GENETICS 603 Exam 3, Dec 2, 2011

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1 GENETICS 603 Exam 3, Dec 2, 2011 Name I. Chromosome IV in Drosophila is a very small chromosome with few genes. Fruitflies that are trisomic for chromosome IV have no apparent phenotypic abnormalities, and they retain their fertility. Among the genes on chromosome IV is one, ey, which if fully recessive produces the eyeless phenotype. A male that is trisomic for chromosome IV and has the genotype ++ey is crossed to a diploid eyeless female with the genotype ey/ey. a. Assuming random segregation of chromosomes takes place during spermatogenesis and that (1) all sperm are viable, or (2) only 1N sperm are viable, what sperm genotypes are expected and in what proportions? : 2 +, ey : 1 +,+ : 1 ey : 1ey b. If these sperm are united with eggs from the eyeless-female, what is the expected ratio of eyeless to normal-eyed flies among the progeny in each case? : 1 eyeless : 1 eyeless II. The five chromosomes illustrated below (a through e) are evolutionarily related. 1) Assuming that (a) is the most recent version of this chromosome propose a lineage that would allow for the stepwise evolution of the modern sequence. For each step, describe the change that occurred at each step during evolution. a) AD CGFGFEH c d b e a b) AD CEFGH c to d is a pericentric inversion, d to b is a deletion of gene B, c) ABC DEFGH b to e is a paracentric inversion, e to a is a duplication of GF d) AD CBEFGH e) AD CGFEH 2) Which step was most likely to have an associated phenotypic effect? D to b, the deletion 3) Show synaptic pairing between an individual with chromosomes c and d. show a pericentric inversion loop at the four stran stage with all alleles paired

2 III. Calculate the coefficient of inbreeding for animals E and H in the cattle pedigree shown below: (A and B are not related) F(E) = (½) 3 (CAD) + (½) 3 (CBD) = 1/4 F(H) = (½) 2 X (1 + F(E)) =1/4 (1.25) =5/16

3 IV. A population is sampled to determine the distribution of alleles for a gene that encodes a protein that functions as a dimer. In electrophoresis, one of the alleles gives a fast migrating polypeptide F while the other is slow or S. Among the 500 randomly selected individuals, isozyme analysis revealed: 172 FF : 220 FS :98 SS. a. What are the genotype frequencies for this population? f(ff) = ; f(fs) = ; f(ss) = 0.2 b. What are the allele frequencies for this population? f(f) = ; f(s) = c. Is this population in Hardy- Weinberg equilibrium? The χ 2 values for critical values at the 5% probability level are for 1 df, for 2 df and for 3 df. Expected for FF = 162.3, for FS = 239.4, for SS = 88.3 The Chi- squared value with 1 df (number of phenotypes number of alleles) = 3.22 which is below the cutoff for significance at the 5% level. d. Whether or not the deviations are significant, what could account for the non- perfect distribution seen? Since both homozygotes were high and the heterozygotes low, inbreeding might be at play. Alternatives could include selection for homozygotes, or sampling errors as in drift; mutation and migration would seem to be less likely e. What is the inbreeding coefficient in the population? 1- obs/exp heterozygous = 0.08 f. Predict the genotype frequencies in the next generation assuming random mating FF : 0,4886 FS : SS g If 80% of the SS embryos die before reproducing, what will the allele and genotype frequencies be in the next generation. (if from original), there would be 172 FF : 220 FS and 20 SS with a f(f) of and of S of : Calculate the p 2 + 2pq + q 2 values using these values giving FF : FS : SS If taken from part g, the values would be p = and q = 0.328

4 V. A 1971 paper in Nature included the following data concerning coat color in feral cats in the city of York, England. Males: 258 black : 76 orange. Females: 258 black : 91 tortoise shell : 15 orange. Recalling that these colors derive from an X linked gene with two alleles, calculate the gene frequencies as seen among a) females; b) males. Is there any evidence of non-random mating or selection in this cat population. From males the frequency of the X-linked B allele = and of the b allele, From females, the values come out f(b) = and f(b) = Using all X s in the population, the f(b) is and of b is While a Chi squared test based on just the female data is not significant, testing the expected frequency of black and orange males from these data is significant. It seem the black allele is favored in females and orange in males if selection is the answer. (authors suggested that) My guess is that the population likely started from a fairly small number of males and females that differed in allele frequency by chance. With sex linked gene the ratio fluctuates in males and females gradually approaching an equilibrium which hasn t been reached in this population VI. The following data have been adapted (for easier math) from a study of Lima beans being tested for suitability for crop production in Nigeria. Pure breeding variety NSWP83 averaged 1,000 kg/ha with a variance of 100 while NSWP46 averaged 400 kg/ha with a variance of 50. The F1 hybrid averaged 800 with a variance of 75 and while the F2 also averaged 800, the variance was 250. a) Calculate heritability based on these data. H 2 = Vg/Vt = (250 75)/250 = 0.7 When F2 individuals with projected averages of 900 kg/ha were self pollinated, the average in the F3 was 850 kg/ha with a variance of 250. B) Calculate heritability based on this added information. h 2 = R/S = ( )/( ) = 0.5 c) Why aren t the two values the same? H 2 includes dominance, epistasis, additive effects etc while h 2 only measures the effects of additive genes d) Can you get an estimate of the number of genes involved? What assumption is needed? About the only way is to assume the 2 parents are at the etremes. Then the difference between the parents divided by 8Vg (from above) says 257 or so genes involved

5 VII. Smith- Lemli- Opitz Syndrome (SLOS) in humans is a relatively rare recessive lethal defect in humans caused by mutations in the gene encoding Δ- sterol reductase. A Canadian study estimated the birth frequency as about 1 in 25,000 while a DNA based sequence analysis of alleles among 1,500 randomly selected individuals suggested 1 in 30 individuals was a carrier of a mutant allele. These two measures do not seem to be in agreement. A) Why not (justify your answer) and what could account for the differences? It would not be surprising if a) many SLOS conceptions abort spontaneously too early to be classified and b) that some combinations of alleles detected by sequencing show a degree of inter- allelic complementation so that compound heterozygotes may not be lethal. B) So far, 35 different mutant alleles have been identified by DNA sequence analysis of SLOS patients. In Poland a base substitution mutation (W151X) is most common while in Britain, a splice site mutation (IVS18-1G- C) is most common. What is the most likely explanation for the difference? Founder effect; different mutation occurred in the two populations and became established by chance VIII. An individual is heterozygous for a reciprocal translocation such that one of the TL chromosomes has genes A- - B C 4 5 and the other is D E. Would a crossover between genes B and C affect the overall rate of sterility predicted for this individual? Justify your answer by showing the meiotic configuration and predicted effect of crossover on gametes generated. (Showing just the XO strands of the tetrad formed, the others will segregate alt OK, etc) A B C A B C D D E E XO between D and C will give A BCDE AND OR ABC45 AND 123DE gametes with adjacent segregation and ABCDE and 123DE or ABC45 AND gametes with alternate segregation. Assuming alternate and adjacent I segregation occur equally, the frequency of functional and non- functional gametes will remain at 50%

6 IX. Cytoplasmic male sterility [CMS] in plants has been exploited to produce hybrid seeds. CMScausing changes in the mitochondrial genome can be suppressed by dominant Rf alleles in the nuclear genome that are specific for each type of CMS. The following stocks are available. [CMS1] rf1/rf1 rf2/rf2, [CMS2] rf1/rf1 Rf2/Rf2, [F] Rf1/Rf1, Rf2,Rf2 Describe a cross, indicating which stock is male and female that would allow you to create a hybrid for use in seed production without the need to detassel the female line. What cytotypes, genotypes and phenotypes do you expect in the F1? If there are F1 plants that are male fertile, what genotypes and phenotypes do you expect in the F2? (not a great question as we could just take the first line as the inbred and pollinate with pollen from the fertile cytoplasm line 3. However most did as intended and set up a cross such as that and then selected out [CMS1] lines which either carried (for pollinator) or did not carry the Rf1 restorer genes (for the female only line) to use for hybrid production. X. List features that are common to and different among transposons, retrotransposons and retroviruses. Common features: Differences: LTRs with inverted repeats for insertion Transposase/Integrase encoded and rqquired for insertion Target sites random and staggered cuts leave footprints on removal Silenced in genome via methylation TNs move conservatively as DNA, other have RNA intermediate and multiple copies can be made and expand numbers rapidly in genome Retroviruses have a coat and can be infective along with the reverse transtriptase and integrative functions that remain in retroposons. XI. A female from Drosophila strain A is crossed to a strain B male. The F1 progeny show reduced fertility. How could you determine if the sterility resulted from a chromosomal aberration, hybrid dysgenesis or a mtdna defect? If P elements are involved, the male line should have P and the female should not: easiest test is via PCR using P element specific primers. The reciprocal cross should not show reduced fertility (same for mtdna though) while a cross of the male to a known M strain should. For mtdna defect, the reciprocal cross (B females by A males) should again give normal fertility in the progeny either line has P elements. Confirmation of maternal inheritance via a mt DNA defect could require sequence analysis. For chromosomal aberrations, the easiest way in Drosophila is to look at the polyten chromsomes in salivary glands of the maggots: they are paired as in meiosis and polytene, so that banding patterns can be seen directly, including inversion loops etc.

7 XII. Nanos and bicoid are both maternal effect genes important in development of Drosophila melanogaster. What do they do and how do they do it? Both are maternal effect genes that come from mrnas in the egg from the female. Bicoid messages have 3 UTR signals that attach to proteins that then attach to dynein to propel the mrna along microtubules to the anterior pole of the egg. When expressed the localization of the mrna creates a protein gradient of bicoid protein which acts as a positive and negative transcription factor to regulate expression of many other genes involved in establishing body segments. It also can affect translation of some other mrnas. Nanos is concentrated at the posterior pole, perhaps by association/protection with the primordial germ cells found there; nanos messages in the rest of the egg are degraded. The polar localization of the mrna establishes a gradient opposite that of bicoid. However, nanos protein is not a transcription factor; it only functions to prevent expression of some other mrnas.