Answer:
Explanation:
The Genotype of the woman is XCbXN
The man XNY
Since the allele for the color blindness is carried on one of the X -chromosomes of the carrier mother, then the male children will be color blind or carrier deopending on the defective gene they inherited. Since they will receive the defective or normal allele from the mothers,but the female will carriers or normal because they will inherit either defective allele or normal allele from the mother,but normal X allele from the father.
Assuming the colorbind gene was inherited by One of the boys, and the girls take the defective genes
1. XNY vs XCbXN.Then the one of the boys will be colorblind and the one of girls will be carriers. XNXCb, XNXN, XCbY,XNY 50 %
2.If the two boys take the effective genes from the mothers. and none of the girls take the defective alelle
Then the two boys will be colorblind and the girls will be carrries XcbY,XcbY and XNXN XNXN.(normal girls).
The proportions that will be the females since the girls can only take the defective allele from one of the normal and defective allele from the mother. Assuming the girls take the defective allele from the mothers and take normal from the father,they will all be carriers,since the X of the father is normal.
They can not have a color blind girl.Since the father is normal It takes two colorblind X-chromosome to give a girl.
Homozygus- same
the PP are the same allele
The genotype would be 100% homozygus and then whatever the P stands for.
Homozygus P
The answers are shown below:
In part A:
coefficient = 20
variable = c
constant = 35 & 23.50
In part B:
savings = 20 (3) +23.5+ 35+35+35 +35 +35 +35+35+35+35+35+35+35
savings = $ 503.5
In part C:
the new expression will be: 25c + 35 + 23.5
