Which process in eukaryotic cells will normally proceed whether O2 is present or absent?
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Answer:
The possible genotype of the normal male: AA or Aa
The genotype of affected female: "aa"
The probability of having a carrier girl child= 1/4
Explanation:
The given trait is autosomal recessive. Let's assume that the recessive allele "a" is responsible for the trait while the dominant allele "A" gives the dominant phenotype. The genotype of the normal male can be AA or Aa while that of the affected female would be "aa"
Their daughter has dominant phenotype and therefore, would be heterozygous for the trait with "Aa" genotype. A cross between carrier daughter "Aa" and her carrier male "Aa" would give progeny in following ratio= 1/4 AA: 1/2 Aa: 1/4 aa. Therefore, they have 1/2 probability of having a carrier child.
The probability of having a girl child is always 1/2.
So, the total probability of having a carrier girl child would be 1/2 Aa x 1/2 XX = 1/4
