Answers to HW #5

4.1: 0.2.

4.2: Females produce gametes of types A B, a b, A b, and a B in the proportions (1-0.08)/2 = 0.46 and (1-0.08)/2 = 0.46 for each of the nonrecombinant types. It is 0.08/2 = 0.04 and 0.08/2 = 0.04 for each of the recombinant types. Since there is no crossing over in male Drosophila, males produce only two gamete types (A B, a b) in equal (0.50, 0.50) proportions.

4.7: The map distance between s and c is 18 cM, but there is only 16% recombination, and the map distance between p and c is 13 cM, but there is only 12% recombination. The fact that the frequency of recombination is smaller than the map distance results from double crossovers.

4.8: The data include 108 progeny with either both mutations or neither, and 92 progeny with one mutation or the other. The first of these groups consists of parental chromosomes; the second consists of recombinant chromosomes.   The chi-square test compares the observed ratio (108:92) to the expected (100:100) under the hypothesis of no linkage. The chi-square value equals 1.28 and there is one degree of freedom. P is approximately 0.26, so there is no evidence of linkage even though both loci are on the X.

4.10: (a) the mutant genes are not alleles because they do not show segregation; if they were alleles, the progeny would show resistance to one insecticide or the other. (b) the genes are not linked. The chi-square of 177 parental:205 recombinants compared to 191:191 (expected) equals 2.05. With one degree of freedom, P is 0.15 (not significant). (c) The two genes must be far apart on the chromosome – too far to be ‘linked’.


(a) The gene in the middle has the alleles D and d.

(b) The triply heterozygous parent has genotype A d B / a D b or B d A / b D a.

(c) The closest genes are B and D.

(d) B−D distance equals (40 + 60 + 10)/1000 = 0.11 = 11 map units = 11 centimorgans (cM).

4.12: (a) parental types = v+ pr bm and v pr+ bm+. Double recombinants = v+ pr+ bm+ and v pr bm. This means that gene v is in the middle. The pr-v recombination frequency is 20.7% and the v-bm frequency is 41.5%. The expected number of double crossovers = 85.9, so the coincidence is 0.90. The interference is thus 0.10. (b) The true map distances are larger than 20.7cM and 41.5 cM because the frequency of recombination between these genes is sufficiently large that there must have been some double crossover events that went undetected in the analysis.


(a)       To solve this type of problem, you should first deduce the genotype of the gamete contributed by the triply heterozygous parent to each class of progeny. In this case the gamete types are, from top to bottom, + + +, + + unc, + dor unc, dpy + +, dpy dor +, dpy dor unc. To determine the order of the genes, you must identify which gene is in the middle. You can do this by comparing the parental type gametes with the double crossovers. The two parental types are dpy dor + and + + unc, because these are the most frequent. In these data there are no observed double crossovers (which means that interference across the region is complete). The missing classes of progeny correspond to the double crossovers. The missing phenotypes of progeny are normal body, deep orange eye, coordinated, corresponding to the genotype + dor +, and dumpy body, red eye, uncoordinated, corresponding to the genotype dpy + unc. Comparing the parental types with the inferred double crossover types reveals that dpy+ and dpy are interchanged relative to the alleles of the other two genes. Hence, the genotype of the F1 heterozygous female is dor dpy unc+ / dor+ dpy+ unc or unc+ dpy dor/ unc dpy+ dor+.

(b)       Analysis of the single recombinants indicates (96 + 110)/1000 = 20.6% recombination between dor and dpy, and (75 + 65)/1000 = 14.0% recombination between dpy and unc.

(c)        Because there are no double recombinants observed, the interference is complete, which means interference = 1.


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