I'm pretty sure it's false if it isn't you can give me a low grade.
When a genetic population follows Hardy-Weinberg Equilibrium (HW), it states that certain biological tenets or requirements must be met. Given so, then HW states that the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimal. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
So the problem states that homozygous dominant individuals (p^2) account for 60%, or 0.60. Thus the square root (sr) of p^2 = p or the dominant allele frequency in the population. So sr(p^2) = sr(0.60) -->
p = 0.775 or 77.5%
Homozygous recessive individuals (q^2) account for 20%, or 0.20. Thus sr(q^2) = q or the recessive allele frequency in the population. So sr(q^2) = sr(0.20) --> q = 0.447 or 44.7%
But since 44.7% + 77.5% = 122.2%, which is not equal to 1, we have a situation in which the allele frequencies do not match up, therefore this population cannot be determined using the Hardy-Weinberg Equation.
Population 11 doesnt belong.
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i hope this helps my dudette
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Zane
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Whats cooler than being cool ?
I think the correct answer from the choices listed above is the last option. Polluted water sources would most likely affect one’s personal health since p<span>athogens would invade the body and cause disease, such as giardia. Pathogens would go with the water we take in our body and would disturb the processes inside causing illnesses.</span>
