The correct answer is true.
Hope that helps, even though I am late.
Answer:
a.The phenotypic proportions obtained after having the genotypes are 50% marbled seeds, 25% spotted and dotted seeds since they are codominant, 25% spotted seeds.
b. Taking into account the F1 genotypes in the previous point, the expected phenotypes for the first crossing are 100% marbled seeds and for the second crossing 100% dotted seeds.
Explanation:
Let's suppose:
Marbled allele: M
Spotted allele: S
Dotted allele: D
Allele for Clear: C
a. Because both crosses were between homozygous parents, the entire F1 genotype is the same.
For the first crossing the descendants have the MS genotype, and for the second crossing the descendants have the DC genotype. It is enough to make a Punnett square to obtain the different combinations of genotypes between the crossing of MS and DC.
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300km is closest to maximum tickness of a plate.
In animals and plants, there are two pairs of chromosomes - one set from a male and the other set from a female. But occasionally, the original fertilized cell doesn't quite divide correctly and more than two sets arises. The fertilized cell continues to divide with the result that all the cells have the extra set of chromosomes. This happens much oftener in plants and the plant will be sterile and can't form seeds.