1. Chapter 23 Using the Hardy-Weinberg Equation

2. p 2 + 2pq + q 2 = 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 q 2 = 0.6

3. p 2 + 2pq + q 2 = 1 p + q = 1 Chapter 23 Using the Hardy-Weinberg Equation q 2 = 0.6 q = √0.6 = 0.77 p = 1 – q = 0.23

4. Chapter 23 Using the Hardy-Weinberg Equation B b B BB Bb b Bb bb

5. Chapter 23 Using the Hardy-Weinberg Equation B b B BB Bb b Bb bb p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 0.60 1.00

6. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 0.60 1.00 The sample of 1000 rabbits includes: 50 homozygous dominant 350 heterozygous 600 homozygous recessive

7. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 0.60 1.00 The sample of 1000 rabbits includes: 50 homozygous dominant 350 heterozygous 600 homozygous recessive 1000 individual rabbits

8. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 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

9. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 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

10. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 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.

11. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 0.60 1.00 B b B BB Bb b Bb bb

12. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 0.60 1.00 B b B BB Bb b Bb bb

13. Chapter 23 Using the Hardy-Weinberg Equation p = 0.23 q = 0.77 1.00 p 2 = 0.05 2pq = 0.35 q 2 = 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

14. Chapter 23 Using the Hardy-Weinberg Equation When this F1 generation breeds, p = 450/1100 = 0.41 q = 650/1100 = 0.59 p 2 + 2pq + q 2 = 1 (.41) 2 + 2(.41)(.59) + (.59) 2

15. Chapter 23 Using the Hardy-Weinberg Equation When this F1 generation breeds, p = 450/1100 = 0.41 q = 650/1100 = 0.59 p 2 + 2pq + q 2 = 1 (.41) 2 + 2(.41)(.59) + (.59) 2 p = 0.41 q = 0.59 1.00 p 2 = 0.17 2pq = 0.48 q 2 = 0.35 1.00

16. 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 p 2 = 0.17 2pq = 0.48 q 2 = 0.35 1.00