Answer:
3/16
Explanation:
<u>According to Mendel's law of independent assortment of genes, when a dihybrid cross involves two genes that assort independently and one of the parents is dominant for the two genes and the other is recessive, the phenotypic ratio of the offspring at F2 would be 9:3:3:1</u>. The proportion of the offspring with the dominant parental traits would be 9/16, those with the recessive parental traits would be 1/16, while those with mixed traits would be 3/16 each.
Assuming the eye color is represented by E and the wing shape is represented by W. At F2
EeWw x EeWw
E_W_ - 9/16 (dominant for both eye color and wing shape)
E_ww - 3/16 (dominant for eye color and recessive for wing shape)
eeW_ - 3/16 (recessive for eye color and dominant for wing shape)
eeww - 1/16 (recessive for both eye color and wing shape)
<em>Hence, the proportion of the offspring with dominant phenotype for eye color and recessive phenotype for wing shape would be </em><em>3/16.</em>
The answer is
A
Behavior isolation is behaviors involved in mating which could also cause it to prevent mating
Answer:
Explanation:
1. Over time more and more molecules of DCPIP would be decreased by electrons taken from plastoquinone (Pq). This statement is true in the sense that DCPIP is an artificial electronic acceptor which is decreased by taking electrons from plastoquinone.
2.) Water splitting by Photosystem II would occur at higher levels than if the drug were administered without DCPIP. This statement is true.
3.)Events in Photosystem I would occur just as they would in the absence of drug and DCPIP. This statement is false because the drug obstructs the electron transport from plastoquinone to cytochrome B. So the events are abnormal because no protein transfer is involved in this mechanism.
