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Mini-Poster Symposium
Peppered moth evolution• gene pool • variation • phenotype • differential reporduction • survival • advantage • environment • allele
Gamehttp://www.techapps.net/interactives/pepperMoths.swf
Cockroach population Evolving?
1) Pick up a whiteboard and marker. Split it in half
2) Draw up a population of cockroaches. 5 dotted, 5 clear. D is the dotted allele and a is the clear allele.
3) On one half put all the cockroaches. Next to each cockroach put the genotype: AA, Aa or aa. THESE ARE INDIVIDUALS. Yell checkpoint
4) On the other side label gene pool just put alleles individually. THESE ARE ALLELE FREQUENCIES
How many A? Total A/TotalHow many a? Total a/Total
What do these symbols mean?
•p2•q2•2pq
•p•q
What are the genotype frequencies?What are the genotype frequencies?
Using Hardy-Weinberg equation
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): 1 - 0.4 = 0.6
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): 1 - 0.4 = 0.6
population: 100 cats84 black, 16 whiteHow many of each genotype?
population: 100 cats84 black, 16 whiteHow many of each genotype?
bbBbBB
p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16
Must assume population is in H-W equilibrium!Must assume population is in H-W equilibrium!
Hardy-Weinberg Equationp=frequency of one allele (A); q=frequency of the other allele (a);
p+q=1.0 (p=1-q & q=1-p)
• p2=frequency of AA genotype; 2pq=frequency of Aa genotype; q2=frequency of aa genotype;
• frequencies of all individuals must add to 1 (100%), so: p2 + 2pq + q2 = 1
G.H. Hardymathematician W. Weinberg
physician
Hardy Problem
• Calculate q2 Count the individuals that are homozygous recessive in the illustration above. Black is recessive to pink. Calculate the percent of the total population they represent. This is q2.
Calculate qQ2=4/16=0.25
q=0.5
p + q = l. You know q, so what is p, the frequency of the dominant allele?
P=0.5
Find 2pq 2pq = 2(0.5) (0.5) = 0.5 , so 50% of the population is heterozygous.
Video on Hardy-Weinberg
• Bozemanscience
Hardy-Weinberg=NO EVOLUTION
• To see what forces lead to evolutionary change, we must examine the circumstances in which the Hardy-Weinberg law may fail to apply.
There are five:• mutation• gene flow• genetic drift• nonrandom mating• natural selection
5 Agents of evolutionary change
Mutation
Gene FlowGenetic Drift
Selection
Non-random mating
Problem 1
• In a certain population of 1000 fruit flies, 640 have red eyes while 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?
• Hint: The first step is always to calculate q2! Start by determining the number of fruit flies that are homozygous recessive. If you need help doing the calculation, look back at the Hardy-Weinberg equation.
Solve the problem below
5 Agents of evolutionary change
Mutation
Gene FlowGenetic Drift
Selection
Non-random mating
Problem 2• The Hardy-Weinberg equation is useful for
predicting the percent of a human population that may be heterozygous carriers of recessive alleles for certain genetic diseases. Phenylketonuria (PKU) is a human metabolic disorder that results in mental retardation if it is untreated in infancy. In the United States, one out of approximately 10,000 babies is born with the disorder.
• Approximately what percent of the population are heterozygous carriers of the recessive PKU allele?
Natural selection NOT in Hardy-Weinberg
Fitness: contribution an individual makes to the gene pool of the next generation
3 types:• A. Directional• B. Diversifying• C. Stabilizing
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