Lecture 8: Population Genetics Flashcards
What is a population?
A group of interbreeding individuals of the same species that inhabit the same space and time.
The key to a population is that they must be able to interbreed.
What is a gene pool?
All of the alleles PRESENT and CARRIED by the population
What is population genetics?
Changes in the frequency of the alleles in a gene pool
over time.
Locus vs Allele
Define both
- Locus:
– location of a gene/marker on the chromosome - Allele:
– one variant form of a marker at a particular locus
Locus vs Allele …draw image of Homologous Chromosomes..
SLIDE 4
What is Hardy Weinberg Equation?
“The Hardy-Weinberg law can be used to find the frequency of disease alleles in a population and the carrier frequency”
Allele Frequencies for genotypes AA, Aa, aa
- Frequency of “A” in a population is called “p”
- Frequency of “a” in a population is called “q”
fA + fa = 1
p + q = 1
p = fAA + ½ fAa
q = faa + ½ fAa
….draw punnet square
Hardy Weinberg Equation
* Frequency of allele “A: in a population is called “p”
- Frequency of allele “A: in a population is called “p”
– In a population of gametes, the probability that both the
egg and the sperm contain the “A” allele is p x p = p2
Hardy Weinberg Equation
– Frequency of allele “a” in a population is called “q”
- Frequency of allele “a” in a population is called “q”
– In a population of gametes, the probability that both the
egg and the sperm contain the “a” allele is q x q = q2
– In a population of gametes, the probability that the egg
and the sperm contain different alleles is
* (p x q) + (q x p) = 2 pq
(p x q) = Male gices “A” and female gives “a”
+ = or
(q x p) = female gives “a” and male gives “A”
Punnet square …MALE (sperm) x FEMALE (Eggs)
EQUATION: 2 x Aa = 2pq
punnet square ..slide 8
What are the HARDY WEINBERG EQUATION? =2
p2 + 2pq + q2 = 1
OR
(p+q)2=1
UNDERSTANDING the Hardy Weinberg Equation…
For a population to be in Hardy Weinberg Equilibrium,
the OBSERVED GENOTYPE frequencies must match those PREDICTED BY THE EQUATION p2 + 2pq + q2
— Determine “p”: equals frequency of AA + ½(frequency of Aa)
— Determine “q”: equals frequency of aa + ½(frequency of Aa)
Then
— p2 = predicted genotype frequency of AA
— q2 = predicted genotype frequency of aa
— 2pq = predicted genotype frequency of Aa
Statistical test required : chi-squared to test observed vs predicted
How can alleles for recessive disorders persist in the population?
1 * Recessive disorders are quite rare
2 * ie. the alleles are relatively rare
3 * most recessive alleles are in heterozygotes
4 * Thus, rare disease-causing recessive alleles persist in the population in heterozygote carriers, even if
they are lethal when homozygous
An Autosomal Recessive Disease…2
- Affects 1/100 people in a population (ie. 1%)
- The frequency of people with the aa genotype is 0.01
– i.e. q2 = 0.01
What is the frequency of carriers (= heterozygotes) ?
1 * q2 = 0.01 (given)
2 * Therefore q = 0.1 (take square root)
3 * p = 1-q = 0.9 (since p+q=1)
4 * 2pq = 0.18 = almost one in every five