Del C. Aaa selfed gives 6 of 18 = 1/3 recessive 4 aa and 2 aaa AAa selfed gives 1/18 aa

Transcription

1 GENE 603 Exam 3, Nov 30, 2018 NAME 1. The following information relates to rii cistron A mutants. Non-reverting mutants A, B and C when crossed gave the following A X B : recombinants A X C: no recombinants found, B X C: no recombinants. Revertable mutants 1-5 gave the pattern shown in this table when ccrossed to mutants A, B and C, where+ and indicates whether or not recombinants were found above the frequency reversion A B C a) What is the significance of reverting vs. non-reverting mutants? Point mutants are single base changes (SNPs) and can revert, deletions do not b) Make a map consistent with the data from all crosses. Pt3 Pt5 Pt 1 Pt4 Pt2 Full cistron (gene) Del A Del B Del C c) How were revertants and recombinants detected? Crosses were made on a permissive strain where mutants can reproduce (E coli B) and progeny tested on K where only wild type r+ can grow (E. coli K) 2. Consider the situation in tomato where each trisomic has a visible phenotype and that disomic eggs can function but pollen with the extra chromosome is inviable. Assuming a single copy of gene A is dominant, how often would recessive progeny appear in the selfed progeny of: 1) Aaa and 2) AAa plants. Aaa selfed gives 6 of 18 = 1/3 recessive 4 aa and 2 aaa AAa selfed gives 1/18 aa How do these ratios compare if gene A is not on the trisomic chromosome and is Aa In that case it would be As X Aa and ¼ progeny would be aa 3. In 2014 a paper in AJHG reported that a common inversion in human 16p11 provides considerable protection against a combination of obesity and asthma, even in heterozygous individuals. The frequency of the inversion was measured at 10 % in an African population and 50% in a Latvian sample, with many others falling in between. In all cases but one, (neither of the two listed) the populations were found to fit Hardy Weinberg equilibrium. a) Is it a pericentric or paracentric inversion? How do you know? Paracentric: it is given as being in the p or short arm

2 b) List 3 plausible ways to detect the presence of the inversion. 1 Inversion loop visible in meiosis 1 2 altered map of genes in and flanking the inversion 3. PCR using primers that flank the breakpoints 4. reduced fertility, especially in males 5. Sequencing could but may be a problem if flanked by repeats or sample from a heterozygote. b) Predict the genotypic frequencies for the inversion in the two populations listed African p= 0.9 and q = 0.1 so 81 NN : 18 NI : 1 II where I is inversion and N is normal Latvian 25 NN 50 NI : 25 II c) A second observation was that in some individuals with the inversion there were missing genes or extra copies of other genes. Does this fit with effects of inversions? Explain: YES- A crossover in the inversion loop of an individual heterozygous inversion leads to gene duplications and deficiencies. d) It was also found that when the inversion is present there is significantly higher or lower expression of a few of the 30 or so genes included in the inversion. Suggest an explanation that could be tested. 1) inverted sequence may place genes under wrong promoters 2) Extra or missing copies of genes may alter level of expression. e) Is there evidence for selection relative to the presence of the inversion? How can the highly varied frequencies be explained? Only in the sense that it is said to protect against obesity and asthma, but we have no data saying reproduction is altered t 4. Shown below are 4 idealized gene maps of chromosome 4R (arms designated L or R) from 4 different species of Drosophila, one of which is thought to be the precursor of the others. A) Which one is the original and what happened in each change. 1. AB HG FE DC RQ MN OP LK JI ST UV 2. AB CD EF GH RQ MN OP LK JI ST UV 3. AB CD EF GH RQ MN OP LK TS IJ UV 4. AB CD EF GH IJ KL OP NM QR ST UV Easiest of # 2 was original and the other 3 came from inversions as indicated: I could be one of the others gave 2 and then it gave the others.

3 b) Show synapsis for these chromosomes for a hybrid of species 1 and 4. This requires a three inversion loop figure that would be simpler to see in 3D 6. What is the coefficient of inbreeding for the individual at the bottom of this pathway? 4 X (I/2) 5 which is 1/8 0r b) What would it be if the top left individual had the same pedigree pattern? Here the first contribution would be (½)5 (1 + 1/8) + 3 X (I/2) 5 or Assume corn borers have a HW population where the frequency of a dominant gene (B) for BT resistance is present at a frequency of The next year when BT corn is planted, only 10% of the bb individuals are able to reproduce. What will the frequency of B and b be the next generation? (Tip: I would find it easier to think of a starting population of 10,000 boll worms.) Here f(b) = 0.02 so BB is or 4 of the 10,000 fb) = 0.98 so f(bb = 2(0.02)( 0.98) = 3092 of the 10,000 f(bb) = 0.98 X 0.98 = 9604 and if 10% reproduce there will be 960 in the next generation, assuming that each genotype that reproduces has the same number of progeny. The new frequency of B among the 1,356 individuals is thus and of b is Inbred sunflower lines that matured on an average of 90 and 110 days after planting at a field location in N. Dakota were crossed. The F1 average maturation time was 100 days. The variance for each of these populations was 3 days. In the F2 the average was still 100 but the variance was 9. a) Calculate heritability based on these figures. Is it H 2 or h 2? Ve = 3 from F1 where all plants have same phenotype; VT = 9 so Vg = 6 and H 2 = 2/

4 b) In a follow up experiment F2 plants that matured at 92 days on average were selfpollinated and the average in their progeny was 95 days. Calculate heritability for this observation. Is it H 2 or h 2? Here h 2 = the response to selection = 5 divided by the selection differential of 8 so is 5/8 c) Is there any evidence for some of the QTLs being dominant? How do you know? None evident since the F1 was exactly betewen the parents. 9. Three types of chromosomal aberrations are considered to play a role in evolution. What are they and what is the basis for the role of each? What factor, if any, contributes to that role in each case? 1. Duplications provide new sources of genetic material and allow one copy to change while the original function is maintained 2. A crossover in an inversions causes semi-sterility in individuals heterozygotes, starting them on a path to being subspecies 3. Adjacent segregation of chromosomes translocation heterozygotes which is likely to occur 50% of the time also creates a separation toward subspecies. 10. Removal of a transposon from one position when it moves to another leaves behind a footprint. What is this footprint? Would you expect different excision events to leave the same footprint? Why or why not? When a tranposase inserts a hairpinned Tn into a site, it creates a staggered gap in the DNA. Repair at either end creates the same flanking sequence at each end, often 8 bp long. If the transposase becomes active and correctly use the terminal repeats to cut out and remove the Tn, end joining will create a duplicated sequence. They would only be expected to conation the same sequence if the transposse recognized a specfic sequence which is rare. Also, often the excieion can lead to repaired chromosomes where one or several bases of the footprint are missing. 11. Defects in the lethal-7 (let7) gene in C. elegans lead to an increased number of identical hypodermal cells being formed between the larval 4 and adult stages of development. (This has been established by the use of a temperature sensitive let7 mutant). The Let-7 functional product has been shown to be a 21 base RNA molecule that is complementary to the 3 untranslated region of several lineage gene mrnas. a) What life form is C. elegans? It's a nematode b) Propose a model to explain these results Let-7 (normal) RNA binds to the 3 end of some mrnas; This will induce RNAI to break down those mrnas or at least prevent their translation since the tail is needed to initiate TL. Thus Let-7 regulates other genes to give normal development.

5 12. Humans have 39 HOX genes found in clusters on 4 chromosomes. As in Drosophila, the genes in each cluster are individually expressed in a head to toe pattern along the axis of development. A defect in a single HOX gene typically causes a relatively minor abnormality. Examples include HOXA1 defects associated with hearing loss, inner ear abnormalities, delayed motor milestones and internal carotid artery malformation, while HOXD13 defects cause fused fingers as a common phenotype. By contrast, inheriting two defective HOX genes causes drastic abnormalities, generally incompatible with life. Propose an explanation of why these observations are true. Each Hox gene product is expressed at different locations along the head to tail axis, setting up and overlapping gradient of Transcription Factors that interact with various promoter elements to establish the correct body patters. Lacking 1 of the 39 has a relatively small effect on the overall gradient but when 2 are missing multiple body segments are likely to be deranged 13. RAG-1 that functions as the catalytic component of rearrangements that occur during immunoglobin and T-cell development has been compared to transposases so far as function, but not in amino acid sequence. a) What do RAG genes do? They remove the DNA in maturing B and T cells so that the last variable selected is connect to one joint region etc. This allows B cells to make only one type of antibody and T cell to make one specific receptor. b) Predict the consequences of defects in RAG-1 Complete immunodeficiency b) What aspect of their function is like that of transposases? They recognize DNA paired in an inverted repeat and cut it out. (The same repeats are found before each V and after each J etc. 14. Cases of meiotic drive, where alleles in a heterozygote do not segregate in a 1 to 1 fashion have been described in a number of organisms. In maize heterozygous for Ab10 (an abnormal chromosome 10 that has a knob with a centromere-like element) and a normal chromosome 10, 3/4ths of the eggs are Ab10. The pollen still has a 1:1 segregation ratio. a) Based on just this information, what would you expect to find with respect to Ab10 and normal chromosome 10 after a long evolutionary period in a random mating species like maize That Ab10 would approach fixation, and all plants would be homozygous b) Population surveys have shown that Ab10 is present in 18 22% of landraces of maize, and within those populations the average frequency of individuals carrying Ab10 is 30%. Assuming this differs from your answer to part a, propose an explanation. There must be detrimental genes very near the centromere or pseudo-centrome that if homozygous significantly lower fitness. If they were scattered elsewhere, crossing over would eliminate dis-equilibrium. Another possibility is deleterious genes are in an inversion is associated with the Ab10 so that XOs do not give recombionants.