Answer:c) glucose
Explanation: The complete balanced chemical equation is C6H12O6 + 6O2 --> 6CO2 + 6H2O + Energy.
The equation above represents cellular respiration. During respiration, one molecule of glucose (C6H12O6) is broken down in the presence of six molecules of oxygen (O2) to produce six molecules of carbon dioxide (6CO2), six molecules of water (6H2O) and energy (ATP).
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Answer:
The correct answer is option d.
Explanation:
The production of Acetyl-CoA takes place by the dissociation of both carbohydrates and lipids in the process of glycolysis and beta-oxidation. It then moves into the TCA cycle in the mitochondria and combines with oxaloacetate to give rise to citrate.
In the given case, no labeling will be found in the acetyl-CoA. The labeled C3 and C4 carbon of glucose signify the carboxyl carbon of pyruvate. In the succeeding reactions of the transformation of pyruvate to acetyl-CoA, the carboxyl carbon gets lost in the form of carbon dioxide. Thus, acetyl-CoA does not comprise any labeled C3 and C4 of glucose.
Answer:
Frequency of allele A1- 0.41
Explanation:
In Hardy weinberg equilibrium,
P refers to the dominant allele
q refers to the recessive allele
The allele frequency will be p+q=1
The genotypic frequency is- P²+q²+2pq=1
P²= genotype of dominant trait ( A1A1)- 77
2pq= genotype of heterozygotes (2pq)- 65
q²= genotype of recessive trait (A2A2)- 123
Total number of offsprings= 77+ 65+ 123
= 265
Now to calculate allele frequency of A1=
= 77/265 + 1/2( 65//265)
= 0.290+ 0.122
= 0.413
Thus, 0.41 is correct.
