Question

MCAT Biology Mini-Tutorial: Hardy-Weinberg Equilibrium

Answer 1

\({\bf{Motivation:}}\) to study the genetic changes within a population, quantitatively

\({\bf{Assumptions:}}\)
1. the population mates randomly
2. the population size is sufficiently large
3. no mutations within the population
4. no gene flow between populations
5. no natural selection

of course, this situation does not actually exist in nature, but it makes the calculations for Hardy-Weinberg possible

for a simple, Mendelian trait with dominant allele frequency p and recessive allele frequency q

p + q = 1 (since the only two alleles under consideration are p and q, their frequencies must add up to 1)

p^2 + 2pq + q^1

p^2 = homozygous dominant
pq = heterozygous
q^2 = recessive

Answer 2

\({\bf{Example:}}\) if 49 out of 100 individuals in a population are recessive for the given phenotype; find the expected # of heterozygous individuals in the population

if q^2 = 0.49 then q = 0.7 (the square root of 0.49)

using the allele-frequency-must-add-up-to-1 rule,

p = 1 - 0.7 = 0.3

therefore the # of heterozygous individuals can be found by calculating 2pq * 100, or

2(0.7)(0.3)*100 = 42 individuals = the sol'n

Answer 3

Anyway, that's the end of my tutorial, I hope it was a helpful resource. Source material is the 2nd Edition Barron's Prep book for the new MCAT