Which one is NOT one of the 5 major parts of a BULB? Root
1 answer:
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Answer: 50% or 25% according to the genotype of the male cat.
Explanation:
The female cat is unaffected, and since this is an autosomal dominant diease, it means this cat does not have any affected alleles. Thereby,<u> its genotype is aa</u>. This is because dominant alleles show their effect even if the organisms only has one copy of the allele. The male cat is affected with this syndrome <u>so its genotype is either AA or Aa</u>. For this disease to manifest itself, only one affected allele is needed. Although both alleles can be affected so the individual can be homozygous dominant or heterozygous.
To find this we have to do a punnett square. Here, we use the gametes produced by each individual. <u>Gametes are sex cells, egg or sperm, which only have one allele of each gene.</u>
Assuming that the cross is between aa and AA, the gametes produced by "aa" are "a", and the gametes produced by "AA" are "A" So we are having a cross between an "a" gamete and a "A" gamete. 100% of the offspring will be Aa so it is a heterozygous, which will develope this syndrome. And, males have one Y chromosome and one X chromosomes, thereby the chances of having a male kitten is 50%. Then, to conclude, there is a 50% chances of having a male kitten with this syndrome.
And, assuming that the cross is between aa and Aa, the gametes produced by "Aa" are "A" but also "a" So we are habing a cross between an "a" gamete and "A" or "a" gamete (See picture) In this example, there is a 50% chances of having a kitten with the syndrome, because only one genotype produced (Aa) has the affected allele. And, as we said before, since there is a 50% chances of having a male kitten, we can said that there is a 25% chances of having an affected male. Because to calculate both probabilities so that they occur simultaneously, it is necessary to multiply them. That is, 0.5 x 0.5 = 0.25 x 100 = 25%.
