1. Population must be large
5 CONDITIONS REQUIRED TO
MAINTAIN GENETIC EQUILIBRIUM
FROM GENERATION TO GENERATION
1. _________________________
2. _________________________
3. _________________________
4. _________________________
5. _________________________
Must be random mating
Population must be large
No movement in or out
No mutations
No natural selection

2. RANDOM MATING http://www.wasatchcomputers.net/gallery/elk_fight.jpg

3. POPULATION MUST BE LARGE

4. NO MOVEMENT IN OR OUT

5. Genetic variation is found naturally in all populations
NO MUTATIONS
Genetic variation is found naturally in all populations

6. Some organisms in a population are
NO NATURAL SELECTION
Some organisms in a population are
less likely to survive and reproduce

7. Hardy & Weinberg
Who?
Godfrey Hardy
Wilhelm Weinberg
They developed an equation that predicted the relative frequency of alleles in a population based on the frequency of the phenotypes in a population.

8. The Hardy-Weinberg Equation
p pq + q2 = 1
p2 = the frequency of homozygous dominant genotype
2pq = the frequency of heterozygous genotype
q2 = the frequency of homozygous recessive genotype
p + q = 1
p= frequency of dominant allele
q= frequency of recessive allele

9. T t T t p + q = 1 T t T T T t t t q ALLELES in population T = _____

10. T t T t pq pp pq qq p2 + 2pq + q2 = 1 p X p = p2 p X q = pq 2 q X q
GENOTYPES in population
Homozygous dominant = ________
Homozygous recessive = ________
Heterozygous = __________
p X p
= p2
q X q
= q2
p X q
= pq 2

11. In a population of pigs color is determined by one gene
In a population of pigs color is determined by one gene. If the black allele (b) is recessive and the white allele (B) is dominant, what is the frequency of the black allele in this population?
p + q = 1
P2 + 2pq + q2 = 1 q2 q p p2
2pq

12. p + q = 1 p2 + 2pq + q2 = 1 Recessive allele = q = 0.5 = 50%
In a population of pigs color is determined by one gene. If the black allele (b) is recessive and the white allele (B) is dominant, what is the frequency of the black allele in this population?
p + q = 1
p2 + 2pq + q2 = 1 q2 q p p2
2pq
4/16=0.25
= 0.5
2(0.5)(0.5)=0.5
(0.5) 2 = 0.25
Recessive allele = q = 0.5 = 50%

13. In a population of 1000 fruit flies, 640 have red eyes and the remainder have sepia eyes. The sepia eye trait is recessive to red eyes. How many individuals would you expect to be homozygous for red eye color?
p + q = 1
p2 + 2pq + q2 = 1 q2 q p p2
2pq

14. In a population of 1000 fruit flies, 640 have red eyes and the remainder have sepia eyes. The sepia eye trait is recessive to red eyes. How many individuals would you expect to be homozygous for red eye color?
p + q = 1
p2 + 2pq + q2 = 1 q2 q p p2
2pq
360/1000=0.36
1 – 0.6 = 0.4
(0.4)2 = .16
2 (0.4)(0.6) = 0.48
HOMOZYGOUS DOMINANT = q2 = 0.36 = 36% individuals
In this population, 1000 X .36 or 360 individuals

15. In a population of squirrels, the allele that causes bushy tail (B) is dominant, while the allele that causes bald tail (b) is recessive. If 91% of the squirrels have a bushy tail, what is the frequency of the dominant allele?
p + q = 1
p2 + 2pq + q2 = 1 q2 q p p2
2pq

16. In a population of squirrels, the allele that causes bushy tail (B) is dominant, while the allele that causes bald tail (b) is recessive. If 91% of the squirrels have a bushy tail, what is the frequency of the dominant allele?
p + q = 1
p2 + 2pq + q2 = 1 q2 q p p2
2pq
(0.7)2 = 0.49
100-91= 0.09
= 0.7
2(0.7)(0.3)=0.42
Dominant allele = p = 0.7 = 70%

17. In the U.S. 1 out of 10,000 babies are born with Phenylketonuria, a recessive disorder that results in mental retardation if untreated. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele?
p + q = 1
p2 + 2pq + q2 = 1 q2 q p p2
2pq

18. In the U.S. 1 out of 10,000 babies are born with Phenylketonuria, a recessive disorder that results in mental retardation if untreated. Approximately what percent of the population are heterozygous carriers of the recessive PKU allele?
p + q = 1
p2 + 2pq + q2 = 1 q2 q p p2
2pq
1/10,000=0.0001
= 0.99
(0.99)2 = 0.98
2(0.99)(0.01)=
Heterozygous = = 2%
individuals