Answers to additional linkage problems.

Transcription

1 Spring 2013 Biology 321 Answers to Assignment Set 8 Chapter 4 Answers to additional linkage problems. Problem -1 In this cell, there two copies of each (homologous) chromosome. Each genome copy (chromosome set) contains two non-homologous chromosomes. Your drawing should show one pair of homologs carrying the A allele (total four copies of A in the duplicated pair). The other pair of homologs should show the B & C allele on one duplicated homolog and the b & c alleles on the other duplicated homolog. Problem 0 2 map units. See also similar Chapter 4 problem. Keep in mind that a single cross-over during meiosis results (after division II) in two daughter cells with the parental combination and two recombinants so only half of the output is recombinant. Problem 1 a. (2x2x2) =8 possible genotypic classes if assorting independently b. 2 genotypic classes (ABC/abc and abc/abc) if tightly linked c. First transform progeny genotypes so you are looking only at the contribution from the triple het parent. Then look at genes pairwise to determine if they are assorting independently or linked. Genes a and b are tightly linked (no recombinant classes observed), and gene c is 10 map units from a and b. Problem 2.1 a. 45% b. 45% c. 5% (vermillion eyes and WT body) d. 5% (sable body and WT eyes) From inspection of the genetic map you can determine that the two genes are 10 map units apart. This tells you that 10% of the gametes produced by the F1 females will be recombinant. The parental combinations are wild-type versus doubly mutant. Since the genes are on the X chromosome, a test cross of the F1 animals is not necessary since the male F2 s will reflect the genotype of the gamete received from their mom. As part of your analysis you should assign allele symbols and write out genotypes for the parent F1 and F2 s. Be sure to use proper notation for linked genes v+ = wildtype eyes v= vermillion eyes s+ = wildtype body s= sable body color Genotype of F1 = v+ s+/ s v Problem 2.2 (i) answer is a (4%) (ii) answer is g (what info is missing?) (iii) answer is c (25%) Recombination will no exceed 50% for two weakly linked genes (see explanation at end of linkage lecture). The 68 map units separating the two genes is an artifact of the way linkage maps are assembled from map distances between closely linked gen 1

2 Problem 3 Since the two genes are only 1 map unit apart, you would only expect 0.5% or 1 in 200 flies to be doubly mutant for the two genes. While might very well have picked up a recombinant among his 100 progeny, he should have scored more progeny to ensure that he obtained the genotype he wanted. [NOTE: For each progeny there is a 199/200 chance of NOT getting a double mutant. For 100 progeny, the probability of not finding a double mutant is (199/200) 100 = 60% chance of not seeing double mutant. NOTE that even if he had scored 200 progeny there is still a (199/200) 200 = 37% chance of not finding a double mutant.] Problem 4 Do a systematic analysis of the data: First, from the description of the crosses, we know that the trait is not X-linked therefore chromosome 1 is eliminated. The short gene must be on chromosomes 2 or 3 or 4. Second: define allele symbols and write out genotypes of each generation. Then determine whether the short gene is assorting independently of the black body gene and likewise for the cd gene. wild-type legs = sh + short legs = sh wild-type body = b + black body = b wild-type eyes = cd + cardinal eyes = cd Next: crunch through genotypes for each generation. Use the nomenclature for unlinked genes, since you don t know at the outset whether the short gene is linked to either of the genetic markers used in the experiment. Phenotype Genotype of gamete from F1 triply heterozygous mom: Wild-type sh + b + cd + Cardinal eye sh + b + cd Short Black sh b cd + Cardinal, short &black sh b cd Analyze the data in the table: write out the genotype of the mother s gamete that produced each of the phenotypic categories. Then determine whether the short gene is assorting independently or is linked to the black body gene. Ditto for the cardinal eye gene. Note, that since the genes are autosomal, we can combine the data from male and female progeny. Conclude: We already know that the cardinal and black genes are on different autosomes and this data confirms that these markers assort independently. The short gene assorts independently from the cardinal gene: all 4 allelic combos in equal frequencies. The short gene is very tightly linked to the black gene. With respect to these two genes, no recombinant progeny are observed in this experiments. How does this problem differ from problem 1? Follow-up: rewrite the crosses using the nomenclature for linked genes. Parents: sh b/ sh b; cd /cd X sh + b + / sh + b + ; cd + /cd + F1 sh + b + / sh b; cd + /cd ETC ; = separates genes on different non-homologous chromosomes / = separates genes(alleles) on a homologous pair 2

3 Problem 5 3

4 Problem 6 ( i) b (8 map units) (ii) c 4

5 Problem 7 h + = wild-type h = hemophilia c + = wild-type c =colorblind a. woman: h + c + / h c husband: h + c + / Y b. [0.03]1/2 = 1.5% Think about this additional question: If the first son is colorblind, what is the probability that he is also has hemophilia? Problem 8 A1 and A2 = SNP a. A1 N/ A2 n b. (last daughter should be A1A1) two (III-1 & III-7) Problem 9 Correct answers are e and f (assuming a single crossover between the disease gene and the site of the polymorphism) D= dominant disease mutation d + = wild-type allele Generation I = D pc/ d + pc X d + pa/d + p? Generation II = D pc/ d + pa X d + pa/d + pc Based on map distance Gametes from Generation II male= Parental-type D pc = 0.4 Parental-type d + pa = 0.4 Recombinant D pa = 0.1 Recombinant d + pc = 0.1 Generation III: Since the son is pc/pc we know that he received either the parental-type D pc allele or the recombinant d + pc allele from his father (see boxes). Since we know that he did not receive the pa allele, there is a 80% probability (4 to 1) that he received the Dp C allele. NOTE: the probability of correctly inferring genotypes using linked molecular markers (that are physically close to but otherwise unrelated to the disease gene) is inversely correlated with the map distance between the disease gene and the molecular marker 5

6 Problem 10 a. A&B are alleles of an SNP site; D = disease d + = normal Genotype of #7 is A D/B d+ Since the child received the A allele from the mother, there is a 90% probability that s/he is heterozygous for the D allele. b. Since #8 is het for the disease allele, there is a 50% chance that her child will be het. The prenatal test is uninformative since #8 was homozygous for the B SNP: B D/ B d+ Problem 11 a. autosomal dominant b. From inspection of the pedigree, the mutant allele must be associated with B2 c. (i) ½ since her mate must be het (ii) D= disease allele d+ = wildtype allele father : B2 D/ B1 d+ mother B2 d+/ B3 d+ child: 96% chance B2 D/ B3 d+ 4% chance B2 d+/ B3 d+ Since we know the kid did not get B1 from dad we have eliminated two of the four possible gametes that he could have contributed. So this is effectively a conditional probability calculation we adjust the probabilities of parental versus recombinant accordingly. Without any conditions on the outcome, the probability of a kid getting B2 D is 48%, B1 d+ = 48%, B1 D = 2% and B2 d+ = 2%. With the knowledge that the kid is B2, the probabilities become 96% and 4% respectively for B2 D and B2 d+ d. The B2 polymorphism in the unaffected granddaughter came from the unaffected father. Problem 13 (a). The polymorphic markers are linked to each other and to the gene causing the disease and are rarely separated by recombination. (b). Autosomal recessive. Unaffected parents produce affected children, and the haplotypes that segregate with the disease must be homozygous for several of the markers. (c). The haplotype analysis indicates that the gene responsible for this disorder is definitely located between markers D17S1866 and D17S960 and probably lies between D17S1308 and D17S938 (d) Sequence the candidate gene in affected and unaffected individuals to see if the affected individuals have mutations NOT shared by a sampling of the general population Other strategies not covered in lecture: generate a mouse model knock-out for the gene (which is very likely conserved) 6

7 see if the mutant phenotype can be rescued in the mouse by a wild type transgene copy (gene therapy in mice) begin gene testing in similar families and possibly also trials for gene therapy in humans. Problem 14 a. dominant (to be consistent with recessive inheritance too many unrelated spouses would have to be het) autosomal inheritance (X-linkage eliminated by male to male transmission) b. BE sure to look at a color version of the pedigree All affected individuals inherited a common segment of chromosome (haplotype) from the male in the first generation. Not X-linked since males have two haplotype copies (two copies of the chromosome in question) c. YES, based on transmission data in pedigree and haplotypes: III-9, III-20, IV-13 d. Looks linked to the lowest two of the haplotype markers. Note the haplotype that segregates with the disease is colored red that may not be the case on an exam question. You should be able to study the pattern of numbers and discover the relevant haplotype. 7