Question

Assume that the genes for tan body and bare wings are 15 map units apart on chromosome II in Drosophila. If a tan-bodied, bare-winged female was mated to a wild type male and the resulting F1 phenotypically wild type females were mated to tan-bodied, bare- winged males. Of 1000 offspring, what would be the expected number of flies that are tan but have normal wings

Answer 1

Answer:

WILD TYPE= 425;

TAN-BARE= 425;

TAN= 75;

BARE= 75

Explanation:

Fifteen map units apart implies that 15% of the offspring are products of recombination.

Out of 1000 offsprings, therefore 15% of 1000= 0.15 × 1000=150

150 offsprings are products of recombination.

100-15 =85%

Therefore, 85% of the offspring are parental.

So, if tan-bodied, ware-winged female was mated with wild-type male resulting in F1 phenotypically wild-type females were mated to tan-bodied, bare-winged males.

The expected number of offsprings that are tanned but have normal wings would be:

Wild type= 50% of 850

0.50 × 850 = 425

Tan-bare= 50% of 850

0.50 × 850 = 425

tan= 50% of 150

0.50×150= 75

Bare= 50% of 150

0.50×150=75

Answer 2

The basic and functional element of heredity composed of DNA is a gene.

The expected number of flies that are tan but have normal wings are:

• WILD TYPE = 425

• TAN BARE = 425

• TAN = 75

• BARE = 75

This can be estimated as:

By 15 units apart it indicates that the 15% of progenies is the product of recombination.

• Total number of offspring = 1000

• Recombinant offspring (R):

$\begin{aligned}\\\text{R} = 0.15 \times 1000\\\text{R} = 150\\\end{aligned}$

• Number of progeny having parental traits (P):

$\text{P} = 100 - 15\% \\\text{P} = 85\%$

Hence, offspring having parental characters are 85%.

When a tan-bodied, bare-winged female is crossed with a wild-type male then the F1 phenotype formed will result in wild type females and tan-bodied, bare-winged males.

The predicted number of progeny that is tanned but has normal wings can be calculated as:

• Wild type (W):

$\text{W} = 50 \% \;\text{of} \; 850\\0.50 \times 850 = 425$

• Tan-bare (TB):

$\text{TB} = 50 \% \;\text{of} \; 850\\0.50 \times 850 = 425$

• Tan (T):

$\text{T} = 50\% \;\text{of}\; 150\\0.50 \times 150 = 75$

• Bare (B):

$\text{B} = 50\% \;\text{of}\; 150\\0.50 \times 150 = 75$

Therefore, the expected number of flies having tan and normal wings will be wild type 425, tan bare 425, tan 75 and bare 75.

To learn more about inheritance follow the link:

https://brainly.com/question/4520267