PS 7 16key - Answer key for Problem Set 7 PDF

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Answer key for Problem Set 7...

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BANNER ID______________________ BIO 47 PROBLEM SET #7 2016 DUE FRIDAY, October 28st BY 1:00 PM PLACE HARD COPY IN DROP BOX OUTSIDE BMC 11 Question 1 (20pts) The following pedigree shows the segregation pattern of a recessive disorder in which affected individuals appear in the general population at 1/400.

(A) (5pts) Use A and a as symbols for the alleles at the disease locus. Assuming that the mode of inheritance is autosomal recessive (AR), what percentage of the recessive alleles are carried in heterozyogtes versus homozyogtes? For AR, affected individuals are found at a frequency of q2, so, q2= 1/400 and therefore, q=1/20. 2pq= 2(19/20)(1/20) = 38/400. p2=(19/20) 2 = 361/400. So, in 400 individuals, there are 38 a alleles in heterozygotes, and only 2 a alleles in homozygotes. For a ratio of 38:2 or 19:1. There are only 5% of a alleles in homozygotes. (B) (5pts) Use A and a as symbols for the alleles at the disease locus. Assuming that the mode of inheritance is X-linked recessive (XLR), what percentage of the recessive alleles are carried in heterozyogtes versus homozyogtes? In XLR, all affected males carry an a allele. Also, q = 1/400 (p=399/400). The only carriers are females, and they are in the proportion 2pq= 2(399/400)(1/400)= 798/160,000. In the equivalent number of males, there are 400/160,000 males. So, there is essentially a 2:1 ratio of alleles heterozyote:homozygote (although, for X linkage, it is more appropriate to say hemizygous. (C) (5pts) Referring to the pedigree and assuming AR inheritance, what is the risk for III-1 being an affected individual? III-2? II-2s risk of being an unaffected carrier is 2/3 since both parents must be carriers. II-1s risk of being a carrier is 2pq=38/400 (see part A). So, the chance that III-1 is affected is 2/3 x 38/400 x ¼= 76/4800. III-2s chance of being an affected individual is the same. (D) (5pts) Referring to the pedigree and assuming XLR inheritance, what is the risk for III-1 being an affected individual? III-2? III-1’s mother would have to be a carrier (since dad is not affected). Her (II-1’s) risk of being a carrier is 798/160,000 (see part B). II-2 (dad) is not affected and has the A allele. So, the chance that III-1 is affected means that he would have to be male. The risk would be 798/160,000 x ½ x ½ = ~200/160,000 =~1/800.

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BANNER ID______________________ III-2’s mother (II-5) has a ½ chance of being a carrier. Father (II-6) is unaffected. So, the only way III-2 could be affected is to be male. Since mom has a ½ chance of being a carrier, the risk is ½ x ½ x ½ = 1/8.

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Question 2 (20 points) Dr. Peig is once again at the forefront of genetic studies in her extended family. A colleague of hers is studying pedigrees that are perplexing and brings them to her attention. One case involves a single locus, Floppy ear trait (F), which makes some comically large floppy ears on those unfortunate enough to carry a rare genetic mutation involving this gene. Her colleague is perplexed because, while the trait “runs in the families”, the genetics seems very non-Mendelian. (A) (6pts) The curious pedigrees show that a floppy-eared father crossed to a wild type mother always has wild-type offspring (case 1). However, a floppy eared mother crossed to a wild-type father always has a 50:50 mix of floppy-eared and wild-type offspring (case 2). Dr. Peig immediately suspects imprinting. Which kind of imprinting does she invoke? Draw out pedigrees in each case (you only need to show offspring classes) indicating the silencing status of the alleles with a X. Be sure to indicate the source of all alleles on chromsomes (M=maternal, P=paternal) (*Hint- you MUST have a loss of function mutation segregating (f)).

She invokes paternal imprinting. Since it is rare, we will assume that affected individuals are heterozygotes. The imprinting shows how affected fathers would NEVER have an affected offspring when mated to a WT female, and why 50% of affected mothers would be imprinted (actually, regardless of the genotype of the father… why?. (B) (2pts) For the above example, briefly explain how the loss of function mutation described above behaves as BOTH a recessive and a dominant loss-of-function depending on the cross. Assuming the allele f is recessive lof to F, case 1 makes it appear truly recessive, while case 2 makes it appear truly dominant.

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BANNER ID______________________ (C) (8pts) A different large pedigree follows two phenotypes, small eyes (E) and big teeth (T). Dr. Peig is no slouch when it comes to genetics, she sees something immediately that suggests that the genes involved in these two traits are both subject to imprinting, but in opposite directions (maternal vs. paternal). Moreover, she has some cytological/molecular evidence from a colleague that the two genes are tightly linked, and that the mutation causing BOTH of these conditions is the same small deletion that spans the two genes, and is therefore a loss of function for both genes. She explains to her colleague that the give-away is the following two pedigree scenarios; (case 1) big-eyed Dads crossed to wildtype moms have big-toothed and wild-type offspring (50:50), and big-toothed Moms crossed to wild-type dads have big-eyed and wild-type offspring (50:50). Again show the pedigrees for these two scenarios (4pts each), remembering that the two wild-type genes are tightly linked, are oppositely imprinted, and there is a deletion mutation segregating that confers the mutant phenotypes. Use the same labeling as part A. (Suggestion- start playing around with the cross scenarios and on scratch paper… this is not intuitive).

So, what works is the paternal imprinting of the E gene, and the maternal imprinting of the T gene. Interestingly, this is ONE mutation (the deletion) of two opposingly imprinted genes that CAUSES two different phenotypes depending on who passes it along (male or female). This is very similar to the story of a pair of diseases tightly linked to their respective genes in humans: Prader-Willi and Angelman syndromes.

(D) (4pts) Following the logic of your answers above, what will be the progeny ratios of a big-toothed Dad crossed to a wild-type Mom?

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BANNER ID______________________ Interestingly in this case, the phenotype is passed on as if it were dominant, with dads affected with the “maternally imprinted disease” having affected offspring of the same phenotype, since they will inherit a maternally imprinted gene from there moms. The trick in this problem, is always remembering that the mutation being passed on is the deletion and the inheritance is determined by the imprinting status of the WT genes.

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BANNER ID______________________ Question 3 (20pts) You are presented with the results of a dihybrid cross of Aa*Bb individuals (see below). * indicates you are not certain about linkage. From looking at the data, there seems to be consensus among the members of your lab that the results look “reasonable” for being close to a 9:3:3:1 ratio expected if the A and B loci are unlinked. You are a skeptic…. A_B_ 590 A_bb 160 aaB_ 160 aabb 90 (A) (12pts) Using the x-square test, decide whether your lab mates are justified in just assuming their hypothesis of unlinked genes. What is your null hypothesis? Determine whether these data are “reasonable”, or whether something else is going on. Null hypothesis= the data fit the dihybrid, unlinked 9:3:3:1 ratio. Therefore, expected #s are (out of 1000), 562.5 : 187.5 : 187.5 : 62.5 (using 9/16, 3/16, 1/16 ratios) Observed 590 160 160 90 Expected 562.5 187.5 187.5 62.5 O-E 27.5 -27.5 -27.5 27.5 2 (O-E) /E 1.34 4.03 4.03 12.1 So, chi-squared = 1.34 + 4.03 + 4.03 + 12.1 = 21.5, with three degrees of freedom. The P value is solidly below P < 0.005, so we reject the null hypothesis and accept that the data are VERY unlikely to occur with this much deviation by chance. (B) (8pts) Show how the data are consistent with the A and B loci being linked at a distance of 40 map units, and the parents of the dihybrid AaBb being AABB x aabb. It will certainly help to think of the starting configuration of chromosomes going into meiosis like this! A-------40mu---------B a-------40mu---------b If the parents were as given, then the heterozygote would be A B/ a b with linkage (P chromosomes AB and ab). The heterozygotes in the cross can make 4 kind of gametes (P)AB : (P)ab : (R)Ab : (R)aB in the ratios 3:3:2:2 (since A and B loci are 40 map units apart, 40% of gametes will be recombinant, divided (20% each) between the reciprocal classes. Everything else (60% non-recombinant) will be divided between the parental input chromosomes (30% each class). There are 16 ways two gametes can combine in this dihybrid cross: Gamete 1 x Gamete 2 = zygote (phenotype) AB (30%) AB (30%) AABB (wt) 9% AB (30%) ab (30%) AaBb (wt) 9%

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BANNER ID______________________ AB (30%) AB (30%) ab (30%) ab (30%) ab (30%) ab (30%) Ab (20%) Ab (20%) Ab (20%) Ab (20%) aB (20%) aB (20%) aB (20%) aB (20%)

Ab (20%) aB (20%) AB (30%) ab (30%) Ab (20%) aB (20%) AB (30%) ab (30%) Ab (20%) aB (20%) AB (30%) ab (30%) Ab (20%) aB (20%)

AABb (wt) AaBB (wt) AaBb (wt) aabb (ab) Aabb (b) aaBb (a) AABb (wt) Aabb (b) AAbb (b) AaBb (wt) AaBB (wt) aaBb (a) AaBb (wt) aaBB (a)

6% 6% 9% 9% 6% 6% 6% 6% 4% 4% 6% 6% 4% 4%

So, WT= 59%, a=16%, b= 16 and ab= 9% which if applied to 1000 progeny matches the ratios seen in part A very well…!

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Question 4 (20 points) A. You are working with a Sordaria fimicola point mutation called tan (t) that causes tan colored spores instead of the wild type, black. t is located 7 map units from the centromere of Sordaria chromosome one. You analyze 100 linear octads following meiosis of a t/T diploid. Draw an example of each type of octad (ascus) you are likely to observe and indicate the number of each type you expect to find. Indicate which octads are the result of segregation of t and T in the first meiotic division and which are the result of second division segregation. Show a calculation supporting your conclusion. Grading Rubric: 3 points for each type of octad 3 points for identifying 1st division segregation 3 points for 2nd division segregants 3 points for the calculation.

B. B is a different Sordaria pigment gene. The B allele confers Black color in spores, spores with the b allele are white. The following spore patterns (A – H) were among those observed in a large analysis of Sordaria asci. The number of asci exhibiting each pattern is indicated below each ascus type. Note: only a subset of possible ascus patterns is shown. A

B

C

D

E

F

12 305 268 508 310 499

G

H

4

250

(2 pts) List the letter(s) of any asci resulting from gene conversion. A,G (4 points)

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BANNER ID______________________ (2 pts) List the letter(s) of any asci resulting from recombination between the spore color marker and the centromere. B,C,E,H (2 points)

(2 pts) Which gene is closer to its centromere, B or T (previous question). Briefly support your answer using data from Question 4 and 5. T is closer. B appears to be so distal from the centromere that it does not show genetic linkage, T is 7 map units away from the centromere (2 points).

Question 5 (20 points) The b allele (previous question) is a T to C transition mutation. Provide a series of diagrams (3-4 will be required) that describe how the observed gene conversion occurs at the molecular level. Your diagrams should be consistent with current models for the molecular mechanisms of meiosis, should include the mutant/wild type nucleotides, and explain how information from one chromatid is transferred to the other during meiosis. You only need to show one possible way this could occur, but your answer should indicate how heteroduplex DNA is formed and whether resolution of meiotic intermediates would result in recombination or not. b B

Use these conventions in your diagrams: new synthesis (wavy line): 3’ 5’ region of inner b chromatid (thin line)

5’ 3

3’ 5’

3’ 5

5’ region of inner B chromatid 3’ (thick line) Indicate important nucleotide on each strand

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Grading Rubric Diagrams showing: DSB 5’ to 3’ degradation initiating at DSB Mutant and wild-type nucleotides at each position D-loop A-G Heteroduplex Resolution of Holliday junction Does/Does not result in recombination

points 4 4 4 4 4 4 2

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