chapter 23 using the hardy-weinberg equation. p 2 + 2pq + q 2 = 1 p + q = 1 chapter 23 using the...
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Chapter 23Using the Hardy-Weinberg Equation
p2 + 2pq + q2 = 1
p + q = 1
Chapter 23 Using the Hardy-Weinberg Equation
A scientist goes out into the field and counts rabbits of different colors.
Out of 1000 rabbits counted, 600 have the recessive trait of white fur.
Abundance = 600/1000 = 0.6
q2 = 0.6
p2 + 2pq + q2 = 1
p + q = 1
Chapter 23 Using the Hardy-Weinberg Equation
q2 = 0.6
q = √0.6 = 0.77
p = 1 – q = 0.23
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
B b
B BB Bb
b Bb bb
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
The sample of 1000 rabbits includes: 50 homozygous dominant
350 heterozygous
600 homozygous recessive
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
The sample of 1000 rabbits includes: 50 homozygous dominant
350 heterozygous
600 homozygous recessive
1000 individual rabbits
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
The sample of 1000 rabbits includes: 50 homozygous dominant
350 heterozygous
600 homozygous recessive
1000 individual rabbits
450 dominant (B) alleles
1550 recessive (b) alleles
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
The sample of 1000 rabbits includes: 50 homozygous dominant
350 heterozygous
600 homozygous recessive
1000 individual rabbits
450 dominant (B) alleles
1550 recessive (b) alleles
2000 alleles
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
Then the environment changes. Global warming causes all the snow to melt. Suddenly the white rabbits stand out against the background and ¾ of them get eaten by predators before they can breed.
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
B b
B BB Bb
b Bb bb
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
B b
B BB Bb
b Bb bb
Chapter 23 Using the Hardy-Weinberg Equation
p = 0.23
q = 0.77
1.00
p2 = 0.05
2pq = 0.35
q2 = 0.60
1.00
The sample of 1000 rabbits now includes: 50 homozygous dominant
350 heterozygous
150 homozygous recessive
550 individual rabbits
450 dominant (B) alleles
650 recessive (b) alleles
1100 alleles
Chapter 23 Using the Hardy-Weinberg Equation
The sample of 1000 rabbits now includes: 50 homozygous dominant
350 heterozygous
150 homozygous recessive
550 individual rabbits
450 dominant (B) alleles
650 recessive (b) alleles
1100 alleles
When this F1 generation breeds,
p = 450/1100 = 0.41
q = 650/1100 = 0.59
Chapter 23 Using the Hardy-Weinberg Equation
When this F1 generation breeds,
p = 450/1100 = 0.41
q = 650/1100 = 0.59
p2 + 2pq + q2 = 1
(.41)2 + 2(.41)(.59) + (.59)2
Chapter 23 Using the Hardy-Weinberg Equation
When this F1 generation breeds,
p = 450/1100 = 0.41
q = 650/1100 = 0.59
p2 + 2pq + q2 = 1
(.41)2 + 2(.41)(.59) + (.59)2
p = 0.41
q = 0.59
1.00
p2 = 0.17
2pq = 0.48
q2 = 0.35
1.00
Chapter 23 Using the Hardy-Weinberg Equation
When this F1 generation breeds, only 35% of the newborn rabbits will be white.
p = 0.41
q = 0.59
1.00
p2 = 0.17
2pq = 0.48
q2 = 0.35
1.00
Let’s try it!
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LAB EIGHT POPULATION GENETICS AND EVOLUTION