what is the relationship between meiosis, sexual reproduction, and genetic diversity? make a claim, cite evidence to support you
1 answer:
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Answer:
(a) 1/2; (b) no
Explanation:
Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.
In contrast to this, her husband is diseased so his genotype will be XᵇY.
The Punnett square diagram related to the cross is attached.
(a) Proportion of their sons expected to be G6PD is 1/2:
They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.
(b) If the husband were G6PD deficient, the answer will not change.
The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.
Answer: 25 of the chicks will be black
Explanation: Since the chickens are intermediate in colour, that is they are not black, they are heterozygous. Let B represent the allele for black colour and B represent the allele for blue colour, therefore the genotype of a chicken with an intermediate colour is Bb. A cross between two Bb chicken will produce 1 BB, 2 Bb and 1 bb chicks. 1 BB will manifest outwardly as black, 2Bb will manifest outwardly as intermediate colored while 1 bb will manifest outwardly as blue colored chicks.
Since one out of four chicks are black colored, if the two chicken produced 100 chicks, the number of chicks that will be black is 1/4 x 100 = 25. Therefore, 25 chicks out of 100 will be black.
See the attached punnet square for more information
