AP Bio Learning Target Illustrate gene pool dynamics
AP Bio Learning Target – Illustrate gene pool dynamics & microevolution Due today – Phylogenetic Trees & Variation & Gene Pools Chapter 21 Reading Guide due Monday AP Biology
There are 5 agents of evolutionary change Mutation Gene Flow Genetic Drift AP Biology Non-random mating Selection
Populations & gene pools a population is a localized group of interbreeding individuals u gene pool is a collection of alleles in a particular population (remember difference between alleles & genes!) u allele frequency is how common that allele is in the population (how many of A or a in whole population) u AP Biology
Evolution of populations § Evolution is a change in allele frequencies in a population u What conditions would cause allele frequencies not to change? u In a non-evolving population you would need to remove all agents of evolutionary change - very large population size (no genetic drift) - no migration (no gene flow in or out) - no mutation (no genetic change) - random mating (no sexual selection) - no natural selection (everyone is equally fit) AP Biology
Hardy-Weinberg equilibrium § In a non-evolving population allele frequencies are preserved (they are said to be in Hardy-Weinberg equilibrium), but: u natural populations are rarely in Hardy-Weinberg equilibrium u However it provides a useful model to measure if evolutionary forces are acting on a population G. H. Hardy APmathematician Biology W. Weinberg physician
Hardy-Weinberg theory § Counting Alleles assume 2 alleles = B, b u frequency of dominant allele (B) = p u frequency of recessive allele (b) = q u § frequencies must add up to 1 (100%), so: p+q=1 BB AP Biology Bb bb
Hardy-Weinberg theory § Counting Genotypes u u u frequency of homozygous dominant: p x p = p 2 frequency of homozygous recessive: q x q = q 2 frequency of heterozygotes: (p x q) + (q x p) = 2 pq § frequencies of all individuals must add to 1 (100%), so: p 2 + 2 pq + q 2 = 1 BB AP Biology Bb bb
H-W formulas § Alleles: p+q=1 B § Genotypes: p 2 + 2 pq + q 2 = 1 BB BB AP Biology b Bb Bb bb bb
Using Hardy-Weinberg equation population: 100 cats 84 black, 16 white How many of each genotype? BB AP Biology First calculate frequency of b from the known number of bb genotypes. Bb bb
Using Hardy-Weinberg equation population: 100 cats 84 black, 16 white How many of each allele? BB AP Biology q 2 (bb)= 16/100 =. 16 q (b): √. 16 = 0. 4 Now work out p p (B)= 1 - 0. 4 = 0. 6 Bb bb
Using Hardy-Weinberg equation population: 100 cats 84 black, 16 white How many of each genotype? p 2 + 2 pq + q 2 = 1 p 2=0. 36 BB AP Biology 2 pq=0. 48 Bb q 2=0. 16 bb q 2 (bb): 16/100 =. 16 q (b): √. 16 = 0. 4 p (B): 1 - 0. 4 = 0. 6 Now work out genotype frequencies
How to Solve H-W Problems (p + q)2 = Frequency of allele types p = Frequency of allele B q = Frequency of allele b p 2 + 2 pq + q 2 = 1 Frequency of allele combinations p 2 = Frequency of BB (homozygous dominant) 2 pq = Frequency of Bb (heterozygous) q 2 = Frequency of bb (homozygous recessive) B Frequency of allele combination BB in the population = p 2 AP Biology b B BB Bb bb Frequency of allele combination Bb in the population (add these together to get 2 pq) Frequency of allele combination bb in the population = q 2
How to Solve H-W Problems § Remember to use proportions in your calculations, not percentages! 1. Examine question to determine what information is given. In most cases this is the frequency of the homozygous recessive phenotype q 2 or the allele q 2. Take the square root of q 2 to find q or multiply q to find q 2 3. Find p by subtracting q from 1 (p = 1 – q) 4. Find p 2 by multiplying it by itself (p 2 = p x p) 5. Find 2 pq by multiplying p x q x 2 6. Check that your calculations are correct by adding values for p 2 + q 2 + 2 pq (the sum should be 1) AP Biology
§ A population of mice has a gene consisting of 90% B alleles (black fur) and 10% b alleles (gray fur). Determine the proportion of offspring that will be black and the proportion that will be gray. Recessive allele q = 0. 1 Dominant allele p= Recessive phenotype q 2 = Homozygous dominant p 2 = Heterozygous 2 pq = AP Biology
§ A population of mice has a gene consisting of 90% B alleles (black fur) and 10% b alleles (gray fur). Determine the proportion of offspring that will be black and the proportion that will be gray. Recessive allele q = 0. 1 Given Dominant allele p = 0. 9 1–q=p 1 – 0. 1 = 0. 9 Recessive phenotype q 2 = 0. 01 q x q = q 2 0. 1 x 0. 1 = 0. 01 Homozygous dominant p 2 = 0. 81 p x p = p 2 Heterozygous 2 x p x q = 2 pq 2 x 0. 9 x 0. 1 = 0. 18 AP Biology 2 pq = 0. 18 Check q 2 + p 2 + 2 pq = 1 0. 9 x 0. 9 = 0. 81 0. 01 + 0. 81 + 0. 18 = 1
§ A population of mice has a gene consisting of 90% B alleles (black fur) and 10% b alleles (gray fur). Determine the proportion of offspring that will be black and the proportion that will be gray. Recessive allele q = 0. 1 Dominant allele p = 0. 9 Recessive phenotype q 2 = 0. 01 Homozygous dominant p 2 = 0. 81 Heterozygous 2 pq = 0. 18 AP Biology 1% of the population will be gray 99% of the population will be black
A population of 134 lizards has 81 individuals with green skin and a gg genotype. The remaining 53 individuals have yellow skin and therefore have either the GG or Gg genotype. What proportion of the population are homozygous dominant? Recessive allele q= Dominant allele p= Recessive phenotype q 2 =. 60 Homozygous dominant p 2 = Heterozygous 2 pq = AP Biology 81/134 =. 60
A population of 134 lizards has 81 individuals with green skin and a gg genotype. The remaining 53 individuals have yellow skin and therefore have either the GG or Gg genotype. What proportion of the population are homozygous dominant? Recessive allele q = 0. 77 √ 0. 60 = 0. 77 Dominant allele p = 0. 23 1 - 0. 77 = 0. 23 Recessive phenotype q 2 = 0. 60 81/134 = 0. 60 Homozygous dominant p 2 = 0. 05 0. 23 x 0. 23 = 0. 05 Heterozygous 2 pq = 0. 35 2 x 0. 77 x 0. 23 = 0. 35 AP Biology 0. 60 + 0. 05 + 0. 35 = 1
Hardy Weinberg problems In humans, the ability to taste the chemical phenylthiocarbamide (PTC) is inherited as a simple dominant characteristic. You find that 360 out of 1000 college students could not taste the chemical. What is the frequency of the allele for tasting PTC? What percentage of students in this population are heterogynous? AP Biology
Hardy Weinberg problems While working with pea plants you find that 24 plants out of 400 exhibit the recessive dwarf trait. What is the frequency of the tall gene? What percentage of the plants have the recessive allele? AP Biology
Hardy Weinberg problems Albinism is recessive to normal pigmentation in humans. The frequency of the albino allele was 10% in a population. Determine the proportion of people that you would expect to be albino. AP Biology
Gene Pool Dynamics & Microevolution Aa AA AA Aa Aa AA AA AA Aa AA AA AP Biology
Gene Pool Dynamics & Microevolution aa Aa AA AA Aa aa A’A aa AA AA AA Aa AA aa AA AP Biology
Gene Pool Dynamics & Microevolution One aspect of gene flow is immigration & emigration – alleles may be gained from or lost to other gene pools aa Aa AA AA Aa aa A’A aa AA AA AA Aa AA aa AA Spontaneous mutations can alter allele frequencies and create new alleles. Important to evolution – original source of variation providing APnew Biology material for natural selection What is the other source of variation in a population?
Gene Pool Dynamics & Microevolution One aspect of gene flow is immigration & emigration – alleles may be gained from or lost to other gene pools aa Aa AA AA Aa aa A’A aa AA AA Aa AA aa AA Selection pressure against certain allele combinations may reduce reproductive success or cause death. Natural selection accumulates and maintains favorable genotypes, reduces genetic diversity within gene pools, and increases differences between populations AP Biology
Gene Pool Dynamics & Microevolution Deme 1 aa Aa AA AA Aa aa A’A aa AA AA AA Aa AA aa AA AP Biology Deme describes a local population that is genetically isolated from other populations. Usually have clearly definable genetic or other character that sets them apart.
Gene Pool Dynamics & Microevolution Gene flow between populations can be the source of new genetic aa variation aa Aa aa AA Aa AA aa aa AA AP Biology Geographical barriers isolate the gene pool and prevent regular gene flow between populations
Gene Pool Dynamics & Microevolution aa Aa aa AA Aa aa aa AA Aa AA aa aa AA AP Biology Mate choice (non-random mating): Individuals may not select mates randomly, seeking particular phenotypes, increasing the frequency of these “favored” alleles in the population.
Gene Pool Dynamics & Microevolution aa Aa aa AA Aa aa aa AA Aa AA Genetic drift: Chance events can cause the allele frequency of Aa AA small populations to change randomly from generation to generation. Can play a significant role in the aa microevolution of AA small populations. The founder effect (small population colonizes new area) and the bottleneck effect (population size dramatically reduced by catastrophic event) AP Biology
Founder effect AP Biology
Bottleneck effect AP Biology
Bottleneck effect Cheetahs - ~ 20, 000 - Very little genetic diversity - Nearly went extinct at end of last ice age - Lack of variation creates problems – sperm abnormalities, decreased fecundity, high cub mortality, sensitivity to disease AP Biology
Application of H-W principle § Sickle cell anemia Caused by inheriting a mutation in the gene coding for haemoglobin § oxygen-carrying blood protein § recessive allele = Hs. Hs w normal allele = Hb u low oxygen level causes RBC to sickle § clogging small blood vessels § depriving tissues of oxygen § damage to organs u AP Biology The condition is often lethal
Sickle cell frequency § High frequency of heterozygotes 1 in 5 in Central Africans = Hb. Hs u unusual for allele with severe detrimental effects in homozygotes u § 1 in 100 = Hs. Hs § usually die before reproductive age Why is the Hs allele maintained at such high levels in African populations? Suggests some selective advantage of being heterozygous… AP Biology
Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells Malaria 1 2 AP Biology 3
Heterozygote Advantage § In tropical Africa, where malaria is common: u homozygous dominant (normal) § die or reduced reproduction from malaria: Hb. Hb u homozygous recessive § die or reduced reproduction from sickle cell anemia: Hs. Hs u heterozygote carriers are relatively free of both: Hb. Hs § survive & reproduce more, more common in population AP Biology Frequency of sickle cell allele & distribution of malaria
Any Questions? ? AP Biology
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