Question

All answers must be typed and in italic tont so we can disth the questions. If you use more than one page they must be stapler togethet or points will be deducted. 1. How many gametes of differing genotypes can a female drosophila of genotype ARCiabe produce? How many gametes of differing genotypes can a male drosophila of gentotype ABC/abe produce? 2. In corn, the genes v (virescent seedlings), pr (red aleurone), and bm (brown midrib) are all on chromosome 5 but not necessarily in the order given. The cross v pr* bm v pr bm produces progeny with the phenotypes listed below: v* pr bm 209 pr+bm Parental Parental sco v pr bm175 v pr bm 181+vbm SCO v' pr bam 69 ++ bm SCO2 76 prv+ pr bm 76 v pr bm V pr bm 36 ++ v pr bm sco2 DCO 41 pr v bm DCo a. Determine the gene order, the recombination frequency between the adjacent genes, the cocfficient of coincidence, and the interference. b. Explain why, in this example, the recombination frequencies are not good estimates of the map distance. 3. Two normal - looking Drosophila (a male and a female) are crossed and yield the following phenoty pes among their progeny: Daughters:

please answer#2 question in the photo. please explain the answers. Thanks

Answer 1

It is three point cross. In data we can observe, parental progeny, single cross overs and double cross overs.

The maximum number individuals are parents and the minimum number individuals are Double cross overs. By comparing parents with double cross overs, we can findout the middle gene. In the data v is the middle gene.

Recombinant frequency between pr & v = (175+181)+1/2 (36+41)/ 1000 =356+38.5/1000 = 0.394 x100 = 39.4 map units.

Similarly recombinant frequency between v & bm = total SCO+1/2 (DCO) / total progeny

= (69+76)+1/2 (36+41) /1000 = 145+38.5 /1000 = 0.183 x 100 = 18.3 map units

Genes map: pr---------39.4mu--------------------v-------18.3mu-----bm

In any cross the total recombinant frequencies should not exceed 50%. But here it is 57%. So in this case the recombinant frequencies are not fair