The answer is stabilizing selection.
<span>Sickle-cell anemia is a recessive disorder caused by the presence of two recessive alleles "s", so genotype is "ss". This disorder is characterized by sickle hemoglobin. In an area with malaria, heterozygous individuals "Ss" (with one dominant allele and one recessive allele) have an advantage. These individuals will have both normal and sickle hemoglobin. But pathogen that causes malaria affect only normal hemoglobin, so heterozygous individuals will have half of the hemoglobin resistant to the pathogen and those individuals are resistant to malaria.</span>
Stabilizing selection favors heterozygotes Ss, disruptive selection favors dominant (SS) and recessive (ss) homozygotes, while directional selection favors dominant (SS) or recessive (ss) homozygote. Since in this example, people with genotype Ss (heterozygotes) are in advantage, then this is an example of stabilizing selection.
I hope you meant, a cross between BbPp and BbPp, else for bbpp x bbpp, all the progeny will be bbpp.
We need to work out a Punnett square to determine the genotypes resulting from the cross between BbPp and BbPp. The dyhybrid cross results in 16 different genotypes, of which a few have the same phenotypes.
It can be seen that there are 4 different kinds of phenotypes present, in the ratio 9:3:3:1.
The energy produced in glycolysis is atp adenosine triphosphate. C10h16N5O13P3
Answer: (A) Genetic changes can be designed to make cows yield more milk.
Explanation: Milk yield per cow has more than <u>doubled</u> in the previous 40 years and many cows now produce more than 20,000 kg of milk per lactation due to genetic changes. <u>This shows that genetic changes to make cows yield more milk has already occurred.</u>
The black horse's genotype would be BB since it is homozygous and black hair is dominant. While the heterozygous black horse would be Bb.
