Answer:
0.0139- frequency of heterozygotes in the population
Explanation:
Let q to the power of 2 represents the frequency of the homozygous recessive (aa) = 1/20000 = 0.00005
This, q = √q^2 = √0.00005 = 0.007
Since p + q = 1
p = 1 - 0.007 = 0.993
Using the formula: p^2 + 2pq + q^2
Where 2pq represents the frequency of the heterozygotes, thing we have
2 x 0.007 x 0.993
= 0.0139
If not parents had a recessive type O allele, they would have a 1/4 or 25% chance of having a child with type O blood
The greater the difference between the dry bulb temperature and the wet bulb temperature, the drier the air is. From the air temperature and the wet bulb temperature, the relative humidity of the air can be easily found.
The correct answer to this question is B. Some members of the Fugate family had a genetic condition called methemoglobinemia, that gave a blue color to their skin. The gene causing this disease is an autosomal recessive gene, so to have this condition you need to be homozygotes for this recessive gene. Gene flow was prevalent in the Fugate family since not all of the children had blue complexion, but only four of them. The three children that did not have blue complexion, had the dominant allele, probably through gene migration from another population nearby. Also, since it is easier nowadays to travel, two separate populations could meet more easily. This increases the incidence of gene flow and explains the reduction of the occurrence of the condition.
Answer:
awnser mine and ill awnser urs
Explanation:
