<span>Statistically, which adolescent is most apt to experience menarche first? The </span> Answer Is ( C)
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https://www.slideshare.net/smullen57/53-classification-and-biodiversitydoc
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All three species share the same taxonomy down to the level of Order: Kingdom Animalia, Phylum Chordata, Class, Mammalia, Order Carnivora.The ferret is in a different family: Mustelidae whereas both the bobcat and domestic cat are in the same family: Felidae and Genus: Felis. Therefore, the bobcat and domestic cat are more related by two branches in the taxonomic hierarchy.</span></span></span></span></span></span><span><span>
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A is the answer I would choose :) amos 12kvng
This is true.
Hope this helps.
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.