Answer:
Therefore, 98% of the US population have no alleles for PKU
Explanation:
The Hardy-Weinberg equilibrium states that the amount of genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors.
The Hardy-Weinberg equilibrium is expressed quantitatively using a mathematical equation known as the Hardy-Weinberg equation. the equation is given below:
p² + 2pq + q² = 1
also, p + q = 1
Given a pair of alleles, S and s with A dominant and a recessive
where p is the frequency of the dominant allele in the population,
q is the frequency of the recessive allele in the population,
p² represents the frequency of the (SS) dominant genotype,
q² represents the frequency of the (ss) recessive genotype,
2pq represents the frequency of the heterozygous genotype
From the given question,
q² = 1/10000 = 0.0001
q = 0.01
from p + q = 1
p = 1 - 0.01 = 0.99
p² = 0.98
Therefore, 98% of the US population have no alleles for PKU
the answer is a <u><em>phospholipid bilayer</em></u>
it's called phospholipid because it's made of a group of lipids that has a phosphate group and bilayer because it's made of hydrophilic heads that are exposed to the solution and hydrophobic tails that are directed inwards
This will be a major concern to transgenic crop manufacturers because
outcrossing can lead to loss of introduced genes from transgenic plants.
Outcrossing involves the process of crossing two different breeds thereby
introducing unrelated genetic material into it.
Crop manufacturers will be concerned if the genes from genetically modified
corn frequently move to non-genetically modified corn plants because as time goes on, the trait which made the genetically modified corn will be lost as different variations and more dominant traits will make such traits recessive and lost.
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If there is no antigen the blood group is 0 !!
extra info ; if both antigen - AB
