Transcript Hardy & Weinberg - local.brookings.k12.sd.us

5 CONDITIONS REQUIRED TO
MAINTAIN GENETIC EQUILIBRIUM
FROM GENERATION TO GENERATION
Must be random mating
1. _________________________
Population must be large
2. _________________________
movement in or out
3. No
_________________________
No mutations
4. _________________________
No natural selection
5. _________________________
RANDOM MATING
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POPULATION MUST BE LARGE
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NO MOVEMENT IN OR OUT
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NO MUTATIONS
Genetic variation is found naturally in
all populations
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NO NATURAL SELECTION
Some organisms in a population are
less likely to survive and reproduce
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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
TT
Tt
t
Tt
tt
ALLELES in population
T = _____
p
p
+
q
q
t = _____
=1
T
t
T
pp
pq
t
pq
qq
p2 + 2pq + q2 = 1
GENOTYPES in population
Homozygous dominant = ________
p X p = p2
Homozygous recessive = ________
q X q = q2
Heterozygous = __________
pXq
= 2 pq
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
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
4/16=0.25
q
p
p2
2pq
1- 0.5 = 0.5 (0.5) 2 = 0.25 2(0.5)(0.5)=0.5
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
360/1000=0.36
q
p
p2
2pq
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
100-91= 0.09
q
p
1 - 0.3= 0.7
p2
2pq
(0.7)2 = 0.49 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
1/10,000=0.0001
q
p
p2
2pq
1-0.01 = 0.99 (0.99)2 = 0.98 2(0.99)(0.01)=
0.0198
Heterozygous = 0.0198 = 2%
individuals