Name. FALL 2005 Exam 3

Transcription

1 GENETICS 603 Name FALL 2005 Exam 3 1) Ten rii mutants of phage T2 were mixed pairwise in tubes of E. coli B and after bacterial lysis, the progeny phage were plated on lawns of E. coli K, with the results shown in the table below. (-) indicates no plaques, r indicates rare plaques (around I/10 7 progeny page plated), + indicates more numerous plaques (> 1 in 10 5 /phage plated and L means complete lysis in complementation tests. Mixed with A cistron tester Mixed with B tester #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #1 - L #2 - L #3 - L #4 - L r #5 - L r #6 - L r #7 - L r + + #8 - L r + #9 L - r # A) Which mutations are in the A cistron? 1-8 & 10 B? 9 & 10 B) Which mutant phage are point mutations and which are deletions? Del: 1-3 & 10; point: 4-9 (all revert as shown in selfs ) C) Make a map that is compatible with these data. A del-1 del-2 del-3 del-10 2) a) What gametic frequencies are expected in a trisomic Aaa plant? 9 B 1A : 1 aa :2Aa::2 a b) if an extra chromosome can passes through eggs and not pollen, predict the genotypic and phenotypic ratios in the progeny following self pollination of the Aaa plant. eggs pollen 1/6 A 1/6 aa 1/3 Aa 1/3 a 1/3 A 1/18 AA 1/18 Aaa 2/18 AAa 2/18 Aa 2/3a 2/18 Aa 2/18 aaa 4/18 Aaa 4/18 aa Thus 6/18 0r 1/3 will be recessive aa or aa

2 3) A man has a flock of sheep where 3/4 th of the rams have horns but 3/4 th of the ewes do not. Not knowing horns is a sex-influenced trait, he decided he would like to have only hornless ewes and horned rams, so sold all the others. a) What are the frequencies of the H and H alleles in his original flock? Since horns is dominant in rams and recessive in ewes, the overall heard must be: 25% H H : 50H H% : 25 HH% and the frequency of H = 0.5and of H = 0.5 (using H H for homozygous horned males) b) What are the allele frequencies in the reconstituted flock? Removing 25 HH males and 25 H H females will not change the frequency of alleles in the flock. ie 0.5 H : 0.5 H c) What will the phenotypic ratios be in the progeny in the next generation, assuming random mating? males now are 25 H H and 50 H H and females are 50 H H and 25 HH Female Gametes Male gametes 1/3 H 2/3 H 2/3 H 2/9 H H 4/9 H H 1/3 H 1/9 H H 2/9 HH Overall 7/9 of the rams will have horns 2/9 will not Overall 7/9 of the ewes will be hormless and 2/9 will have horns 4) What is the difference between homologs and homoeologs? Use examples from an appropriate organism to aid in your definition. Homologs refers to homologous chromosomes (or genes) as would be found on the two A1 chromosomes of wheat. Homeologs would be the B chromosome (gene) equivalents tothose in the A or D genomes 5) There has been considerable work and some success in developing a transposon called Sleeping Beauty (it was awakened from a fish genome after long being inactive) as a vector for use in cancer gene detection (mouse) and for human gene therapy. Predict the properties that make these applications possible and tell how it might be used in gene therapy. inverted repeat ends for recognition by integrase ability to code for, or respond to co-expressed transposase and integrase outward promoters could be useful for increasing transcription of adjacent genes enough capacity to carry selectable marker and or functional gene ability to regulate the movement frequency not present in human genome etc?

3 6) When my wife was a graduate student working with Neurospora, she collected the following data from a cross between an adenine auxotroph (ad3), with normal colony morphology (pk+) and a peak mutant (ad3+, pk): (The genotypes of each spore pair in an ascus are shown across the rows of the table) # of asci with this pattern Ascus type (PD, NPD, TT) ad3 pk+ ad3 pk+ ad3+ pk ad3+ pk 57 PD ad3 pk+ ad3 pk ad3+ pk+ ad3+ pk 18 TT ad3 pk ad3 pk+ ad3+ pk ad3+ pk+ 13 TT ad3 pk ad3+ pk ad3+ pk+ ad3 pk+ 6 TT ad3 pk+ ad3+ pk ad3 pk+ ad3+ pk 4 PD ad3 pk ad3 pk ad3+ pk+ ad3+ pk+ 2 NPD a) Fill in the ascus type column with respect to the two genes. b) Before making any calculations, are the two genes on the same chromosome? Tell how you know. Many more PDs than NPDs ( should be equal if on different chromosomes) c) How far is each gene from it s centromere? 10 show 2 nd div segr for A/a, so 5 c 35 show second div for pk/pk+, so 17.5 cm d) Make a map showing the most likely gene & centromere arrangement(s). All the spores in the NPD and half those in the TT are recombinant for ad pk, or 20.5 cm Thus ad3 5cM--* cM---pk, where * is the centromere 7) The average days to flowering in a population of sunflowers was 100 days. The h 2 value for flowering time was estimated to be 0.2. In an effort to decrease the generation time, a breeder selected only early flowering plants (average = 90 days). What would the average days-to flowering in the progeny of the selected plants? Show the formulas used. h 2 = R/S = M*-M/M s -M; h 2 is given as 0.2 and M is 100 and M s is 90. Solving for M* gives 98 days

4 8) Calculate F for individual A in the following pedigree: A F = ( ) 7 + (1/2) 7 = 1/64 = Would individual A be at siginificantly (> 5%) increased risk for a recessive trait that occurs in 9 % of the infants born in a random mating population? Show your work. The easiest way to do this is to see how much inbreeding increases the frequency versus random mating (q 2 ). Inbreeding adds pqf, ie half the heterozygosity lost from the 2pq heterozygotes, to each homozygous class. If q 2 is 9/100, pqf will be 0.7 X 0.3 X or , less than 5% 0f 0.09 What if the frequency of homozygous recessives was 9 in a million? Here q is 0.003, p is about 1, so we can say pqf is about X or 4.7 X this is way bigger than 9 x ) Many traits like schizophrenia and cleft palate in humans are thought to be polygenic, even though the trait may be all or none in phenotypic expression. In these cases, there seems to be a threshold: when the number of contributing genes/alleles is greater than the threshold, the trait is present. Typically, H 2 for these traits is near Several years ago a study reported in Nature from families in Iceland said their pedigrees suggested a single dominant gene on chromosome 5 was the cause of schizophrenia. Within a few weeks another group from Italy suggested the first group was in error; their data suggested a gene on chromosome 11 was responsible. As a geneticist, what is your analysis of these claims? With the polygenic model, it is possible that in some families one gene may be segregationg that causes recipients of a contributing allele to exceed the treshold, while in another family, another gene could be responsible. For example if seven recessives are essential for the phenotype, one family could be aa,bb, cc and segregating for D or d, while another could be aa, bb, Cc dd etc. In other words, we would expect to see families where different genes tip the scales.

5 10. a) Could any of the following 4 plant types be maintained as pure lines by selfing? [Brackets indicate the cytotype.] Check if yes: X Rf/Rf [s]; rf/rf [s] X rf/rf [F] Rf/rf [F] b) Use a stylized map to describe a typical mammalian mitochondrial genome, with respect to structure and gene content. (Specific gene locations are not important!) The map is a circle: Features include rrna, an origin or D-loop, left and right promoters, genes that encode subunits of ATPase and other energy complexes, trna units in clusters that separate encoded proeins 11. Maize plants homozygous for a reciprocal translocation between chromosomes one and two are crossed to a plant with the normal chromosome arrangement but homozygous for a recessive virescent (vir) mutation on chromosome 2. The F1 is then backcrossed to the vir/vir parent. A) What fraction of the F1 gametes are expected to be fertile? 1/2 B) Diagram synapsis in the hybrid, using visual keys to indicate chromosomes 1 and 2. chrom 2 vir+ vir chrom 1 C) Results of the cross were 231 fully fertile, virescent: 287 partial sterile, normal: 39 partial sterile, virescent: 31 fully fertile, normal. Account for these data. Only alternate segregation results in functional gametes: 2 and 3 together in the backcross progeny are fertile virescent, 1 and 4 together give partial sterile, vir+.. The rare cases are the result of a crossover showing that the vir gene locus is about 12 cm from the breakpoint.

6 12. Explain why pericentric inversions might play a greater role in species isolation than paracentric inversions. Crossovers in pericentric inversions cause duplications and deletion gametes, but unlike paracentric they to not create bridges. In females of most species, the prospective gametes with bridges end up in polar bodies rather than the egg, meaning oly males are semisterile. 13. How can the number of copies of one specific transposon increase in a genome? How about retrotransposons? Transposopons: transpositon of one replicated copy to a region yet to replicate will add a copy Retrotransposons encode reverse transcriptase that can make multiple copies of themselves which may then integrate into host DNA 14. Suppose that in a random mating population, selection could suddenly be imposed against heterozygotes. A. What will happen to gene frequencies and the population structure? (Hint: try different examples where p and q differ by a good deal.) Removing heterozygotes will decrease the frequency of the rarer allele and increase the frequency of the predominant allele. The population will have more homozygotes and fewer heterozygotes exs if p= 0.9 HW will give : 81 AA : 18 As and 1 aa; removing the heterozygotes leaves only 2 a alleles for the next generation. conversely if the population was 1 AA : 18 Aa : 81 aa; the relative frequency of A alleles would decline to 2 in 164 and a would become 162 of 164. B) Can you provide examples where selection against heterozygotes would be expected? invesrsions and translocation heterozygotes make good examples as show reduced fertility 15. If 1% of a population in Hardy-Weinberg equilibrium is homozygous recessive, what fraction of the recessive alleles are found in heterozygotes? What if only 1/10 6 is homozygous recessive? The square root of 1 in 100 is 1 in 10, or 0.1. In a population of 100 we would expect to see 81 AA : 18 Aa : 1 aa. Thus 18 of the 20 total a alleles (90%) are in the heterozygotes. For the second part q = and since p is so close to one we can just say that 2pq is essentially From a population of 1 million, there would be 2000 a alleles in heterozygotes and 2 in the homozygotes. Thus, 2000 of 2002 or 99.9% of the recessive alleles are in the heterozygotes.

7 16. In elephant seals, one male often mates with all the females in the herd. Is it true that no matter how many females are in the herd, the effective population (Ne) will never reach the Ne of a random mating population where there are only 2 males and 2 females? Show the rationale for your answer. Ne = 4 Nm *Nf dovoded by (Nm + Nf) For 2 males and 2 females, Ne is 4 x 2 x 2 divided by or 4. Where Nm is one the formula becomes 4 x Nf divided by (1+ Nf) so will always be less than 4. Thus it is true. 17. Female Drosophila homozygous or heterozygous for esc + (extra sex combs) produce progeny with normal larva no matter the genotype of the male. However, in larva that develop from esc - /esc - females all segments appear to develop into A8 segments. What kind of a gene is esc in terms of its role in development? This must reflect something deposited into the egg by the female that affects development all the way to the stage where segments are expected to develop different body parts. It is thus an example of a maternal effects gene. It is not a cytoplasmically inherited trait, or we would not talk about homozygous and heterozygous females!