For this question, we will utilize a population of martians that is in hardy weinberg equilibrium. the dominant martian phenotype is the possession of 2 antennae. in this population, 84 of the martians have 2 antennae, while 16 lack antennae. what is the frequency of heterozygotes in this population?
a. o.16
b.0.4
c.0.6
d.0.24
d.0.48

d.0.48

Explanation:

When a population is in Hardy Weinberg equilibrium the genotypic frequencies are:

freq (AA) = p²

freq (Aa) = 2pq

freq (aa) = q²

p is the frequency of the dominant A allele and q is the frequency of the recessive a allele.

In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.

Therefore:

q²=16/100

q² = 0.16

q=√0.16

q = 0.4

And p+q=1, so:

p = 1 - q

p = 1-0.4

p = 0.6

The frequency of heterozygotes is:

freq (Aa) = 2pq = 2 × 0.4 × 0.6

freq (Aa) = 0.48

answer; /// i believe the correct answer is (thrombocytopenia); (a)

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.