1. HARDY WEINBERG

2. What is Hardy-weinberg Population?
a population that is in genetic equilibrium (not evolving) due to (five conditions) …
no mutations
large population size
no migration
random mating
no natural selection
**if any one of this conditions is not met, the population will evolve (allele frequencies will change from generation to generation)

3. What is Hardy-weinberg Population?
is represented by two mathematical equations…
p2 + 2pq + q2 =1 and p + q = 1
p= frequency of dominant allele
q= frequency of recessive allele
p2= frequency of homozygous dominant
2pq= frequency of heterozygous
q2= frequency of homozygous recessive

4. Problem one
A study on blood types (M and N are codominant) in a population found the following genotypic distribution among the people sampled: 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N and the three genotypic frequencies.
Start with …
How many people total are in the population?
= 3100

5. Allele frequency Shows the percentage of alleles in the group
p = % of M alleles
q = % of N alleles
the total alleles in the population = number of individuals in the population X 2
p = 1101 X / X 2 = .59
q = 503 X / X 2 = .40
Now…
Calculate the genotypic frequency for MM, MN and NN.

6. Genotype frequency
% of MM= 1101/3100= .36 or (.59)2 =.35
% of MN= 1496/3100= .48 or 2(.59)(.40) = .47
% of NN= 503/3100 = or (.40)2 = .16
Genotype frequency determines the % of Homo dominant (MM), heterozygous (MN) and homo recessive (NN)
MM= p2
MN = 2pq
NN = q2

7. Problem 2
The compound phenylthiocarbamide (PTC) tastes very bitter to most persons. The inability to taste PTC is controlled by a single recessive gene. In America, about 70% can taste PTC while 30% cannot (are non-tasters). Estimate the frequencies of the Taster (T) and nontaster (t) alleles in this population as well as the frequencies of the genotypes.
Hint: start by calculating q2 (think about why)

8. Allele frequency 30% cannot taste PTC, so 30% of population is aa
aa= q2 = .30
so q = .5477
p + q = 1
p =
p = .4523

9. Genotype frequency p = .4523 So, AA = p2 = .2045
We know that aa is .30
So, Aa = 2pq =
20.5% of population is AA
30% of population is aa
49.6% of population is Aa

10. Problem 3
In another study of human blood groups, it was found that among a population of 400 individuals, 230 were Rh+ and 170 were Rh-. Assuming that this trait (i.e., being Rh+) is controlled by a dominant allele (D), calculate the allele frequencies of D and d. How many of the Rh+ individuals would be expected to be heterozygous?

11. Allele Frequency p + q = 1 We know that Rh- is recessive
There are 170 people who are Rh- out of 400
So, 170/400 = 42.5% of population are aa
So q2 = .425
So q = .652
So p = = .348
So 65.2% of the alleles are recessive while 34.8% of the alleles are dominant

12. How many of the Rh+ individuals would be expected to be heterozygous
2pq= Aa
2 (.348) (.652)= Aa
So 2pq = .453
So 45.3% of population is Aa
.453 (400 people) = 181 people in the population are Aa

13. Problem one
A study on blood types (M and N are codominant) in a population found the following genotypic distribution among the people sampled: 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N and the three genotypic frequencies.

14. Problem 2
The compound phenylthiocarbamide (PTC) tastes very bitter to most persons. The inability to taste PTC is controlled by a single recessive gene. In America, about 70% can taste PTC while 30% cannot (are non-tasters). Estimate the frequencies of the Taster (T) and nontaster (t) alleles in this population as well as the frequencies of the genotypes.
Hint: start by calculating q2 (think about why)

15. Problem 3
In another study of human blood groups, it was found that among a population of 400 individuals, 230 were Rh+ and 170 were Rh-. Assuming that this trait (i.e., being Rh+) is controlled by a dominant allele (D), calculate the allele frequencies of D and d. How many of the Rh+ individuals would be expected to be heterozygous?