Problem 5. A male Drosophila from a wild-type stock is discovered to have only s
ID: 207089 • Letter: P
Question
Problem 5. A male Drosophila from a wild-type stock is discovered to have only seven chromosomes rather the normal 2n-8. Karyotyping of the fly reveals translocation of one chromosome IV to distal end of a chromosome II thus accounting for the reduction in chromosome number (see figure to the 5. If this male mates with a female with a normal chromosome composition who is homozygous for the recessive chromosome IV mutation eyeless (ey), what will be the phenotypic ratio of the eyeless flies in the progeny, assuming all chromosome compositions survive? A. 0 B. 1/6 C. 1/4 D. 1/3 E. 1/2 F. 2/3 G. 3/4 H. 1Explanation / Answer
C. 1/4 is a answer
Lets consider the genotype of wild type male as +/+ and of the recessive female is ey / ey.
The individual will have eyeless phenotype only when it is in recessive combinations. E.g. ey/ey or -/ey.
With the translocation of Chr. 4 in male it will have 7 chromosomes and therefore can lead to production of two possible alleles. One with 4 chromosomes and another with 3.
It can have two possible fatesof producing gametes.
1. One allele with 4 normal chromosomes. And one with three chromosomes out of which chromosome 2 will have translocated chr. 4.
If these gametes fuse with the normal ey/ey female gametes it will lead to 2 normal offsprings. Therefore eyeless flies in this combinations = 0/2
2. In second possibility, wild male will give one one alleles with translocated Chr. 2 along with other 3 normal chromosomes. Giving genotype ++. Other gamete will have 3 chromosomes without any part of Chr. 4.
When ++ fuses with ey female it will give normal individual. While Other gamete on fusion with ey gamete of female will give eyeless phenotype.
Therefore no. Of eyeless individuals in this possibility will be 1/2.
In total we will have 1/4 eyeless individuals.