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Population Genetics Unit 4 AP Biology

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Population Genetics Study of the frequency of particular alleles and genotypes in a population Ex. Suppose you want to determine the % of alleles and genotypes for Alu in the PV92 region.

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**Sample Data- Alu in the PV92 region**

100 AP Biology students total 45 students are +/+ 20 students are +/- 35 students are -/-

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**Allelic Frequencies & Genotype Frequencies**

Genotype Frequency % of a particular genotype that is present in a population Allelic Frequency % of particular allele in a population

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**Sample Problem 100 AP Biology students total**

45 students are +/+ 20 students are +/- 35 students are -/- Calculate the genotype frequencies Calculate the allelic frequencies

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**Solution- Genotype Frequencies**

Genotype frequency of +/+ 45 (+/+) / 100 students total = Genotype Frequency of +/- 20 (+/-) / 100 students total = Genotype Frequency of -/- 35 (-/-) / 100 students total =

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**Solution- Allelic Frequencies**

Step 1: Calculate the total # of alleles 100 students x 2 alleles each = 200 total alleles

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**Solution- Allelic Frequencies**

Allelic Frequency of “+” allele Each student with +/+ contributes 2 “+” alleles each 2 x 45 = 90 “+” alleles Each student with +/- contributes 1 “+” allele each 1 x = 20 “+” alleles = 110 “+” alleles Allelic Frequency = 110 “+” alleles /200 total =

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**Solution- Allelic Frequencies**

Allelic Frequency of “-” allele Each student with -/- contributes 2 “-” alleles each 2 x 35 = 70 “-” alleles Each student with +/- contributes 1 “-” allele each 1 x = 20 “-” alleles = 90 “-” alleles Allelic Frequency = 90 “-” alleles /200 total =

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**Practice Problem #1 325 ants total:**

140 ants have the genotype GG 75 ants have the genotype Gg 110 ants have the genotype gg Calculate the genotype and allelic frequencies.

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**Solution #1 Genotype Frequencies: GG = 140 / 325 = 0.43**

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**Solution #1 How many alleles total? Allelic Frequencies:**

325 X = alleles total Allelic Frequencies: G : ( (140 x 2) + 75) / = g : ( (110 x 2) ) / =

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**Question… What should all of the allelic frequencies add up to?**

They should add up to 1 What should all of the genotype frequencies add up to?

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**Gene Pool All the alleles in a population**

Alleles in the gene pool can change as a result of several different factors (mutations, immigration, natural selection)

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Evolution If the genotype and allelic frequencies are NOT changing from generation to generation then the population is NOT evolving.

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**Hardy Weinberg Equilibrium**

Mathematical model to describe a nonevolving population In real life, populations are almost never in Hardy Weinberg Equilibrium 2 variables: p = allelic frequency of one allele q = allelic frequency of other allele

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**Conditions of Hardy Weinberg**

Very large population size No Gene Flow (no moving in or out) No Mutations (no new alleles introduced) No Sexual Selection No Natural Selection (no allele causes the individual to survive better or worse) Not meeting these conditions would cause the allelic and genotype frequencies to change

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**Hardy Weinberg Equations**

p and q are frequencies of alleles in a population p + q = 1 (the 2 alleles make up 100% of the alleles) From this, there are 3 genotypes possible: Homozygous dominant (Ex. HH) Homozygous recessive (Ex. hh) Heterozygous (Ex. Hh)

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**Hardy Weinberg Equations**

Using the allelic frequencies (p and q), genotype frequencies can be calculated for each genotype p2 = genotype frequency for homozygous dominant (HH) 2pq = genotype frequency for heterozygous genotype (Hh) q2 = genotype frequency for homozygous recessive (hh)

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**Hardy Weinberg Equations**

p pq q2 = 1 Why is it 2pq? Because there are two heterozygous combinations possible (Hh and hH)

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**Hardy Weinberg Equations**

Based on probability p = 0.8 (allelic frequency for CR) q = 0.2 (allelic frequency for CW) Chance of CRCR = 0.8 x 0.8 = 0.64

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**Population in HW Equilibrium?**

You can use calculations to determine if a population is in Hardy Weinberg Equilibrium Let’s take practice problem #1: Actual allelic frequencies are: p = q = Actual genotype frequencies are: 0.43 (GG), (Gg), (gg)

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**If the population is in HW Equilibrium…**

If the population is in Hardy Weinberg equilibrium, then the calculated expected genotype frequencies should match the actual genotype frequencies. Using the Hardy Weinberg Equation: Expected GG Genotype frequency = p2 Expected Gg Genotype frequency = 2pq Expected gg genotype frequency = q2

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**Expected Genotype Frequencies**

p = , q = Expected Genotype frequencies: GG = p2 = (0.55)2 = Gg = 2pq = 2 (0.55)(0.45) = 0.50 gg = q2 = (0.45)2 = Actual genotype frequencies are: 0.43 (GG), (Gg), (gg) The actual genotype and expected genotype frequencies do not match the population is NOT in HW Equilibrium.

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**One more Practice Problem**

You are studying a ferret population for their nose sizes. B is the allele for a long nose, b is the allele for a short nose. 78 ferrets are BB 65 ferrets are Bb 21 ferrets are bb What are the actual allelic and genotype frequencies? Is the population in HW Equilibrium?

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**Solution Actual genotype frequencies: BB = 78 / 164 = 0.47**

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**Solution Actual Allelic frequencies: Total alleles = 2 x 164 = 328**

Allelic frequency of B = p = ( (2 x 78) + 65) / = Allelic frequency of b = q = ( (2 x 21) + 65) / =

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**Solution Expected genotype frequencies:**

BB = p2 = (0.67)2 = 0.45 Bb = 2pq = 2(0.67)(0.33) = 0.44 bb = q2 = (0.33)2 = 0.11 These do not match the actual frequencies, so the population does not appear to be in HW Equilibrium.

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