[c] 50% and 9%

[f] No. This is one of those questions that’s all about interpreting graphs. Start by looking at the triangle graph on the right. Between 10am and 2pm, the tallest trees are receiving at least 75% of full sunlight. Now look at the graph on the left. At 75% percent of full sunlight (X axis), what percentage of photosynthesis can a plant perform? For the bushes, the % of full sunlight might be as low as 12%. But look at how much photosynthesis a plant can do with 12% of full sunlight.

[c] 50% and 5%

[f] No. This is one of those questions that’s all about interpreting graphs. Start by looking at the triangle graph on the right. Between 10am and 2pm, the tallest trees are receiving at least 75% of full sunlight. Now look at the graph on the left. At 75% percent of full sunlight (X axis), what percentage of photosynthesis can a plant perform? For the bushes, the % of full sunlight might be as low as 12%. But look at how much photosynthesis a plant can do with 12% of full sunlight

[c*] 100% and 60%

[f] Excellent! Between 10 and 2 pm, the tallest trees will be performing 100% of the photosynthesis that they can perform, and the bushes will be at about 60%.

[q json=”true” multiple_choice=”true” unit=”Animal_Systems” dataset_id=”144Qs|92560944e0334″ question_number=”134″] As shown in the diagram below, eyestalk retraction in crabs is controlled by a sensory-motor reflex. Note that the sensory and inhibitory neurons synapse with the axon of the motor neuron.

Based on the wiring of this reflex, which of the following is most likely?

[c] retraction of the eyestalk when the sensory and inhibitory nerves are stimulated at the same time at locations S and T, respectively.

[f] No. The key concept here is summation: post synaptic neurons are going to add together their input in order to compute a response. If the inhibitory nerve is stimulated at the same time as the sensory nerve, then the two inputs will cancel each other out. The result is that the motor neuron won’t fire an impulse, and the eyestalk won’t contract.

[c*] retraction of the eyestalk if the sensory hair is stimulated.

[f] Yes. Based on the wiring of this reflex, stimulation of the sensory hair should stimulate the motor neuron, which should fire off an impulse that will result in contraction of the retractor muscle.

[c] retraction of the eyestalk if the inhibitory neuron is stimulated after cutting the sensory nerves at site J.

[f] No. If the sensory nerve is cut at site J, then no excitatory impulse will reach the motor neuron. The inhibitory neuron still synapses with the motor neuron, but its effect on the motor neuron is inhibitory (as you can tell by the “-” sign at the synapse).

[c] no retraction of the eyestalk if the sensory and inhibitory nerves are stimulated after cutting the inhibitory nerve at site K.

[f] No. If the sensory nerve is stimulated, it’s going to send an excitatory impulse to the motor nerve. If the inhibitory nerve is cut at K, then any inhibitory impulse sent from that nerve won’t arrive at the motor neuron. As a result, the sensory nerve will induce the motor neuron to fire, and the eyestalk will retract.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|9246c1b6d7f34″ question_number=”135″] Which of the following is responsible for holding water molecules together as they get pulled up through the stem of a plant?

[c] surface tension

[f] No, but you’re thinking about this the right way. Surface tension is what’s generating some of the negative pressure (also known as tension) that sucks water up from the root, through the stem, and up to the leave. But what’s keeping those water molecules together?

[c] heat of vaporization.

[f] No. Heat of vaporization is the amount of energy required to turn a liquid into a gas. You’re looking for the name of the interaction between water molecules that keeps them in an unbroken chain as they get pulled up the stem during transpiration.

[c*] cohesion

[f] Fabulous. That force that’s keeping water molecules together is cohesion.

[c] adhesion

[f] No.  Adhesion is the force that enables water molecules to adhere to the sides of the tubes they’re being pulled through. What force enables the water molecules to stick together. Here’s a hint: you’re just one syllable away from the answer.

[c] transpiration

[f] No. Transpiration is the process that allows moves water through plants. It depends on the properties of water, and the property that you’re looking for is the name of the interaction between water molecules that keeps them in an unbroken chain.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|9238346c86f34″ question_number=”136″] The process of transpiration, which pulls water up from the roots through the plant’s stem and out through the leaves, only works because water molecules stay connected to the walls of xylem as they get pulled up through the stem of a plant. What’s the name for the force behind that connection.

[c] cohesion

[f] No, but you’re very close (only a syllable away). Cohesion describes how water molecules stick to one another. What describes how water sticks to other molecules?

[c] surface tension

[f] No. Here’s two things to know about this question. 1) Surface tension is what happens at the water/air interface of a body of water. It’s the force that lets you float a paperclip on the surface of a cup of water. 2) Xylem are the thin tubes in the roots, stems, and leaves of plants through which water flows up during transpiration. You’re looking for the name of the property that describes how water molecules “stick” to other polar molecules (such as those making up the walls of the xylem).

[c] heat of vaporization.

[f] Sorry, that’s not correct.

[c*] adhesion

[f] Excellent! Adhesion is correct! Water molecules adhere to the walls of the xylem during transpiration. Along with cohesion, that’s how these water molecules can get pulled up the length of a plant’s stem.

[c] evaporation

[f] No. Evaporation is, ultimately, generating the tension that’s pulling water molecules up the stem of a plants. But you’re looking for the name of the property that describes how water molecules “stick” to other polar molecules (such as those making up the walls of the xylem).

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|922981e177b34″ question_number=”137″] The beta-pleated sheet and alpha helix result from

[c] hydrogen bonds that form between the side chains of non-polar amino acids

[f] No. Interactions between side chains are what underlie tertiary (third-level) protein structure. You’re looking for the interactions that underlie secondary structures (like beta-pleated sheets and alpha helices).

[c*] hydrogen bonds that form between the N-H and C=O groups of a protein’s polypeptide backbone

[f] Fabulous! Hydrogen bonds between amino and carbonyl groups of a proteins polypeptide backbone are what lead to secondary structures (like beta-pleated sheets and alpha helices).

[c] hydrogen bonds that form between the side chains of polar amino acids.

[f] No. You’re looking for hydrogen bonds, but not between side chains. What other parts of a polypeptide (a chain of amino acids) could form hydrogen bonds with one another?

[c] ionic bonds that form between oppositely charged amino acid side chains

[f] No. Ionic bonds between oppositely charged amino acid side chains are a force involved in tertiary protein structure. You’re looking for the interactions that underlie secondary structures (like beta-pleated sheets and alpha helices)

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|921a84d4ebf34″ question_number=”138″] Four different biological molecules are represented in the diagram below. Molecule A is glucose. Molecules B, C and D are composed of glucose subunits that are covalently linked together.

All of the molecules shown above are

[c] lipids

[f] No. In a lipid, you typically see a long hydrocarbon chain. Here’s a hint: this is a class of macromolecules that includes monosaccharides (like the glucose shown above), disaccharides, and polysaccharides.

[c] proteins

[f] No. Proteins (such as oxygen-carrying hemoglobin or the contractile protein myosin) are polymers of amino acids. Here’s a hint: this is a class of macromolecules that includes monosaccharides (like the glucose shown above), disaccharides, and polysaccharides.

[c*] carbohydrates

[f] Excellent. All of the molecules shown above are carbohydrates.

[c] nucleic acids

[f] No. Nucleic acids, such as DNA or RNA, are polymers of nucleotides. Here’s a hint: what’s shown above is a class of macromolecules that includes monosaccharides (like the glucose in the diagram), disaccharides, and polysaccharides.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|920520a7ad334″ question_number=”139″] Four different biological molecules are represented in the diagram below. Molecule A is glucose. Molecules B, C and D are composed of glucose subunits that are covalently linked together.

Molecules “C” and “D” are

[c] monosaccharides

[f] No. Monosaccharides are the monomers of carbohydrates. The are the simplest of sugars, and consist of only one monomer. “C” and “D” consist of many sugar monomers linked together. Here’s a hint: what’s the prefix for “many?”

[c] disaccharides

[f] No. Disaccharides are sugars that consist of two monosaccharids linked together, as is shown in “B.””C” and “D” consist of many sugar monomers linked together. Here’s a hint: what’s the prefix for “many?”

[c*] polysaccharides

[f] Correct! C and D are polysaccharides.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|91f3850dc0f34″ question_number=”140″] Four different biological molecules are represented in the diagram below. Molecule A is glucose. Molecules B, C and D are composed of glucose subunits that are covalently linked together.

If molecule “C”  a) originates in a plant and b) consists of glucose monomers connected by bonds that allow that that molecule to be broken down by humans for energy, then that molecule must be

[c] cellulose

[f] No. Cellulose is the polysaccharide that makes up plant cell walls. We ingest cellulose everytime we eat fruit or a vegetable, but we lack the enzymes to break apart cellulose into glucose monomers from which we could derive energy. Foods high in cellulose, as a consequence, are very low calorie (think of lettuce or celery). What’s a polysaccharide that we can break down for energy.

[c] glycogen

[f] No. Glycogen is a polysaccharide that can be broken down by humans for energy, but it doesn’t come from plants. Mammals synthesize glycogen when they eat sugar, and store it in liver and muscle tissue for later use. What’s a polysaccharide made by plants that we (and other animals) can break down for energy?

[c] albumin

[f] No. Albumin is a protein (found in egg white, among other places). You’re looking for a molecule that’s a carbohydrate, and more specifically, a polysaccharide made by plants that we (and other animals) can break down for energy?

[c*] starch

[f] Excellent!

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|91e51d042e334″ question_number=”141″] Four different biological molecules are represented in the diagram below. Molecule A is glucose. Molecules B, C and D are composed of glucose subunits that are covalently linked together.

If molecule “C” were a component of plant cell walls, then it would most likely be

[c*] cellulose

[f] Correct! Cellulose is the polysaccharide that makes up plant cell walls.

[c] glycogen

[f] No. Glycogen is a polysaccharide, but one that’s used for energy storage in animals.

[c] albumin

[f] No. Albumin is a protein. You’re looking for a polysaccharide.

[c] starch

[f] No. Starch is a polysaccharide, but one that’s used for energy storage in plants.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|91d6453860b34″ question_number=”142″] Four different biological molecules are represented in the diagram below. Molecule A is glucose. Molecules B, C and D are composed of glucose subunits that are covalently linked together.

The process most directly responsible for creating molecule A (glucose) is

[c] glycolysis

[f] No. Glycolysis is the first step in cellular respiration, and therefore involved in breaking glucose apart (as opposed to creating it). Here’s a hint: the process happens in plants.

[c] the krebs cycle

[f] No. The Lrebs cycle is one of the major reactions involved in cellular respiration. That makes it more involved in breaking glucose apart than building it. Here’s a hint: the process happens in plants.

[c] the light reactions of photosynthesis

[f] No, but you’re very close. The light reactions create the NADPH and ATP that’s required for the next phase of photosynthesis to take in carbon dioxide and reduce it to carbohydrates such as glucose. What is that next phase called?

[c*] the Calvin Cycle

[f] Nice job. The Calvin cycle, also called the light independent phase of photosynthesis, is the process directly responsible for creating carbohydrates such a glucose.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|91c6b328dbf34″ question_number=”143″] The image below shows myoglobin, an oxygen-carrying globular protein found in the cytosol of muscle tissue. In globular proteins, one would expect to find

[c] only hydrophobic amino acids

[f] No. The key thing to focus on here is that this protein is globular. Globular means like a sphere. In living systems, which are watery, that means that the outside of the protein will be interacting with water, while the inside could be (depending on the structure) relatively water free. Thinking about globular proteins this way, where would you expect to see hydrophobic amino acids, and where would you see hydrophilic ones?

[c] only hydrophilic amino acids

[f] No. The key thing to focus on here is that this protein is globular. Globular means like a sphere. In living systems, which are watery, that means that the outside of the protein will be interacting with water, while the inside could be (depending on the structure) relatively water free. Thinking about globular proteins this way, where would you expect to see hydrophobic amino acids, and where would you see hydrophilic ones?

[c] hydrophobic amino acids on the surface of the protein

[f] No. The key thing to focus on here is that this protein is globular. Globular means like a sphere. In living systems, which are watery, that means that the outside of the protein will be interacting with water, while the inside could be (depending on the structure) relatively water free. Hydrophobic amino acids would tend to avoid water (and not be on the outside of such a molecule). What kind of amino acids would more likely be on the outside of a globular protein.

[c*] hydrophilic amino acids on the surface of the protein

[f] Great job. In a globular protein, you’d most likely find hydrophilic amino acids on the outer surface, where they’d be interacting with the surrounding water molecules.

[q json=”true” multiple_choice=”true” unit=”Biochemistry” dataset_id=”144Qs|91b3334546734″ question_number=”144″] Amino acids differ from one another because

[c*] they have side chains with distinct chemical properties.

[f] Correct! Amino acids all have a central carbon, connected to an amino group and a carboxyl group. They differ in their side chains.

[c] some have greater catalytic activity than others.

[f] No. Enzymes, which are almost always made of amino acids, differ in their catalytic activity. But amino acids don’t really have catalytic activity on their own. Next time you see this question, think about the structure of amino acids.

[c] some have more phosphate groups than others.

[f] No. No phosphate groups on amino acids.

[c] some have more amino groups than others.

[f] No. Every amino acid has one amino groups (and one carboxyl group).

[/qwiz]