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.
A bottleneck is very narrow, which means that only a portion of the contents of the bottle cat get out (at the time).
From this effect, people use the word bottleneck when only a few individuals of a population survive:
the answer is: I<span>nsecticide spraying eliminates all but a few of the beetles on an island.</span>
Answer:
1. C.) Their cell walls contain chitin
2. C.) Chloroplast
Although it's rare, with the odds of getting struck in your lifetime being roughly 1 in 12,000, every now and then a human will provide an attractive target for lightning bolts to unleash their power. And of the roughly 500 people who are struck by lightning each year, about 90 percent survive.
<span>F- allele for freckles
f- </span><span>allele without freckles
1) The man is heterozygote and has freckles, its indicating that the allele for freckles is dominant.
A cross between him and a woman who is also </span><span>heterozygote: Ff x Ff
it would result in the following probabilities:
- 1/4 - homozygote with freckles: FF
- 2/4 - </span><span>heterozygote with freckles: Ff
- 1/4- </span><span>homozygote without freckles:ff
Their son would have a probability of 75% of being born with freckles.
2) The cross resulted in this probabilities:
</span><span><span>- 1/4 - homozygote with freckles: FF
- 2/4 - </span><span>heterozygote with freckles: Ff
- 1/4- </span><span>homozygote without freckles:ff
So, the chance of being born heterozygote for this gene is 2/4, which is the same as half (50%).
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