Before meiosis a cell has the genotype AaBb . Which of the following genotypes is POSSIBLE for one of the daughter cells after both divisions of meiosis are finished?
A, AABB
B. AB
C. Aa
D. AaBb
Stopping the Calvin Cycle stops the Light Dependent Reactions. Why?
A. The Calvin cycle produces NADP+ which is used in light-dependent reactions
b. because the light-dependent reactions are needed any longer
c. the Calvin cycle produces sugar, which is used in the light-dependent reactions
d. the Calvin cycle produces light, which is used in the light-dependent reactions
1.The meiosis of the given parent cell will result in only one set of chromosomal genes i.e. the generated daughter cells will be haploid.So A and D not possible, Aa is also not possible because both genes and B should be present in daughter cells so correct answer is AB i.e. B answer.
2.The calvin cycle which involves fixation of carbon dioxide to form the final hexose sugar molecules utilses ATP and NADPH synthesized in Z cycle and so releases the NADP which have to be utilized by Z cycle or light dependent reactions.So if no NADP the cycle or reactions will not be able to occur.
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