Answer:
Parents: Yellow (Aarr) and Grey (AaRr)
Explanation:
Given:
allele A = yellow
allele R = black,
Heteroozygous = gray
Genotypes of the parents:
yellow (Aarr) - female
gray (AaRr) - gray
cross between these
Parents: Yellow (Aarr) and Grey (AaRr)
Gametes: (Ar, ar) and (AR, Ar, aR, ar)
F1 (Punnet square)
----|----- AR ------|------- Ar ------|------ aR -----|----- ar
Ar | AARr (gray) | AArr (yellow) | AaRr (gray) | Aarr (yellow)
ar | AaRr (gray) | Aarr (yellow) | aaRr (black) | aarr (cream)
Ratio: 3/8 yellow : 3/8 gray : 1/8 cream : 1/8 black
Answer:
O blood is a recessive trait, so none of the children will have O blood. The genes for either type A or Type B will be determined by the egg/sperm being fertilized with the reproductive cell of the other parent. the probability of receiving either of the gene versions is 1/2, so half of the children will have A blood and the other half will have B blood
A during cytokinesis
<span>Cleavage furrow forms during cytokinesis.</span>
In the matrix of mitochondria the reactions known as the citric acid or Krebs cycle produce a chemical called NADH. NADH is then used by enzymes embedded in the mitochondrial inner membrane to generate adenosine triphosphate (ATP). In ATP the energy is stored in the form of chemical bonds.
