Answer:
Half would have the genotype Aa.
Explanation:
Given that the genotype of both of the parents was "Aa"; both parents would make two types of gametes. The 50% gametes from each parent would carry allele "A" while the rest 50% gametes from each parent would carry allele "a".
The random fusion of gametes of two parents would give the progeny with genotype ratio 1/4 AA: 1/2 Aa: 1/4 aa.
Answer:
The middle carbon atom of pyruvate in the TCA cycle becomes the carboxyl carbon of acetate and hence the newly added (upper) carboxyl group in citrate
Explanation:
Before entering the TCA cycle, pyruvate is converted into Acetyl-CoA by oxidative decarboxylation of pyruvate. 1st two carbons of glucose to be fully oxidized. This reaction is catalyzed by Pyruvate dehydrogenase.
The middle carbon atom of the pyruvate in the TCA cycle becomes the carboxyl carbon of the acetate as the carboxyl group is removed from the pyruvate , releasing CO2. NAD is reduced to NADH. Acetyl group is transferred to coenzyme-A , resulting in acetyl coA
