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Chapter 23 – Part 1 Part 2 After Break. Slide 2 of 15 Context Evolution appears to be the typical situation Sometimes evolution is not occurring Hardy-Weinberg.

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Presentation on theme: "Chapter 23 – Part 1 Part 2 After Break. Slide 2 of 15 Context Evolution appears to be the typical situation Sometimes evolution is not occurring Hardy-Weinberg."— Presentation transcript:

1 Chapter 23 – Part 1 Part 2 After Break

2 Slide 2 of 15 Context Evolution appears to be the typical situation Sometimes evolution is not occurring Hardy-Weinberg Equilibrium Conditions that specify if evolution is NOT happening Equilibrium = no evolution More of a theoretical case than a real test But affords comparison to evolution

3 Slide 3 of 15 Definitions Population Genetics – study of how populations change genetically over time Population – Group of same species individuals that can and do interbreed successfully Gene pool – all of the alleles at all loci in all the members of a population Fixed – only one allele exists for a locus in the population More fixed alleles = lower species diversity

4 Slide 4 of 15 Hardy-Weinberg If gene pools are NOT Evolving, can use Hardy- Weinberg Describes a population that is NOT evolving Allelic frequencies will remain constant Gene frequencies remain constant throughout the generations

5 Slide 5 of 15 Conditions for Hardy-Weinberg 1. No mutations 2. Random mating 3. No natural selections 4. Large population size 5. No gene flow Immigration or emigration Genes coming into or leaving the population

6 Slide 6 of 15 Math in Biology? p = Frequency of the dominant allele = f(A) q = Frequency of the recessive allele = f(a) There are only 2 alleles (dominant + recessive), so p + q = 1 square both sides:(p + q) 2 = 1 2 expand the binomial:p 2 + 2pq + q 2 = 1

7 Slide 7 of 15 Hardy-Weinberg Math (Page 2) p 2 = pp = f(aa) = Frequency of homozygous dominant pq = f(Aa) = frequency of heterozygote However, remember from Punnett Square, there were 2 heterozygotes for the F1 generation; So 2pq = f(Aa) q 2 = qq = f(aa) = frequency of homozygous recessive

8 Slide 8 of 15 In Summary Summary: p 2 + 2pq + q 2 = 1 f(AA) + f(Aa) + f(aa) = 1 f(Homo. Dom.) + f(Hetero.) + f(Homo. Recess.) = 1 Are there any other genotypes? Makes sense?

9 Slide 9 of 15 Problem (Part 1) 1. A trait has two characters, dominant (A) and recessive (a). In a population of 500 individuals in Hardy- Weinberg equilibrium, 25% display the recessive phenotype (aa). a) What is the frequency of the dominant allele in the population? b) What is the frequency of the recessive allele in this population?

10 Slide 10 of 15 Answer f(aa) = 0.25 OR q 2 = 0.25 q = (0.25) = 0.5 OR f(a) = 0.5 So the frequency of the recessive allele = 0.5 From before: p + q = 1 AND q = 0.5, so p = 0.5 = f(A) So the frequency of the dominant allele = 0.5

11 Slide 11 of 15 Part 2 of Problem What are the frequencies of a) homozygous dominant genotype? b) Heterozygous genotype? c) homozygous recessive genotype?

12 Slide 12 of 15 Part 2 of Problem (Answer) What are the frequencies of a) homozygous dominant genotype? Since p = 0.5, p 2 = (0.5) 2 = 0.25 b) Heterozygous genotype? Since p = 0.5 & q = 0.5,2pq = 2(0.5)(0.5) = 0.5 c) homozygous recessive genotype? Yep, it was given, but if we must: q = 0.5, q 2 = (0.5) 2 = 0.25

13 Slide 13 of 15 Part 3 of Problem How many individuals have the a) homozygous dominant genotype? b) Heterozygous genotype? c) homozygous recessive genotype?

14 Slide 14 of 15 Part 3 of Problem (Answer) How many individuals have the a) homozygous dominant genotype? 500 members & p 2 = 0.25, (0.25)(500) = 125 b) Heterozygous genotype? 500 members & p = 0.5, q = 0.52pq = 0.5 So 500 (0.5) = 250 heterozygotes c) homozygous recessive genotype? 500 members & q 2 = 0.25, (0.25)(500) = 125

15 Slide 15 of 15 Part 4 of Problem How many individuals have the a) Dominant phenotype? b) Recessive phenotype?

16 Slide 16 of 15 Part 4 of Problem (Answer) How many individuals have the a) Dominant phenotype? = (Homo. Dom.) + (Hetero.) = = 375 b) Recessive phenotype? = Homo. Recess. = 125


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