Answer:
1. The parents genotypes could have been BO and AO
2. wire-hair
Explanation:
There are four possible blood types which are type A, B, AB, O. blood group is the classification of blood based on the presence or the absence of inherited antigenic substances on the surface of the red blood cells. They have hereditary basis and also rely on a series of alternative genes sometimes used in solving dispute of parental heritage. With the four possible blood groups, there are six possible genotypes and these are:
Blood type possible genotypes
Type A AA, AO
Type B BB, BO
Type AB AB
Type O OO
Thus, for parents with blood type B and A to give birth to a child with blood type O, it means their genotype could have been both BO and AO for them to be able to produce a child with OO. a cross between these two could give rise to OO.
Question 2
Wire hair is dominant (S) to smooth (s), thus wire hair could be in the homozygous (SS) and heterozygous form (Ss) and the smooth hair can only be expressed in the homozygous recessive form (ss).
thus, in a cross between homozygous wire haired and smooth haired, we will have:
homozygous wire haired homozygous smooth haired
P gen SS x ss
F1 gen. Ss
phenotype: wire haired
The correct genotypes of the parents are ggrr for yellow pods with wrinkled seeds and GgRr for <span>heterozygous for green pods with round seeds.
If the heterozygous individuals for both traits express have green pods and round seeds that tells us that these traits are dominant.
In the gross presented below, you can see that the offspring will have 4 different genotypes, all present in an equal percentage:
</span><span>gGrR 25%
</span><span>gGrr </span>25%
<span>ggrR 25%
</span><span>ggrr 25%</span>
An allele. It’s a variant of a cell.