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2. In a newly-discovered species of insect early emergent pupae (E) are dominant

ID: 191391 • Letter: 2

Question

2. In a newly-discovered species of insect early emergent pupae (E) are dominant to late emergent (e). Fuzzy antennae (F) are dominant to smooth antennae (f), and green eyes (G) are dominant to black eyes (g). All three genes are linked. A testcross on an individual with the genotype EeFfGg was done, giving the results listed below. Phenotypes Number of Offspring EFg efG 95 85 748 800 125 135 -ear rtye-lat e eFg EFG ef Efg - Smooth rtenq black Total 2000 a) What is the linear order of the genes and the map distances between them? There are two map distances to determine. What is the parental genotype (i.e. allele arrangement)? (10 pts.)

Explanation / Answer

The best approach to solve these problem is to begin with step wise process. 1st is to identify the parental genotype. The most frequently found genotypes are the parental genotypes. Next process is to know the order of genes. The gametes of double-crossover are present in the lowest frequency. Then, the next point is to focus that an event of double-crossover moves the middle allele of one sister chromatid to the other.

In this given question, EfG and eFg is the parental type because it is present in large number. D.C.O. gamete is Efg and eFG due to least frequency. hence we get the order of genes that should be EGf and egF. We can go for calculating the linkage distances between E and G, and G and f.

Recombination frequency (RF)=Recombinants / Total offspring ×100%

The linkage distance is the total number of recombinant gametes divided by the total gametes. So the distance between genes E and G is 95 + 85 + 5 + 7 / 2000 * 100 = 192 / 2000 * 100 = 9.6%. a 1% recombination frequency (RF) =1 centimorgan or 1 map unit. therefore, distance between E and G = 9.6cM.

Similarly, distance between G and f = 125 + 135 + 5 + 7 / 2000 * 100 = 272/2000 * 100 = 13.6% = 13.6cM.

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