Unformatted text preview: Name _______________________________________ PID _____________________ BICD100 Final Exam 3/17/05
1.) In a field of transgenic corn expressing a certain Bacillus thuringiensis toxin, European Corn Borers are killed unless they have become resistant to the toxin. When resistance alleles arise, they are rapidly amplified by selection. You are a population geneticist making recommendations to the EPA regarding the effectiveness of planting an area of non transgenic corn around B.t. corn fields (a so-called "non B.t. corn refuge") as a strategy for slowing the growth of the resistant insect population. In a particular test field, a new dominant resistance allele arose in the European corn borer (call it R). Shortly afterwards, sampling of the population showed that 2% of the insects were resistant to B.t. (assume that all of these were heterozygous Rr) and the remaining 98% were B.t. sensitive (rr). a. At the time this sample was taken, what was the frequency of R alleles?_________[call this p = f(A)] (4 pts) r alleles?_________[call this q = f(a)] b. If this were a non B.t. cornfield so there was no selection for resistant insects, what would be the frequency of resistant insects in the next generation? (Assume that mating is random, there is no migration in or out of this population, and no new resistance alleles arising through mutation 6 pts). c. In the B.t. cornfield where all sensitive insects die before reproducing, what will be the new frequency of resistant insects after one generation (assume mating is random and no new R alleles arise through mutation 10 pts)? d. If no new insects migrated into the B.t. cornfield, what would be the frequency of resistant insects after another round of reproduction (as before, assume mating is random and no new R alleles arise through mutation 8 pts)? e. Due to the presence of a non B.t. corn refuge around the field, the population in c (that is, the population as it stands after one generation of selection) is actually joined by an equal number of B.t. sensitive immigrants before reproducing again. What will be the actual frequency of resistant insects in the next generation? (12 pts) 1 2.) Time to flowering is an agronomically important trait in corn, as for other crop plants. In Northern climates with short growing seasons, a short time to flowering is a useful trait. To identify genetic factors controlling time to flowering in corn, a corn geneticist crossed Gaspay Flint (an inbred line with a very short time to flowering of 35 days) to another inbred line called N28 (with a long time to flowering of 52 days). In the F1 hybrids, time to flowering was 43 days (standard deviation 3 days). F1 s were crossed together to produce an F2 generation whose mean time to flowering was 44 days (standard deviation 8 days). a. What value for broad sense heritability of time to flowering in corn is indicated by this data? (10 pts) F2 progeny were scored individually for time to flowering and also analyzed to determine their genotypes for molecular marker loci distributed throughout the genome. In the graph below, r values are plotted on the y axis for several marker loci on chromosome 8, whose map locations in cM are indicated on the x axis. In this analysis, r values greater than 0.05 indicate a significant correlation between marker genotype and time to flowering. b. Which of the following can be concluded from the data shown? (Circle all that apply 12 pts) i. There is a minimum of one gene on chromosome 8 controlling time to flowering in corn. ii. There are a minimum of two genes on chromosome 8 controlling time to flowering in corn. iii. There are a minimum of five genes on chromosome 8 controlling time to flowering in corn. iv. 20% of the variance in time to flowering is accounted for by QTLs on chromosome 8. v. 20.8% of the variance in time to flowering is accounted for by QTLs on chromosome 8. vi. The corn homeobox gene ZmHox1a (located at 82 cM on chromosome 8 and used as a marker in this analysis) has a modest but significant effect on time to flowering in corn. 2 c. Sequencing of BAC clones spanning the interval between UMC89a and DGG9 showed that one of the genes in this region encodes a transcription factor homologous to one that has been previously identified as a regulator of time to flowering in the model species Arabidopsis thaliana. To determine whether this gene might be controlling time to flowering in corn, the Gaspay Flint allele for this gene (coding region + promoter/regulatory region) was introduced into the N28 inbred line using the particle bombardment method described in class. Number the steps below to indicate the order in which they must be carried out. Label any step that is not needed with an "N" (12 pts). _____ Grow cells in the presence of a selective agent (e.g. kanamycin or herbicide) to favor the growth of transgenic cells. ______Splice the gene of interest into the T-DNA region of a Ti plasmid that also contains a selectable marker gene such as kanr. _____ Bombard maize cells with metal particles coated with a plasmid carrying the transcription factor gene and a selectable marker gene (eg., herbicider). _____ Splice two copies of the gene of interest in an inverted repeat configuration into the T-DNA region of a Ti plasmid that also contains a selectable marker such as kanr. _____Regenerate plants from transgenic cells. ____ Infect corn cells with Agrobacterium containing a Ti plasmid carrying the transcription factor gene and a selectable marker gene (eg., kanr). d. When introduced into N28, the Gaspay Flint allele of the transcription factor gene caused about a 15% decrease in time to flowering. Sequencing of Gaspay Flint and N28 alleles of the transcription factor gene showed no differences that would change the amino acid sequence of the gene product, but revealed the presence of about 200 bp in the promoter/regulatory region of the Gaspay Flint allele not present in the N28 allele. Circle each of the following statements that are consistent with these results (9 pts). i. Shorter time to flowering in Gaspay Flint compared to N28 is due partly to a gain of function mutation in the promoter region of the Gaspay Flint transcription factor gene allele. ii. Longer time to flowering in N28 compared to Gaspay Flint is due partly to a loss of function mutation in the promoter region of the N28 transcription factor gene allele. iii. The difference in flowering time between Gaspay Flint and N28 cannot be accounted for by differences in their alleles for the transcription factor gene, since the proteins encoded by these two alleles are identical. 3 3.) Insects of the order Hymenoptera (which includes bees and wasps) have an unusual mode of sex determination: males are haploid and females are diploids. Females make haploid eggs through normal meiosis; males make haploid sperm through a mitosis-like process that does not reduce chromosome number. When mating occurs, the female fertilizes some of her eggs with the male's sperm to produce diploids; these will develop as females. The remaining eggs are not fertilized; these haploid eggs develop into males through a process called parthenogenesis. Thus, males have a single copy of every gene that they inherit from their mothers; females have two copies of every gene, one inherited from the mother and one from the father. Posterior to the head, wasps have 3 thoracic segments followed by 10 abdominal segments. The second thoracic segment (T2) bears a pair of midlegs, two pairs of wings, and a pair of spiracles (breathing holes). The first and third thoracic segments have no wings or spiracles, but T1 bears a pair of forelegs, and T3 bears a pair of hindlegs. As a first step to investigate the genetic control of development in wasps, a geneticist screens for mutations affecting anterior-posterior patterning in the jewel wasp (Nisonia vitripennis). Among the mutants recovered, there are 4 different wingless mutants in which the middle thoracic segment lacks wings and spiracles and bears a pair of forelegs instead of midlegs. a. To analyze one of these mutants, the geneticist crosses a wingless mutant female with a male from a true-breeding wild-type strain. Female progeny from this cross are wild type, while male progeny are wingless. Using wg1 to represent the mutant allele and wg1+ to represent the wild-type allele, diagram this cross, showing the genotypes of the parents and progeny (both male and female)(8 pts). b. Is the wg1 mutation dominant or recessive? (4pts) c. Based on your answer to part b, what can you conclude about the function of the wg1+ gene in wasp development and in which embryonic segment do you expect the gene to be expressed? (6 pts) d. Two other wingless mutations (wg2 and wg3) are found to be recessive. When you cross a male wg2 mutant with a female wg3 mutant, you find that the male progeny are all mutant, while the female progeny are all wild-type. Would you conclude from this result that the wg2 and wg3 mutations affect the same gene or different genes (6 pts)? 4 e. Suppose the geneticist determined that wg2 and another wingless mutation wg4, affect different genes and are both recessive. She crosses a wg2 mutant male (wg2; wg4+) with a wg4 mutant female (wg4/wg4; wg2+/wg2+), and finds that the F1 females are all wild-type. Examination of 1,600 F2 male progeny of these F1 females showed that 1,192 were mutant and 408 were wild-type. Use a chi squared test to determine whether wg2 and wg4 are linked, indicating the hypothesis you are testing, the value of chi squared, the p value, and the conclusion (assume that wg2 wg4 double mutant males would be viable and show the same phenotype as single mutants). See chi squared equation page 8 and chi squared table page 3. (12 pts) f. The wg1 gene is isolated and found to encode a homeobox gene. To further analyze the function of this gene, the geneticist wants to produce transgenic wasps ectopically expressing the wg1 gene in the T1 segment. To do this, she produces a synthetic wg1 gene consisting of a promoter/regulatory region from a gene normally expressed in T1 coupled to the wg1+ coding region. Which of the following will be necessary to produce a transgenic line of wasps that have a heritable version of this synthetic wg1 gene? Circle each of the following that are necessary. 12 pts i. ii. iii. iv. v. vi. Transformation of Agrobacterium tumefaciens with a Ti plasmid carrying the synthetic wg1 gene and a selectable marker gene. Addition of a thymidine kinase gene to the plasmid carrying the synthetic wg1 gene. Integration of the synthetic wg1 gene into a wasp chromosome via recombination. Replacement of the endogenous wg1 gene with the synthetic wg1 gene via homologous recombination. Cleavage of a double-stranded RNA transcript from the synthetic wg1 gene by the enzyme "dicer". Incorporation of cells having a chromosomal copy of the synthetic wg1 gene into the wasp germ line. g. Assuming that a transgenic line of wasps ectopically expressing the wg1 gene in the embryonic T1 segment is successfully produced, what changes in the body plan do you expect these transgenic wasps to have? (Before answering, refer back to your answer to part c) (8 pts). 5 4.) Phenylketonuria (PKU) is a genetic disease in humans caused by a loss of function mutation in the phenylalanine hydroxylase gene, which encodes an enzyme that converts one amino acid, phenylalanine, into another amino acid, tyrosine. Affected individuals accumulate high levels of phenylalanine, causing mental retardation. However, this disease can be controlled by eliminating phenylalanine from the diet. The mutation causing albinism in humans (lack of melanin pigment in the skin, eyes, and hair) affects the same biosynthetic pathway. Albinos have a loss-of-function mutation in the gene for tyrosinase, which converts tyrosine to melanin. A simplified scheme describing the biosynthetic pathway affected by both diseases is as follows: phenylalanine hydroxylase tyrosinase phenylalanine -------------------------------------------- > tyrosine ------------------------------ > melanin Although phenylalanine hydroxylase acts upstream of tyrosinase, people affected by PKU are not albino so long as they get enough tyrosine in their diet for synthesis of normal amounts of melanin. The pedigree for a very unusual family is shown below the first generation is made up of two unrelated couples, both of which consisted of one parent with PKU and another with albinism! Both mutations are recessive and fully penetrant. generation I
affected by PKU 1 2 3 4 generation II
1 2 3 4 5 6 affected by albinism affected by both PKU and albinism generation III
1 2 3 4 5 6 7 8 A human geneticist studying this fascinating family obtains DNA samples from each individual in the pedigree. Each sample is digested with a restriction enzyme and analyzed via Southern blotting with RFLP markers from chromosome 3 and chromosome 12. The results are as follows: (see questions, next page) 6 a. Are either the albinism or PKU mutations X-linked? If so, which one (5 pts)? b. Do PKU and albinism mutations appear to be tightly linked to each other? Provide a one sentence explanation of why or why not (6 pts). c. Which individuals in this pedigree can make gametes that can provide information about recombination frequencies between these RFLP markers and these mutations (4 pts)? d. Do either of these mutations appear to be linked to either of the RFLP markers? If so, which mutation(s) and marker(s) are linked (10 pts)? e. For the marker/mutation combination(s) showing linkage, which individual(s) in the pedigree appear to have inherited one or more recombinant chromosomes, if any (8 pts)? f. The first line of the table below reiterates information from the pedigree showing the phenotype of each individual in generation III when fed a diet with sufficient amounts of tyrosine for biosynthesis of normal amounts of melanin. Fill in the second line of the table, indicating the expected phenotype with respect to both PKU and albinism for a child fed a diet completely lacking tyrosine (8 pts). child: Tyr in diet no Tyr in diet 1 2 3 4 PKU not PKU not PKU PKU pigmented pigmented pigmented albino 5 6 7 not PKU not PKU PKU pigmented albino pigmented 8 not PKU pigmented 7 Write your name and PID here: _______________________________________________________ Tear it off from the rest of the exam and hand it in with your exam as proof that you took the test! EQUATIONS
POPULATION GENETICS BASICS f(A) = p and f(a) = q p+q=1 and f(AA) + f(Aa) + f(aa) = 1 p = f(A) = f(AA) + 1/2f(Aa) q = f(a) = f(aa) + 1/2f(Aa) If mating is random, f(AA) = p2 f(Aa) = 2pq f(aa) = q2 so p2 + 2pq + q2 = 1 SELECTION AGAINST A RECESSIVE TRAIT W is fitness - for fittest genotype = 1 ; W for lethal genotype = 0 s (selection co-efficient) = 1 W For a recessive mutation a under negative selection (ie., W for AA and Aa = 1 but W for aa < 1): after one round of selection q' = (q sq2 )/(1 sq2 ) MUTATION Rate of conversion of A a is (mutation rate) q' = q + p qmut = p MUTATION/SELECTION BALANCE (EQUILIBRIUM) 1. Recessive mutation (where W for AA and Aa = 1 but W for aa < 1) qequil = /s (where s is selection co-efficient for aa) 2. Overdominance (where Aa is fittest genotype, ie., W for Aa > W for AA or aa) pequil = s2 /(s1 + s2 ) and qequil = s1 /(s1 + s2 ) (where s1 is selection co-efficient for AA and s2 for aa) MIGRATION Let pm = f(A) for migrants and pi = f(A) for initial population before arrival of migrants Let p' = f(A) for new population made when migrants join initial population If m is the proportion of individuals in the new population that were migrants, p' = mpm + (1 m)pi p = m(pm pi) m = (p' pi)/(pm pi) NON-RANDOM MATING F is inbreeding co-efficient = probability of being homozygous by descent f(aa) = qF + q2 (1-F) f(Aa) = 2pq(1-F) f(AA) = pF + p2 (1-F) F for a population = 1 (H/2pq) where H is the actual frequency of heterozygotes QUANTITATIVE GENETICS EQUATIONS HB (broad sense heritability) = Vg /Vt (where Vt = Vg + Ve) (Vt = total phenotypic variance, Vg = genotypic variance, Ve = environmental variance) HN (narrow sense heritability) = Yo - Y/ Yp Y (Yo = mean phenotype score for offspring, Y= mean phenotype score for original population, Yp = mean phenotype score for selected parents). If solving for Yo = Y+ HN(Yp -Y) CHI SQUARED: 2 = [(O-E)2 /E] 8 ...
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This test prep was uploaded on 04/07/2008 for the course BIBC 100 taught by Professor Nehring during the Winter '07 term at UCSD.
- Winter '07