In a population of 525 flowering plants, the red-flower allele is dominant over the white-flower allele. in this population, 365 of the plants have red flowers and 160 have white flowers. use the hardy–weinberg equation to determine the allele frequency for the recessive, white-flower allele in the population. round your response to two decimal places.

The allele frequency of the white-flower allele is 0.55

OR

f(a) = 0.55

Explanation:

Let red=A and white=a; A>a

We know the the Hardy-Weinberg equation to be:

p²+2pq+q²

But we are not given the individual number of plants that are homozygous for red (AA) and heterozygous for red coloring (Aa).

f(AA)+f(Aa) = 365/525 = 0.695

f(aa) = 160/525 = 0.305

q²=f(a)²=f(aa)= 0.305

∴ q = f(a) = = 0.55

And we know p+q = 1

∴  f(A) = p = 1-q = 1-0.55 = 0.45

f(Aa) = 2pq = 2(0.45)(0.55) = 0.495

To make sure your answer is correct: p²+2pq+q²=1

(0.45)²+2(0.45)(0.55)+(0.55)=1

1=1

both anaerobic and aerobic respiration use glucose (or other carbohydrates), proteins and lipids to produce atp and nadh. cell uses those energy molecules, obtained from food and   breaks them down during the chemical reactions of respiration. in anaerobic respiration, nadh is regenerated to nad+ by enzymes lactate dehydrogenase (lactate is produced, e.g. in muscle cells) or by alcohol dehydrogenase where ethanol is produced (alcohol fermentation), while in aerobic respiration, electrons from nadh are deposited in the mitochondria to regenerate nad+ (electron transport chain. anaerobic respiration produces 2 atp per molecule of glucose compared to aerobic respiration where about 38 atp molecules are produced.