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

RANDOM MATING http://www.wasatchcomputers.net/gallery/elk_fight.jpg http://ideiasemdesalinho.blogs.sapo.pt/arquivo/Peacock%20with%20its%20tail%20fanned%20out_Tony%20Ruta.jpg

POPULATION MUST BE LARGE http://www.sturgisrallydaily.com/gallery/full/crowd.jpg

NO MOVEMENT IN OR OUT http://whiteafrican.com/wp-content/Wildebeast%20Migration-East%20Africa.jpg

Genetic variation is found naturally in all populations NO MUTATIONS Genetic variation is found naturally in all populations http://www3.nationalgeographic.com/animals/images/primary/zebra-herd.jpg

Some organisms in a population are NO NATURAL SELECTION Some organisms in a population are less likely to survive and reproduce http://www.cartoonstock.com/newscartoons/cartoonists/rrs/lowres/rrsn69l.jpg

Hardy & Weinberg Who? Godfrey Hardy 1877-1947 Wilhelm Weinberg 1862-1937 They developed an equation that predicted the relative frequency of alleles in a population based on the frequency of the phenotypes in a population.

The Hardy-Weinberg Equation p2 + 2pq + 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

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

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

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

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   1- 0.5 = 0.5 2(0.5)(0.5)=0.5 (0.5) 2 = 0.25 Recessive allele = q = 0.5 = 50%

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

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

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

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   1 - 0.3= 0.7 2(0.7)(0.3)=0.42 Dominant allele = p = 0.7 = 70%

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

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   1-0.01 = 0.99 (0.99)2 = 0.98 2(0.99)(0.01)= 0.0198 Heterozygous = 0.0198 = 2% individuals