BIO Class 2 Corrections

Huntington's disease is due to CAG repeats in the HTT gene. An individual with 26 or fewer CAG repeats will not develop the disease and will not pass the trait onto any offspring. A patient with 27-35 repeats will not be affected but has a small chance of passing the trait onto their offspring due to further repeat expansion. An individual with 36-39 repeats in one copy of HTT may or may not develop Huntington's disease but has a 50% chance of passing this trait onto offspring. A patient with 40 or more CAG repeats in one copy of HTT will develop the disease and has a 50% chance of passing it onto their children. Which of the following is true? A) Huntington's disease displays autosomal dominant inheritance and a patient with 37 repeats will have partial penetrance of phenotype development. B) Huntington's disease displays autosomal dominant inheritance and a patient with 41 repeats will have partial penetrance of phenotype development. C) Huntington's disease displays autosomal recessive inheritance and a patient with 37 repeats will have enhanced penetrance of phenotype development. D) Huntington's disease displays autosomal recessive inheritance and a patient with 41 repeats will have full penetrance of phenotype development.

A. A patient with 40 or more CAG repeats in one copy of HTT has a 50% chance of passing Huntington's disease onto their children. This individual is heterozygous; since the offspring have a ½ chance of being affected, Huntington's disease must display autosomal dominant inheritance (eliminate choices C and D). A patient with 37 repeats may or may not develop the disease, which is a type of partial penetrance (another reason to eliminate choice C). A patient with 41 repeats will develop the disease, which is full penetrance (eliminate choice B). Thus, choice A is the correct answer.

Which best describes the role of inorganic phosphate in the reaction catalyzed by glycogen phosphorylase? A) Nucleophile B) Electrophile C) Acid D) Base

A. According to the passage inorganic phosphate is used to assist in the hydrolysis of the α1 → 4 linkage and results in the release of glucose-1-phosphate. This means that the phosphate attacked the glucose residue on the glycogen chain and is now linked to the freed glucose molecule at the 1 position. Nucleophiles are the species that use their electrons to attack other atoms and attach themselves to the atoms. Thus, choice A is the correct answer. The anomeric carbon (carbon 1) of the glucose residue attacked by the phosphate is the electrophile, not phosphate itself, making choice B incorrect. There is no mention in the passage of a proton being either donated or abstracted by inorganic phosphate and so it does not act as either an acid or a base, eliminating choices C and D.

Which of the following is NOT true regarding muscle and liver? A) Muscle can release glucose into the bloodstream. B) The liver converts pyruvate into glucose. C) Liver and muscle both synthesize glycogen from glucose. D) Muscles rely on glycolysis for energy during activity.

A. Since muscle lacks the enzyme glucose-6-phosphatase (this was stated in the first main paragraph), it is unable to convert glucose-6-P (produced during muscle glycogen breakdown) to uncharged glucose capable of crossing the muscle cell membrane. Thus muscle cannot release glucose to the bloodstream (choice A is false and is the correct answer choice). Note that this is important in ensuring that muscle has an adequate energy source, especially important in cardiac muscle. The liver can convert pyruvate to glucose; this is gluconeogenesis (choice B is true and can be eliminated). Liver and muscle can both synthesize glycogen from glucose; this is where they get the glycogen that can be broken down in the pathways discussed in the passage (choice C is true and can be eliminated). Muscle does rely on glycolysis for energy during exercise; hence the buildup of lactic acid (and the drop in pH mentioned in the passage) in muscle tissue during exertion (choice D is true and can be eliminated).

Which of the following is true regarding the control of glycogen metabolism? A) Futile cycles, i.e., glycogen synthesis and degradation running simultaneously, are prevented by reciprocal regulation. B) Regulation depends on the amount of synthesis of metabolic enzymes, i.e., it takes place at the level of transcription. C) The glycogen stored in muscle is chemically distinct so that glucose stored in muscle glycogen cannot be exported to the blood stream. D) The primary storage depot for glycogen is adipose tissue.

A. There is no point in running reciprocal cycles (cycles with opposite end results) simultaneously (choice A is correct). While regulation of many pathways occurs at the transcriptional level, glycogen breakdown is primarily regulated via hormonal (glucagon) and ionic (calcium and phosphate) triggers (choice B is wrong). Muscle glycogen is identical to all other glycogen found in the body; the export of glucose is prevented by the absence of glucose-6-phosphatase and the subsequent inability to dephosphorylate glucose-6-P (the product of glycogen breakdown, choice C is wrong). The primary storage depot for glycogen is liver; adipose tissue stores fat (choice D is wrong).

Which of the following best describes the main site of ketogenesis? A) Liver cell mitochondria B) Brain cell mitochondria C) Liver cell cytoplasm D) Brain cell cytoplasm

A. While ketones are utilized by the brain, they are not produced in the brain (choices B and D can be eliminated). Ketogenesis occurs in the mitochondria of liver cells (choice A is correct), and not in the cytoplasm (choice C is incorrect).

Which of the following reactions is NADPH likely to participate in? Conversion of GSH to GSSG Conversion of H2O2 to H2O Conversion of O2 to •O2- A) I only B) II only C) I and III only D) II and III only

B. Since all statements appear twice in the answer choices, there is no best place to begin, so we will start with Item I. Item I is false: NADPH is a reducing agent, so we expect it to participate in reduction reactions and in reactions that reduce radical-forming species, as noted in the passage. The conversion of GSH (reduced form) to GSSG (oxidized form) is an oxidation so we would expect NADP+ to participate in this, not NADPH (eliminate answer choices A and C). Item II is true: conversion of H2O2 (a radical forming species) to H2O (a non-radical-forming species) is a reduction reaction, so we would expect NADPH to participate in this reaction. Item III is false: Conversion of O2 to •O2- generates additional radicals; considering the passage states that NADPH is used to "rid the cell of radical forming species," we would not expect it to generate more radicals, thereby creating more oxidative stress (eliminate answer choice D; answer choice B is correct).

The missing karyotype data from Table 2 should read: A) 46 autosomes, 46 centromeres, 46 telomeres, 21 medium size chromosomes B) 44 autosomes, 92 centromeres, 92 telomeres, 21 medium size chromosomes C) 44 autosomes, 46 centromeres, 92 telomeres, 21 medium size chromosomes D) 44 autosomes, 46 centromeres, 92 telomeres, 20 medium size chromosomes

C. HC7W is a human cell line, and should therefore have 46 total chromosomes; 44 autosomes and 2 sex chromosomes (choice A can be eliminated). Each chromosome should have one centromere (for a total of 46, choice B can be eliminated) and two telomeres (for a total of 92). Table 2 lists 10 large chromosomes and 15 small chromosomes. If there are 46 chromosomes total, there must be 21 medium size chromosomes (choice C is correct, and choice D is wrong).

Which of the following will activate glycogen phosphorylase? Muscle contraction Muscle relaxation Phosphorylation A) I only B) II only C) I and III only D) II and III only

C. Item I is true; the passage states that glycogen phosphorylase can be activated by increased intracellular Ca2+. Since intracellular Ca2+ rises during muscle contraction, this would activate glycogen phosphorylase. Item II is false; intracellular Ca2+ levels fall upon muscle relaxation. Item III is true; the passage also states that glycogen phosphorylase is activated by phosphorylation.

The hypoglycemia caused by von Gierke's disease is due to: A) conversion of glucose to pyruvate in the liver. B) elevated glucagon levels. C) inability to obtain glucose from glycogen breakdown. D) inability of synthesized glucose to leave liver cells.

C. Since glucose-6-phosphatase is missing, the phosphorylated glucose (glucose-6-P) obtained from glycogen breakdown cannot be dephosphorylated to produce glucose. The absence of this source of glucose leads to hypoglycemia (choice C is correct). Choice D is wrong; glucose itself has no problem leaving the liver cells, the problem is that no glucose is synthesized because of the missing enzyme. Glucose is normally converted to pyruvate in the liver (in fact, in all cells during glycolysis), however this is not the cause of the hypoglycemia (choice A is wrong). Glucagon is a pancreatic hormone that leads to an increase in blood glucose levels; elevated levels of glucagon would not lead to hypoglycemia (choice B is wrong).

The Bcl-XL protein has an important role in programmed cell death. Which of the following could be the role of Bcl-XL? A) Promotes apoptosis, thus causing cell growth and proliferation B) Promotes mitochondrial permeability, thus promoting apoptosis C) Prevents the release of cytochrome c, thus negatively regulating apoptosis D) Promotes the release of cytochrome c, thus inhibiting apoptosis

C. Since two answer choices say promoting apoptosis (or programmed cell death) and two say inhibiting apoptosis, this is a typical 2X2 question. The insertions in Bcl-XL in Figure 1 all go in the forward direction. This means Bcl-XL is a likely oncogene and is promoting tumor growth (see explanation to question 2). To do this, it must block apoptosis, not promote it (eliminate choices A and B). Cytochrome c is an important component of the electron transport chain. It would only be released from the mitochondria when the mitochondria is made permeable during apoptosis. Therefore apoptosis must involve the release of cytochrome c. To inhibit apoptosis, Bcl-XL protein must inhibit this release (choice C is correct, eliminate choice D).

The Cdh1 gene codes for a protein called E-cadherin, a transmembrane protein involved in adhesion. Transposon insertion in this gene most likely has which of the following effects in the skin? A)Altered gap junction function due to decreased E-cadherin expression B) Altered tight junction function due to increased E-cadherin expression C) Altered desmosome function due to decreased E-cadherin expression D) Altered cell morphology due to increased E-cadherin expression

C. The key to working with this passage (and to answering questions 2, 3, 5 and 6) is to figure out how to use the data in Figure 1 based on information in the passage. The passage says that transposon insertion direction can give information on whether gene expression is decreasing (as would be the case for tumor suppressors) or increasing (as would be the case for proto-oncogenes and oncogenes). It also describes that the only way to increase gene expression is to have the transposon insert in the forward direction only, where both forward and reverse direction insertions can decrease gene expression. Looking at the data in Figure 1, you can see that Cdh1, PTEN and Arid2 have insertions scattered throughout the gene and in both directions. This means the SB transposon is likely causing decreased expression of these genes in the tumors and that their normal proteins inhibit cancer growth or progression. These are likely tumor suppressors. Bcl-XL, Kit and Ros1 have only insertions in the forward direction and close to the 5' end of each gene. This means the SB transposon is likely driving expression of these genes, increasing their expression, indicating they are likely proto-oncogenes or oncogenes (note that this matches the function of Ros1 as a proto-oncogene, as stated in Question 1). Epithelial cells in the skin are connected by desmosomes, not gap junctions (eliminate choice A) or tight junctions (eliminate choice B). Figure 1 shows that the insertions in Cdh1 occur in both forward and reverse directions. This means the transposon insertions are most likely interrupting gene expression of Cdh1, leading to less protein expression (choice C is correct, eliminate choice D). Also note that two answer choices say decreased E-cadherin expression and the other two say increased E-cadherin expression, so this is a 2x2 question.

Michelle is a scientist who studies trinucleotide repeat disorders. She is working with two genomic DNA samples. Sample 1 is from a 67-year-old man with Friedreich's ataxia. Sample 2 is from a 12-year-old girl with spinobulbar muscular atrophy. Which of the following most likely describes the characteristics of these two genomic samples? A) Sample 1: short segment of GAA repeats in the gene X25, leading to a late onset polyglutamine disease; Sample 2: long segment of CAG repeats in the gene AR, leading to an early onset polyglutamine disease. B) Sample 1: short segment of GAA repeats in the gene X25, leading to a late onset non-polyglutamine disease; Sample 2: long segment of GAC repeats in the gene AR, leading to an early onset polyglutamine disease. C) Sample 1: short segment of GAA repeats in the gene X25, leading to a late onset non-polyglutamine disease; Sample 2: long segment of CAG repeats in the gene AR, leading to an early onset polyglutamine disease. D) Sample 1: long segment of GAA repeats in the gene X25, leading to a late onset non-polyglutamine disease; Sample 2: short segment of CAG repeats in the gene AR, leading to an early onset polyglutamine disease.

C. The passage says that CAG is one of two codons that code for glutamine. For the amino acids that are coded by only two codons, the first letter of the codon is always the same. It is therefore not possible that a GAA repeat would lead to a polyglutamine disease (eliminate choice A). Using similar logic, CAG repeats (as in the second half of choice B) also cannot lead to a polyglutamine disease (eliminate choice B). Choices C and D are quite similar, but have different details for the length of repeats in the two patient samples. The second paragraph in the passage says that trinucleotide repeat disorders are a set of genetic disorders caused by expansion, meaning unstable DNA repeat sequences multiply. This leads to anticipation, where severity increases with each successive generation. Because the first patient is older than the second patient, it is most likely that Sample 1 contains a shorter repeat sequence than Sample 2 (choice C is correct, eliminate choice D).

Which of the following is most likely to be decreased in an individual with untreated type 1 diabetes? A) Acetyl-CoA B) Ketone bodies C) PCO2 D) Blood glucose

C. While an individual with untreated type 1 diabetes has an excess of glucose in their blood (choice D is wrong), the glucose will not enter the cells and cannot be used for energy. These individuals would rely more on fatty acid oxidation and protein catabolism for energy, both of which produce acetyl-CoA, thus acetyl-CoA is unlikely to be decreased (choice A is wrong). By similar reasoning, ketone bodies are unlikely to be decreased, since ketones would be produced to fuel the brain in the absence of glucose (choice B is wrong). The excess fat breakdown and ketogenesis would lower blood pH since ketones are acidic; to compensate, ventilatory rate would increase in an effort to eliminate CO2 and bring the blood pH back to normal (choice C is likely to be decreased and is the correct answer choice).

There are many naturally occurring and non-problematic repeats in the human genome. Which of the following structures are rich in repeats? A) Euchromatin, centrosomes and telomeres B) Euchromatin, centromeres and telomeres C) Heterochromatin, centrosomes and telomerase D) Heterochromatin, centromeres and telomeres

D. Heterochromatin is rich in repeats, but euchromatin is not (eliminate choices A and B). Centrosomes are organelles that serve as the main microtubule organizing center (MTOC) in animal cells. They are made of two centrioles as well as proteins, but not DNA (eliminate choices A and C). Telomeres are rich in repeats, but telomerase is an enzyme that maintains telomeres in the germ line, embryonic stem cells, some white blood cells and some cancer cells (another reason to eliminate choice C). Overall, choice D is correct and lists genomic regions that are rich in repeats.

Based on the data in Figure 1, would a protein product be made in the tumor cells that had a transposon insertion between exons 2 and 3 of Kit? A) No: the start codon of a gene is always found in the first exon and without a start codon, no protein can be made. B) No: the insertion is in the opposite direction of the gene and so must be stopping gene expression. C) Yes: all insertions, both forward and reverse, that occur in introns will cause protein expression. D) Yes: the start codon and splice donor in the transposon will allow expression of exons 3 through 17 of Kit.

D. The insertions in Kit are all going in the forward direction. This means Kit expression is likely increased in tumor samples (see explanation to question 2). The promoter, start codon and splice donor in the transposon are in the forward direction, and allow expression of a downstream gene after insertion. This means a modified version of Kit protein is likely being made (eliminate choices A and B). Also note that choice A is a false statement. The start codon can be in the first exon of a gene, but does not have to be. In addition, the transposon contains a start codon, so the endogenous start codon is not required here. There is no information in the passage to suggest that all insertions that occur in introns will cause protein expression. In fact, the passage says that transposons can insert forwards or backwards to inhibit protein expression (eliminate choice C). By process of elimination, the correct answer must be choice D and this statement also matches the information in the passage.

Which of the following would be LEAST likely to occur in an individual with alcoholic ketoacidosis? A) Decreased blood pH B) Increased β-oxidation C) Increased protein catabolism D) Increased insulin secretion

D. This is a LEAST likely question, so the correct answer here is the answer choice that is a false statement. By definition, ketoacidosis involves a decrease in blood pH (choice A would occur and can be eliminated). As is stated in the passage, individuals with alcoholic ketoacidosis have difficulty maintaining blood glucose levels and rely instead on the breakdown of fats and proteins for energy (choices B and C would occur and can be eliminated). However, since blood glucose is lower than normal, increased insulin secretion would not be expected (choice D would not occur and is the correct answer choice).