Suppose you cross heterozygous Drosophila females of wild type red eye appearanc

Question

Suppose you cross heterozygous Drosophila females of wild type red eye appearance and normal wing length with males showing two autosomal recessive traits: purple eyes and vestigial wings. In the offspring, you observe the following combinations of traits:

Wild type (red eyes, normal wings)                      441

Purple eyes, vestigial wings                                  484     

Red eyes, vestigial wings                                      30  

Purple eyes, normal wings                                    45

1) If 1000 progeny of this cross are distributed in the phenotypic classes shown above, is this ratio diagnostic of two linked genes, and if so, why?

2) What is the genetic distance between the gene loci for eye color and wing length described above?

Choose one of the choices below for part 2:

75 map units

0.75 map units

7.5 map units

750 map units

They are not linked

Explanation / Answer

1)

Based on the given data, the phenotype of females (heterozygous): red eye and normal wing (wild) type, hence the genotype is: RrWw. The phenotype of male is: purple eyes and vestigial wings, hence the genotype is rrww.

Therefore, the cross is made between the RrWw× rrww:

The obtained observed data is as follows:

So, the ratio is: 1:1:1:1, yes this ratio will support diagnostic of two linked genes. Here, both genes are linked the progeny of recombinants is less than progeny of parents. If you perform Chi-square test to the data, you know the data is supportive.

2)

There is 30 + 45 = 75 progeny show recombination between genes R and W.

Thus, the gene distance between R and W. is 7.5 map units (m.u).

Hence, the correct option is (c) 7.5 map units.

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