chapter 23 using the hardy-weinberg equation. p 2 + 2pq + q 2 = 1 p + q = 1 chapter 23 using the...

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