Quiz #4 Answers

correct answers in bold

#1 (5 points) How can the abundance and function of a gene’s products (RNA, protein) be regulated inside the cell?

A. At the level of transcription, synthesis of a RNA molecule from coding DNA

B. At the level of translation, synthesis of a protein product from a template mRNA

C. RNA stability and structure, post-transcriptional regulation of RNA

D. All of the above

#2 (5 points) You have a lacIS (‘super repressor’) mutant of coli. The LacI protein of this mutant cannot bind lactose, but can bind the operator. The lac operon structural genes (lacZ, lacY, lacA) will be expressed:

A. Only when there is lactose in the growth media.

B. Only when there is not lactose in the growth media

C. Always, regardless of whether or not there is lactose in the growth media

D. Never, regardless of whether or not there is lactose in the growth media


#3 (5 points) You are a population geneticist studying a diploid species of salamander. The gene B influences salamander coloration. Two alleles of this gene exist, allele B (dominant) results in yellow salamanders, allele b (recessive) results in brown salamanders. The frequency of allele B is 0.07. What is the frequency of the b allele?

A. 0.93

B. 0.14

C. 0.07

D. 0.01


#4. (5 points) In which of the following populations is genetic drift expected to be strongest, in terms of its effects on change in allele frequencies over time?


A. 10,000 rabbits

B. 1,000 rabbits

C. 100 rabbits

D. 10 rabbits

Week 8 Reflections

We started the week discussing gene regulation – how cellular systems regulate the amounts of protein products (RNA, protein). The famous lac operon system was introduced; this set of genes in E. coli provided the foundation for much of our understanding of regulation at the transcriptional level. We then turned to discuss the similar gal system in yeast as a model for eukaryotes before considering some post-transcriptional forms of regulation such as alternative splicing and RNAi. On Thursday we turned to population and evolutionary genetics. Much of the discussion centered on the Hardy-Weinberg Principle of population genetics which provides important avenues for calculating genotype frequencies and allele frequencies. The class discussed some of the key underlying assumptions of the H-W Principle as well as two major implications of that principle for population genetic processes. Lecture material transitioned to a bit about evolutionary genetics with special emphasis on the forces of evolution. Mutation, although the most fundamental of the forces that provides the key ‘variation substrate’ for other evolutionary forces to act upon, is also a very weak evolutionary force because it is not able to cause rapid change in allele frequencies (unless population sizes are very, very small). Genetic drift was also discussed – the role of ‘sampling error’ of gametes from one generation to the next. When population sizes are small, there is a high probability that one particular allele might be completely lost (or completely fixed) in the population simply because that allele happened not to be ‘sampled’ purely due to chance from one generation to the next. An analogy used to help make drift more understandable is flipping a coin – with one million coin flips, you are very likely to end up with something very close to 50% heads and 50% tails (and almost certainly heads and tails getting sampled at least once) whereas with a much smaller number of coin flips (six, for example) there is a much greater chance of not getting a ‘50/50’ result and not either getting heads or tails at all in the series of coin flips.

We also discussed Muller’s Ratchet – a theoretical evolutionary concept not covered in the book. The idea behind the ratchet is that if you have a small isolated population that reproduces asexually, you might expect to lose the most-fit class of individual in that population (individual with fewest deleterious alleles) by drift. On top of this, because the population is small, mutation is a stronger force – individuals are accumulating new deleterious mutations on top of this. With no input of genetic variation from migration and no recombination, such populations would be expected to over time become less and less fit (individuals harboring increasing numbers of deleterious alleles) until it reaches extinction. How optimistic! This might seem rather unrealistic, but the theory has served as a model for helping evolutionary geneticists understand why other evolutionary forces (such as recombination) came about. Also, there are some asexual genetic components of endangered populations (e.g., mitochondrial DNA) that might be subject to the ratchet. Why ‘ratchet’? The idea here is that every time the population gets worse (loses most-fit individual) due to drift, this is a turn of the ratchet toward extinction.


Week 9 Sneak Peek: We will finish up Thursday’s lecture material, and then continue our population/evolutionary-genetic unit by discussing natural selection – Darwin’s key premises and deduction leading to his theory, and the many different sub-types of selection that are possible. We will focus on one test for the effects of natural selection on protein-coding sequences. I will also be posting some supplemental reading on this topic – it is a primary research article that I will be covering. Reading this paper is not absolutely required, but might help you understand things a bit better. Remember: no class on Thursday, and no recitation all week!

Answers to HW #8

Chapter 9:

9.3: (a) One mutation that might cause the constitutive phenotype is a mutation in an operator region of the enzyme-coding gene that makes it insensitive to repression. The second is a mutation that impairs the structure of the repressor. Such mutations can range from deletion of the repressor gene to subtler mutations that impair binding of the repressor to arginine, to the DNA, or to both. (b) A mutation in an arginine biosynthetic enzyme, not sufficient to cause a requirement for arginine but enough to reduce the amount of arginine in the cell, could activate a regulatory response by a normal regulatory system and induce constitutive synthesis.

9.6: The mutant gene should bind more of the activator protein, or have more efficient binding of the activator. Thus, the mutant gene should be induced with lower levels of activator protein or expressed at higher levels, compared to wild-type.

9.10: (a) The phenotype of cells with mutant Gal4p would be non-inducible; the mutant gene would be recessive because the wild-type Gal4p would still function normally. (b) The phenotype of cells with mutant Gal80p would be constitutive; the mutant gene would be recessive because wildtype Gal80p would bind Gal4p in the normal way.

9.18: (a): Yes, the repressor is functional, and the presence of lactose activates transcription of the lac genes; (b) and (d): Yes, at 42C, the repressor cannot bind the operator, which means that the lac operon is transcribed whether or not the inducer is present. (c): No, at this temperature, the repressor functions normally – because lactose is absent, the lac operon is in a repressed state.

Chapter 14:

14.1. The frequency of A1 equals 0.35.

14.2. The expected frequency of A2 in the next generation is the same as its frequency in the current generation: 0.65.

14.4. (A) no; (B) yes.

14.8. The frequency of homozygous recessives is q2 = 0.16; this implies that q=0.4. The frequency of the dominant allele is therefore p=0.6.

14.9. Two issues need to be considered. First, recessive alleles are maintained in heterozygous individuals and so are not exposed to selection. Second, new mutations in each generation replenish the number eliminated by selection in the homozygous recessives.

14.16. The numbers of alleles are as follows: A, 8+10+2 = 20; B, 10+48+20=78; C, 20+20+2 = 42. The total number of alleles is 140, so the allele frequencies are as follows: A, 20/140 = 0.14; B, 78/140 = 0.56; C, 42/140 = 0.30. The expected numbers among the 70 plants are:

AA: 1.42, AB: 11.14, BB: 21.72, BC: 23.40, CC: 6.30, AC: 6.00.

Answers to HW #7

8.4: MALISASY (in single-letter amino acid code)


8.9: 5’-AUU = Isoleucine at amino terminus of protein; UUA-3’ = Leucine at carboxy terminus


8.10: This encodes the alternating polypeptide Cys-Val


8.18: There are three possible reading frames, each encoding a different repeating polymer. One has a repeating Val (GUC), one has a repeating Ser (UCG), and one has a repeating Arg (CGU).

Answers for questions from last year’s Midterm #2


D, dosage compensation


D, 0.0001


Cyclins tether specific target proteins and bring them to the complex; CDKs phosophorylate those target proteins to change their structure/function.


(0.10 x 0.005) x 10,000 = 5; coeff. coinc. = 2/5 = 0.4

i = [1-(0.4)] = 0.6

A few example questions from last year’s Midterm #2

Here are a few questions from last year’s Midterm #2 to help with your studying.  Answers will appear on the blog in a few days.

#7. (3 points) In domestic cats, an X-linked gene influences coat coloration. One allele results in black coloration and a second allele causes orange coloration. Female cats with both black and orange spots are heterozygous at this X-linked locus. What is the explanation for this genotype-phenotype relationship?

a) Incomplete dominance

b) Codominance

c) Epistasis

d) Dosage compensation


#13. (3 points) A genetic distance of 0.01 cM corresponds to what recombination frequency?

a) 1

b) 0.01

c) 0.001

d) 0.0001


#20. (6 points) Cyclins and cyclin-dependent kinases (CDKs) form heterodimers that function to regulate the cell cycle. Briefly describe the specific, molecular function played by cyclins in this process, as well as the specific molecular function of CDKs.


#21. (6 points) You are studying three linked genes: D, E and F. There are two alleles for each gene (D and d, E and e, F and f). E is the gene in the middle. The genetic distance between D and E is 0.5 cM. The genetic distance between E and F is 10 cM. You perform a cross between a triply heterozygous organism and a triply homozygous recessive tester and analyze 10,000 progeny.   Among the progeny, only two double-recombinants were observed. What is the degree of crossover interference (i)?

Hint:   i = 1 – (coefficient of coincidence)

= 1 – (obs. # double recombinants/exp. # double recombinants)

Week 6 Reflections

This week we wrapped up a little bit of recombination, then turned our focus toward mutation and then cancer genetics.  We turned attention to the molecular mechanisms of DNA recombination, overviewing a few early (but ultimately incorrect) models before learning about the double-stranded (ds)DNA break pathway, which is currently held as the correct model.  dsDNA break-mediated recombination can give rise to both recombinant as well as non-recombination chromosomes depending on how chi structures are resolved, and can also result in gene conversion events.  The example of yeast mating-type switching provided an example of dsDNA break-mediated recombination occurring between repetitive DNA sequences on the same molecule (between hidden and silenced ‘mating cassettes’ and the expressed MAT gene itself).

Next, we first learned about different pathways by which DNA can mutate, ranging from base substitution changes resulting from the natural fluctuations in electron distributions in nucleotides (for example, the keto versus enol form example in thymine) to the significant genome alterations that can result from mobile genetic elements such as retrotransposons. We also learned about how mutations can be classified in a variety of ways.  Our discussion of mutation concluded with the topic of mutation rates, covering two different methods for how such rates can be experimentally estimated (reporter gene versus mutation-accumulation line approach). Reporter gene approaches rely on the phenotypes associated known to occur when the gene is mutated. The phenotypic change (for example, from blue to white colony of E. coli cells) reports the occurrence of a mutation. With mutation-accumulation lines, on the other hand, organisms are propagated across many (usually hundreds) of generations in the lab, and then their DNA is directly sequenced afterwards to count mutations at the DNA-level. This strategy offers a much more direct avenue (not relying on assumptions about reporter gene phenotypes) to understanding the mutation process. On Thursday, our attention turned toward cancer genetics. This lecture also highlighted the utility of temperature-sensitive mutants in genetic analysis with the yeast model organism. We learned about how cyclin-CDK complexes regulate cell cycle progression.

Week 7 Sneak Peek: We will wrap up cancer genetics on Tuesday, and transition to some basic discussion about gene expression (for example, transcription basics). Thursday will be midterm #2… Study hard! Bring your calculators! The basic format will be the same as the first midterm – about 45% of the points coming from multiple choice and about 55% coming from short answer. This exam will cover all material up through (and including) next Tuesday’s lecture content. Material covered in-between Midterm #1 and #2 will be the main emphasis of the questions, though you will still need the knowledge base of material presented early in the class to succeed. Also, reminder that next week on Friday is Veteran’s Day, and OSU will not be in session.

HW #6 answers

12.1: (b) Formation of crosslinked thymine dimers.

12.2: Such people are somatic mosaics, a condition that can be explained by somatic mutations in the pigmented cells of the iris of the eye or in their precursor cells.

12.5: There are 18 depurinations per minute and 1440 minutes per day, so the total number of depurinations per cell per day is 18 x 1440 = 25,920.


13.2: It is a “genetic disease” in that genetic mutations take place in cells and make them undergo uncontrolled growth and division. However, most cancers are caused by somatic mutations in the cells of affected individuals; hence, they are not inherited from the parents (familial) but rather are due to new somatic mutations.

13.11: A defect in the G1/S checkpoint allows cells with damaged chromosomes to enter DNA synthesis, when replication and repair can cause chromosome rearrangements or replication errors leading to aneuploidy.

13.19. Effectively, HPV removes all the brakes on cellular proliferation, turns on all the stimuli, and removes the infected cell’s ability to induce apoptosis to kill itself. The cell will effectively be p53−, because E6 is present to ensure that levels of p53 do not rise. This means that synthesis of Bax will not be induced, and cells will not undergo apoptosis. In addition, synthesis of p21, GADD45, and 14-3-3s will not be accelerated, so cells will not arrest in the cell cycle. Because E7 disables RB, E2F will be active at all times, resulting in new rounds of DNA synthesis and continued cellular proliferation.

13.20. For generations I and II, the band at position c is associated with the disease. In generation III, individuals 2 and 7 are at risk because of parent 5, and the individuals 21 and 22 are at risk because of parent 20. However, individual 13 is not at risk because the parent 14 with a band at position c is not affected; band c in this individual evidently originates from a nonmutant allele.

Extra Question #1:

Endogenous mutagens originate from sources inside the cell or organism; exogenous mutagens come from outside (or, environmental) sources.

Endogenous mutagen sources include nucleotide tautomerism (enol/keto), replication errors, and metabolic byproducts (such as 8-oxoguanine).

Exogenous mutagen examples include UV light, chemical base analogs (such as 5-bromouracil), and diesel fumes.