Allele Frequencies: Staying Constant Chapter 14
What is Allele Frequency? How frequent any allele is in a given population: Within one race Within one nation Within one town/school/research project Calculated by genotyping a large sample of the population Or estimated by phenotype frequency in entire population (recessive disease only)
Example: PKU allele
Population Genetics Considers all alleles within a given population: Allele is the version of the gene that a person carries (Allele frequency) Gene pool = all alleles that are possible within population s gametes Genotype frequency = proportion of the population that has each type of genotype Phenotype frequency = percentage of population that have phenotype
Bi-allelic Gene In bi-allelic gene there are only two alleles possible T or t for tall or short pea plants R or r for wrinkled or round seeds p = frequency of the more common of the two alleles q = frequency of the less common of the two alleles
Commonly Many genes have more than two alleles Most common diseases/disorders are multifactorial: More than one gene each with more than two alleles Environment and genetics Therefore phenotype will not equal genotype
Rare cases are still useful Even though most genes and most diseases don t follow these rules we are about to learn There are still many cases where these rules are important and useful for genetics Next class we ll learn some of the complications
Hardy-Weinberg Equilibrium Where the allele frequencies stay constant from one generation to the next Often calculated with a bi-allelic gene (p and q) Therefore p and q remaining constant
Changing Allele Frequencies 1. Mutation introduces new alleles into population 2. Natural Selection specific alleles are more likely to be passed down because they are somehow advantageous 3. Non-random Mating individuals of one genotype are more likely to mate with individuals of same genotype Think of an example of this happening?
Changing Allele Frequencies 4. Migration individuals with specific genotypes move in or out of a population 5. Genetic Drift random changes in allele frequencies Caused by random sampling of specific genotypes Often seen in small, isolated populations Can you think of why? Nothing to do with natural selection
Hardy-Weinberg Equilibrium Requires that none of these things are happening in a population: No mutation No selection No migration No genetic drift Random mating Large population Obviously this is VERY rare in real life
Hardy-Weinberg Equilibrium 1908 Hardy an English mathematician Weinberg a German physician Both derived, independently, an algebra calculation for what happens to allele frequencies within a population Assuming all those false conditions
Hardy-Weinberg Equilibrium 1. If there are only two alleles then the following must be true: p + q = 1 The frequency of the two alleles added together must equal the entire population (a frequency of 1)
Hardy-Weinberg Equilibrium 2. The genotype frequencies can also be calculated: p 2 + 2pq + q 2 = 1 The frequency of each homozygote equals the frequency of the allele squared The frequency of heterozygote is 2 times p times q Product Rule These three Product genotypes and Addition must add Rules to one
Hardy-Weinberg Equilibrium 1. Allele frequencies add to one: p + q = 1 2. The genotype frequencies can be calculated from the allele frequencies: p 2 + 2pq + q 2 = 1
Hardy-Weinberg Equilibrium 1. Allele frequencies add to one: p + q + r = 1 2. The genotype frequencies can be calculated from the allele frequencies: p 2 + 2pq + 2pr + 2rq + q 2 + r 2 = 1
How it was derived: A (p) A (p) AA (pp) a (q) Aa (pq) Frequencies: Allele A Allele a = p = q a (q) Aa (pq) aa (qq) Genotype AA = p 2 Genotype Aa = 2pq Genotype aa = q 2
Let s work through HWE: Autosomal recessive trait (middle finger is shorter than 2 and 4) All we know is this: In 100 individuals there are 9 that show the recessive shorter finger Use HWE to figure out: Both allele frequencies All three genotype frequencies
Let s work through HWE: Know: 9/100 show recessive phenotype Calculate: p = q = Homozygous Dominant = Heterozygous = Homozygous Recessive = 0.09
Let s work through HWE: 9/100 = recessive phenotype Know this is an autosomal recessive trait Therefore: Recessive phenotype = qq genotype q 2 = 0.09 Therefore: q = 0.3 p must equal 1 - q = 1 -.3 = 0.7
Let s work through HWE: p = 0.7 and q = 0.3 Homozygous Dominant = p 2 (.7)(.7) =.49 or 49% Heterozygous = 2 pq 2(.7)(.3) =.42 or 42% Homozygous Recessive = q 2 (.3)(.3) =.09 or 9% (which is what we based all of these calculations on)
Solved HWE: Know: 9/100 show recessive phenotype Calculate: p =.7 q =.3 Homozygous Dominant (p 2 ) = 49% Heterozygous (2pq) = 42% Homozygous Recessive (q 2 ) = 9% What about in the next generation?
Practical Applications of HWE 1. Genotyping error If your genotypes are grossly off of the expected from HWE calculations 2. Artificial Selection 3. Population Genetics Determining genetic risk in different populations 4. Disease risk 5. Forensic Biology
Genotyping Error Are genotypes present in expected proportions with allele frequencies? Allele freq. Genotype Observed Expected HWE Calculations: p(1) = 0.62 1/1 123 116.9 CHI2: 2.207 q(2) = 0.38 1/2 131 143.2 df: 1 2/2 50 43.9 p-value: 0.137 304 304 Expected: 1/1 = p 2 * total # genotypes 1/2 = 2pq * total # genotypes 2/2 = q 2 * total # genotypes
Artificial Selection This is the human act of purposely selecting certain traits over others: Changing phenotype frequencies Agriculture What examples can you think of? Pure breed dogs (other animals) HWE calculations will tell you: How many mating pairs to set up How many generations to get desired result
Population Genetics Estimate genotype frequency from phenotype frequency Based on known percentage of population that shows a recessive phenotype That percent must be homozygous for the recessive allele right? (q 2 ) What are problems here? Multifactorial, more than two alleles, etc
Disease Risk Couple wants to know their risk of having a child with a specific disease If one (or both) parents have phenotype in question run genetic tests If neither have phenotype then question is about being a carrier (2pq) Based on population genetics calculations and therefore their assumptions
Disease Risk Couple wants to know their risk of being carriers for disease (2pq) Population genetics tells us how frequent phenotype is in population That s q 2 Square root calculate q Calculate p Calculate 2 pq That s carrier frequency
Carrier Frequencies:
X-linked is different Females follow same HWE formula: p 2 + 2pq + q 2 = 1 Males however only carry one allele: Therefore In males phenotype frequency is allele frequency (not genotype) Therefore frequency of recessive phenotype gives you q, not q 2
Forensic Biology Using biology to add to the forensics of a crime scene Although we all share 99.9 percent of our DNA with every other human That 0.1 % equals about 3 million base pairs of difference Product rule means that this can ID more than there are humans on the planet
Forensic Biology Identifying individuals with DNA: 1. Genotype a few polymorphisms (~10) 2. All on different chromosomes 3. All highly polymorphic More than two alleles More alleles, more information per polymorphism 4. Match to crime scene Or body, or baby to father in paternity
Forensic Biology Identifying individuals with DNA: Match to crime scene Or body, or baby to father in paternity Calculating the chance of seeing the DNA profile is calculated based on Hardy-Weinberg Equilibrium Exact calculations depend on how frequent alleles are in given population How likely genotypes will be
Polymorphisms: SNPs: Single Nucleotide Polymorphism More common, but only have two alleles RFLPs: Restriction Fragment Length Polymorphism Many alleles Microsatellites: Polymorphic repeats in non-coding sequence Most alleles possible (avg. around 8)
How is HWE involved? Determine genotype of individual Use HWE to calculate probability of seeing specific genotype: Het. = 2pq = 2(.6)(.4) = 0.36 Homo = q 2 = (.25)(.25) = 0.0625 Then use product rule to calculate final probability that another person has the same combination of genotypes: (.36)(.06) = 0.0225 or 2.25% chance
HWE and Product Rule: Genotype Five Bi-allelic Polymorphism Het = 2pq = 2(.6)(.3) =.36 Het = 2pq = 2(.5)(.3) =.30 Het = 2pq = 2(.15)(.8) =.24 Homo = q 2 = (.2)(.2) =.04 Het = 2pq = 2(.80)(.18)=.29
HWE and Product Rule: =.36 =.30 =.24 =.04 =.29 (.36)(.3)(.24)(.04)(.29) = 0.00031 Or 1/3,226 Therefore, the chance of this matching the wrong person is 1/3,226
Summary Hardy-Weinberg Equilibrium(HWE) states: p + q = 1 p 2 + 2pq + q 2 = 1 HWE is unlikely to exist in a real population But it is still useful for many fields of genetics know how and when to use it Know how to calculate it for biallelic genes
Next Class: Read Chapter Fifteen Homework Chapter Fourteen Problems; Review: 1, 2, 3, 5 Applied: 1, 2, 6, 7, 11 Happy Halloween!