in a population of 1,000 mice, there are 400 mice that show the dominant trait of being dark-haired. using hardy-weinberg, determine the number of mice that are in each of the three genotypes for this trait.

If D is dominant and d is recessive, 160 with DD, 360 with Dd, and 480 with dd

Explanation:

if p= the frequency of the dominant allele (D for dark hair, perhaps?)

and q= the frequency of the recessive allele (d for not dark hair?)

remember that p + q = 1

400/1000= frequency of D= 0.4

0.4 + q = 1

q= 0.6 = frequency of d

remember also that p^2 +2pq + q^2 = 1

p^2 = (0.4)(0.4)= 0.16 = DD

q^2 = (0.6)(0.6) = 0.36 = Dd

2pq= (2)(0.4)(0.6)= 0.48 = dd

We now have the frequencies of each genotype and can apply them to the population size.

(0.16)(1000 mice) = 160 mice with DD

(0.36)(1000 mice) = 360 mice with Dd

(0.48)(1000 mice) = 480 mice with dd

the answer that best describes the relative refractory period in the neuron is the resting membrane potential is restored, but concentrations of sodium and potassium are not restored to their original state.

the refractory period is the time during which another stimulus given to the neuron will not lead to a second potential. thus, because na+ channels are inactivated during this time, additional depolarizing stimuli do not lead to new action potentials.  

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