In drosophila, the genes for withered wings (whd), smooth abdomen (sm) and speck body (sp) are located on chromosome 2 and are separated by the following map distances: whd .. female with withered wings and a smooth abdomen was mated to a male with a speck body. the resulting phenotypically wild-type females were mated with males that had the mutant phenotype for all three traits, producing 1000 offspring. part acalculate the number of expected double crossover progeny. express your answer to the nearest whole number. part bfor the remainder of this problem, assume that the interference for these genes is 0.3. calculate the number of observed double crossover progeny. express your answer to the nearest whole number. part chow many offspring are expected to have the following phenotypes? withered wings, speck bodyorsmooth abdomenexpress your answer to the nearest whole number. part dhow many offspring are expected to have the following phenotypes? withered wings, speck body, smooth abdomen or wild typeexpress your answer to the nearest whole number.

A) 47; B) 33; C) 272; D) 122

Explanation:

The three genes are linked.

The female with withered wings and a smooth abdomen has the genotype whd sm sp+/whd sm sp+.

The male with a speck body has the genotype whd+ sm+ sp/whd+ sm+ sp.

Both individuals are homozygous for all genes, so each of them only produces one type of gamete. The resulting F1 therefore has the genotype whd sm sp+/ whd+ sm+ sp, heterozygous for all genes and with a wild-type phenotype.

The females of the F1 were mated with homozygous recessive males (test cross): whd sm sp/whd sm sp.

If we assume interference is 0, the probability of crossing over happening between the genes whd and sm is independent from the probability of crossing over happening between sm and sp.

The distance = frequency of recombination × 100, so the frequency of recombination (RF) between genes whd and sm is 0.305 and the RF between genes sm and sp is 0.155.

The expected double crossover progeny among the 1000 offspring will be:

RF whd-sm × RF sm-sp  × 1000 =

0.305  × 0.155 × 1000 = 47 individuals will be double crossover.

Interference is 0.3

The interference is calculated as 1- coefficient of coincidence (cc).

cc = observed double crossover/expected double crossover

Therefore:

I = 1 - cc

cc = 1 - I

cc = 0.7

Observed DCO / 47 = 0.7

Observed DCO = 0.7  × 47

Observed DCO ≅ 33

The parental gametes are whd sm sp+ and whd+ sm+ sp (the genotype of the F1 female is known).

Looking at them and at the gene map we can tell that the gametes that give rise to withered wings, speck body (whd sm+ sp) and smooth abdomen (whd+ sm sp+) phenotypes are the result of recombination occurring between genes whd and sm.

To calculate the expected number of individuals with those phenotypes among the 1000 progeny we need to determine the frequency of recombination between the genes whd and sm considering there's interference.

The distance whd-sm = RF x 100

The recombination frequency is the sum of the single crossover between whd and sm and the double crossovers.

The frequency of DCO is 33/1000=0.033.

Distance whd-sm/ 100 = SCOwhd-sm + DCO

0.305 - 0.033 = SCO whd-sm

Frequency of SCO whd-sm= 0.272

And the expected number of individuals with those phenotypes will be 0.272 x 1000 = 272.

The gametes that originate the phenotypes withered wings, speck body, smooth abdomen (whd sm sp) and wild type (whd+ sm+ sp+) are the result of recombination between genes sm and sp.

Distance sm-sp/ 100 = SCOsm-sp + DCO

0.155 - 0.033 = SCOsm-sp

Frequency of SCO sm-sp= 0.122

And the expected number of individuals with those phenotypes will be 0.122 x 1000 = 122.

answer; polycythemia is the correct answer;