In garden peas a pure line that is tall with purple flowers is crossed to another pure line that is short with white flowers. the f1s are all short and white. you cross the f1s with each other and get 900 short white, 315 short purple, 307 tall white, and 99 tall purple. you allow 90 of the short purple f2 plants to self-fertilize and then select a single seed from each one. what phenotypes would you expect to see in the plants reared from these seeds and in what numbers? why? a human female "carrier" who is heterozygous for the recessive, sex-linked trait red color blindness, marries a normal male.

a. what proportion of their female progeny will show the trait?
b. what proportion of their male progeny will show the trait?
c. if one of their daughters marries a normal male, what is the probability that the first son of this marriage will show the trait?
d. if one of their daughters marries a normal male, what is the probability that the first child of this marriage will show the trait?

See the explanation

Explanation:

Answer 1.

As given that in F1 all are short and white then it can be said according to Mendel's law that short and white are dominant over tall and purple.  

Let S for short s for tall and W for white and w for purple allele.

So the genotype of short purple will be Ssww or SSww. So In first case Ssww self crossed then resultant offsprings will be,

Sw sw

Sw SSww Ssww

sw Ssww ssww (tall and purple)

So from this 1/4 will be tall and purple while 3/4 will be short and purple.

In second case SSww only short purple progeny will appear.

Answer 2.

2. a) The female progeny will not show any trait because there are two X chromosome in females , so female offspring can be carrier but not show any trait in case of X- linked trait.

2. b) Half of male offsprings show trait because X is inherited from mother. So the chance of having X-linked recessive allele is 1/2.

2. c) The chance of having X linked affect allele in daughter is 1/2. So the chance of inheriting that X to son will be 1/2 so in total there is chance of 1/4 that son will be affected.

2. d) the chance of first child show this trait will be 1/4 in case of male offspring while 0 in case of female offspring.

Hope this helps!