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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)

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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.

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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.

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Quiz #3 answers

Correct answers in bold

  1. (5 points) The figure to the right shows a chromosome during mitotic metaphase. Which term best describes the structure of this chromosome? (Note: see bottom left panel of textbook Figure 5.5 for image)

a) metacentric

b) acrocentric

c) chiasma

d) compensatory

 

2. (5 points) The Y chromosome in humans is thought to have evolved from a homologous partner to which other chromosome?

a) Chromosome 21

b) Chromosome 22

c) The Z Chromosome

d) The X Chromosome

 

3. You are studying the rb gene (rb indicates recessive mutant allele, + indicates wild-type) and cv gene (cv indicates recessive mutant allele, + indicates wild-type) in Drosophila. Both genes are on the X chromosome. You set up a testcross between a doubly heterozygous female and a male tester fly. You infer the following genotypes from the phenotypes observed in the male progeny of this cross:

 

Genotype                                Number Observed

+   + / Y                                              90

+ cv / Y                                              11

rb + / Y                                             9

rb cv / Y                                             90

 

Part A (5 points): Does this data suggest that the rb and cv genes are linked?

a) Yes

b) No

c) Not Enough Information Provided

 

Part B (5 points): What is the genetic distance (in cM) between rb and cv?

(9+11)/200 = 20/200 = 0.10 recombination freq. (x100) = 10 cM

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Answers to HW #5

4.1: 0.2.

4.2: Females produce gametes of types A B, a b, A b, and a B in the proportions (1-0.08)/2 = 0.46 and (1-0.08)/2 = 0.46 for each of the nonrecombinant types. It is 0.08/2 = 0.04 and 0.08/2 = 0.04 for each of the recombinant types. Since there is no crossing over in male Drosophila, males produce only two gamete types (A B, a b) in equal (0.50, 0.50) proportions.

4.7: The map distance between s and c is 18 cM, but there is only 16% recombination, and the map distance between p and c is 13 cM, but there is only 12% recombination. The fact that the frequency of recombination is smaller than the map distance results from double crossovers.

4.8: The data include 108 progeny with either both mutations or neither, and 92 progeny with one mutation or the other. The first of these groups consists of parental chromosomes; the second consists of recombinant chromosomes.   The chi-square test compares the observed ratio (108:92) to the expected (100:100) under the hypothesis of no linkage. The chi-square value equals 1.28 and there is one degree of freedom. P is approximately 0.26, so there is no evidence of linkage even though both loci are on the X.

4.10: (a) the mutant genes are not alleles because they do not show segregation; if they were alleles, the progeny would show resistance to one insecticide or the other. (b) the genes are not linked. The chi-square of 177 parental:205 recombinants compared to 191:191 (expected) equals 2.05. With one degree of freedom, P is 0.15 (not significant). (c) The two genes must be far apart on the chromosome – too far to be ‘linked’.

4.11:

(a) The gene in the middle has the alleles D and d.

(b) The triply heterozygous parent has genotype A d B / a D b or B d A / b D a.

(c) The closest genes are B and D.

(d) B−D distance equals (40 + 60 + 10)/1000 = 0.11 = 11 map units = 11 centimorgans (cM).

4.12: (a) parental types = v+ pr bm and v pr+ bm+. Double recombinants = v+ pr+ bm+ and v pr bm. This means that gene v is in the middle. The pr-v recombination frequency is 20.7% and the v-bm frequency is 41.5%. The expected number of double crossovers = 85.9, so the coincidence is 0.90. The interference is thus 0.10. (b) The true map distances are larger than 20.7cM and 41.5 cM because the frequency of recombination between these genes is sufficiently large that there must have been some double crossover events that went undetected in the analysis.

4.20.

(a)       To solve this type of problem, you should first deduce the genotype of the gamete contributed by the triply heterozygous parent to each class of progeny. In this case the gamete types are, from top to bottom, + + +, + + unc, + dor unc, dpy + +, dpy dor +, dpy dor unc. To determine the order of the genes, you must identify which gene is in the middle. You can do this by comparing the parental type gametes with the double crossovers. The two parental types are dpy dor + and + + unc, because these are the most frequent. In these data there are no observed double crossovers (which means that interference across the region is complete). The missing classes of progeny correspond to the double crossovers. The missing phenotypes of progeny are normal body, deep orange eye, coordinated, corresponding to the genotype + dor +, and dumpy body, red eye, uncoordinated, corresponding to the genotype dpy + unc. Comparing the parental types with the inferred double crossover types reveals that dpy+ and dpy are interchanged relative to the alleles of the other two genes. Hence, the genotype of the F1 heterozygous female is dor dpy unc+ / dor+ dpy+ unc or unc+ dpy dor/ unc dpy+ dor+.

(b)       Analysis of the single recombinants indicates (96 + 110)/1000 = 20.6% recombination between dor and dpy, and (75 + 65)/1000 = 14.0% recombination between dpy and unc.

(c)        Because there are no double recombinants observed, the interference is complete, which means interference = 1.

 

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Week 5 Reflections

It was all about recombination! We learned about the inheritance patterns of linked genes nearby one another on a chromosome, most importantly how to quantitatively express the incidence of co-inheritance of linked markers in terms of recombination frequencies (or, genetic distance as measured in cM). In addition, we gained an understanding of how patterns of genetic distance between marker genes with clear associated phenotypes led to the creation of genetic maps: maps revealing the relative ordering of genes along chromosome physical space. These genetic maps have been enormously helpful to both scientists learning about genetic and other biological processes in their study systems, and also to plant and animal breeders interested in producing high-value crops and livestock in the world of agriculture. The three-point testcross – appearing in lecture, homeworks, and recitation – illustrates the basic logic that underlies the construction of genetic maps.

We also learned about recombination rates (not covered in the book).  This is an expressed as the genetic distance (in cM) divided by the physical distance (in Mb).  Looking at recombination this way, it becomes very apparent that different regions of DNA can vary extensively in term of how ‘recombinogenic’ they are.  A human chromosome example was provided, showing lots of recombination hotspots, and coldspots, across the length of the chromosome space.

We started to talk about the molecular mechanism of homologous recombination on Thursday and I discussed things a bit differently than the treatment in the book.  Understanding the molecular DNA-level actions and movements of DNA during recombination has been a central area of genetics research for decades. This was an area of great controversy for a very long time, until the development of the double-stranded DNA break and repair (DSBR) model was established, which has been well-accepted and experimentally supported every since. The first model was the Holliday Model which posited that the whole process began with a pair of single-stranded DNA nicks at homologous positions at the two participating DNA molecules. After nicking, reciprocal strand invasion occurred (forming Holliday or chi structure) followed by branch migration followed by endonucleolytic resolution. Though many of the details were right, the Holliday model was ultimately discounted because there was no known enzymatic activity that could cause ssDNA nicks at the same exact (homologous) sites in two DNA molecules – still nothing known to this day that can do that. The Meselson-Radding (M-R) Model modified things by suggesting that a single-stranded nick could start the process – the free end of the nicked strand (“donor strand”, gray in diagram) could then carry out strand invasion of the “acceptor” DNA molecule (pink in diagram) which would then cause a displacement loop (or, D-loop) strand from the acceptor that could undergo base-pairing with homologous sites in the invading molecule. This model was widely regarded as essentially correct, but there was still one puzzle related to recombination – gene conversion – that could not be explained by the M-R model. Gene conversion, originally only observed in certain species of fungi, is a process whereby there is allelic “conversion” during meiosis that is in opposition to predictions of basic Mendelian genetics (e.g. one ‘a’ allele is “converted” into a ‘A’ allele in the example shown). Since this phenomenon was only observed in gametes, it was presumed to be closely associated with a meiosis-specific process such as recombination. This led researchers to develop the double-strand break & repair (DSBR) model. The DSBR model has been strongly supported over the last few decades in all lab experiments and is also able to effectively account for gene conversion. At the end of the DSBR mechanism there are tracts of DNA at the end of the process where there is base pairing between nucleotides from the “donor” strand and the “acceptor” strand. There is the possibility of nucleotide mispairing between these two strands (because they originally came from different dsDNA molecules) – these mispairings are recognized by the mismatch DNA repair enzymatic machinery that ultimately removes one of the mismatching nucleotides and replaces it with a correct complementary base. Thus, there is the possibility in this DSBR path for one allele to be converted into the other allele! Since this was all worked out, we’ve come to discover that gene conversion outcomes occur much more frequently than recombinant chromosome outcomes of the molecular recombination process in virtually all eukaryotic life forms, from fungus to human.

Week 6 Sneak Peek: Next week we will quickly recap recombination, and then transition to talk about the mutation process (Ch. 12) on Tuesday. On Thursday we will talk about the molecular genetics of cancer (Ch. 13), and how yeast (a single-celled organism…) provided key insights into the genetic underpinnings of cancer in humans.

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Week 4 Reflections

Tuesday was the first midterm – we are about halfway done with grading and aim to return them to students next week during lecture.

On Thursday, we moved into the world of mammalian and human chromosome biology, discussing some basic terminology and conventions used in karyotype studies. We discussed some interesting chromosome evolution stories in the human lineage, demonstrating the utility of nonrecombining chromosomes (e.g., Y) for certain genetic applications. We learned about methods for studying chromosome biology, such as chromosome painting and G-banding. After the not-thinking break, we started to move into the topic of recombination…

Week 5 preview: The upcoming week will be dedicated to the topic of recombination, linkage, and gene mapping. We will cover chapter 4 – make sure you look at the material ahead-of-time.  It is important to wear your ‘thinking caps’ for Chapter 4 material, and to be engaged in lectures and recitations.

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Answers to Homework #4

Chapter 5

5.1, 5.2, 5.12, 5.13

5.1: The term “inactive” is not quite correct because about 15% of the genes in this chromosome are transcriptionally active to some degree, producing 15−50% of the RNA transcripts as are produced from the active X chromosome.

5.2: The genetic consequence of the obligatory crossover is that alleles in the pseudoautosomal region in the X chromosome can be interchanged with their homologous alleles in the pseudoautosomal region in the Y chromosome, yielding a pattern of inheritance in pedigrees that is indistinguishable from that of ordinary autosomal inheritance.

5.12: The mother has a translocation that joins the long arm of chromosome 21 with the long arm of another acrocentric chromosome (a ‘Robertsonian’ translocation). The affected child has 46 chromosomes. This karyotype differs from the usual karyotype for trisomy 21 because the extra chromosome 21 is not a free chromosome.

5.13: Because the X chromosome in the 45, X daughter contains the color-blindness allele, the 45, X daughter must have received the X chromosome from her father by way of a normal X-bearing sperm. The nondisjunction must, therefore, have occurred in the mother, resulting in an egg cell lacking an X chromosome.

Additional Question not in your textbook:

#1. In Drosophila, waxy wings are inherited as a sex-linked recessive trait, and hairy body is inherited as an autosomal dominant trait. Indicate the F1 and F2 phenotypic ratios expected from a cross between a waxy wing male and a hairy body female.

Approach this problem by first denoting the genotypes at both genes of the parents:

P          waxy male                 X          hairy female

wY h+h+                                                    w+w+ HH             

 Then, proceed to the F1 generation. Remember that males get their X chromosome from their mother and their Y chromosome from the father. Females get an X chromosome from each of the mother and father.

F1         males are:                 w+Y Hh+

females are:                         w+w Hh+

 Next, indicate the F2 generalized genotypes that will result in each phenotype. Calculate the probability of obtaining the genotype at each gene pair and multiply the individual probabilities. This give the expected ratio of the corresponding phenotype as indicated below:

F2         w+_ H_                        probability = ½ X ¾ = 3/8 hairy females

w+_ h+h+         probability = ½ X ¼ = 1/8 wildtype females

w+Y H_                        probability = ¼ X ¾ = 3/16 hairy males

w+Y h+h+         probability = ¼ X ¼ = 1/16 wildtype males

wY H_             probability = ¼ X ¾ = 3/16 waxy, hairy males

wY h+h+           probability = ¼ X ¼ = 1/16 waxy males