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Softa [21]
3 years ago
10

Hemophilia and colorblindness are both X-linked, recessive traits. A hemophiliac woman marries a colorblind man. Assume that bot

h are otherwise normal. If meiosis occurs normally for both parents, what phenotypes are possible among their offspring?

Biology
1 answer:
Gre4nikov [31]3 years ago
4 0

Answer:

Normal daughters and Hemophiliac son

Explanation:

Females exhibit hemophilia and colorblindness in homozygous genotype only.  

The genotype of hemophiliac woman= X^hX^h

The genotype of colorblind man= X^cY

A cross between X^hX^h and X^cY would produce progeny in the following phenotype ratio= 1 Normal but carrier daughters:1 Hemophiliac son

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A Pea plant that is homozygous for inflated pods are heterozygous for round seeds is crossed with a plant that is heterozygous f
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The answer is False!

P - dominant allele for inflated pods
p - recessive allele for inflated pods
PP - homozygote with inflated pods
Pp - heterozygote with inflated pods
pp - homozygote with inflated pods

S - dominant allele for round seeds
s - recessive allele for wrinkled seeds
SS - homozygote with round seeds
Ss - heterozygote with round seeds
ss - homozygote with wrinkled seeds

1. <span>A Pea plant that is homozygous for inflated pods are heterozygous for round seeds: PPSs
2. A</span><span> plant that is heterozygous for inflated pods and a homozygous for wrinkled seeds: Ppss

Let's look at their genotypes separately, and cross them that way:

Parents:              PP       x      Pp
Offspring:     PP      PP      Pp     Pp
All of the offspring will have the same phenotype (100% = 1). So, this trait does not affect phenotypic ratio

</span>Parents:              Ss       x      ss
Offspring:     Ss      Ss      ss     ss
Half of the offspring will have wrinkled seeds and half of them will have round seeds: 50% : 50% = 1 : 1

So, the phenotypic ratio is not  1 : 3
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