Answer:
Hi elmo!
Explanation:
ANd i think the answer is C!
hope this helps c:
4 is the correct answer hope you get it right!!!
Answer:
1500
Explanation:
Let's assume that the allele for yellow seed color is "Y" and the allele for green seed color is "y". Genotype of pure breeding yellow seeded plant would be "YY" and that of the green seeded plant would be "yy". A cross between YY and yy gives all heterozygous yellow seeded plants (Yy) in F1 progeny. Self pollination of two F1 plants (Yy x Yy) obtains F2 generation in 3 yellow: 1 green ratio.
The total population size of F2 generation = 2000
The proportion of yellow seeded plants in F2 generation = 3/4 (since the F2 phenotype ratio is given 3 yellow: 1 green)
Therefore, total number of yellow seeded plants in F2 progeny = 3/4 x 2000= 1500
Answer:
3
Explanation:
even if you calculate is still going to be 3
Phosphorylation of glucose adopt the pentose phosphate pathway.
<h3>What Phosphorylation of glucose do?</h3>
The phosphorylation of glucose maintains the downhill gradient for metabolism of glucose and extramitochondrial glucose flux through the use of pentose phosphate pathway. Induction of the pentose phosphate pathway and the generation of NADPH may also contribute to protection against death of cells.
Learn more about glucose here: brainly.com/question/397060