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 GeneticsConsiders 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 EquilibriumWhere 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 Frequencies1. Mutation – introduces new alleles into

population2. 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 Frequencies4. Migration – individuals with specific

genotypes move in or out of a population5. Genetic Drift – random changes in allele

frequencies – Caused by random sampling of specific

genotypes– Often seen in small, isolated populationsCan 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 Equilibrium1908• 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 Equilibrium1. 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 Equilibrium2. The genotype frequencies can also be

calculated:p2 + 2pq + q2 = 1

The frequency of each homozygote equals the frequency of the allele squared

The frequency of heterozygote is 2 times p times q

These three genotypes must add to one

Product Rule

Product and Addition Rules

Hardy-Weinberg Equilibrium1. Allele frequencies add to one:

p + q = 1

2. The genotype frequencies can be calculated from the allele frequencies:

p2 + 2pq + q2 = 1

Hardy-Weinberg Equilibrium1. Allele frequencies add to one:

p + q + r = 1

2. The genotype frequencies can be calculated from the allele frequencies:

p2 + 2pq + 2pr + 2rq + q2 + r2 = 1

How it was derived:

A (p) a (q)

A (p)

a (q)

AA Aa(pp) (pq)

Aa aa(pq) (qq)

Frequencies:Allele A = pAllele a = q

Genotype AA = p2

Genotype Aa = 2pqGenotype aa = q2

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 frequenciesAll 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 traitTherefore:• Recessive phenotype = qq genotype• q2 = 0.09Therefore:• 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 = p2

(.7)(.7) = .49 or 49%• Heterozygous = 2 pq2(.7)(.3) = .42 or 42%• Homozygous Recessive = q2

(.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 = .7q = .3Homozygous Dominant (p2) = 49%Heterozygous (2pq) = 42%Homozygous Recessive (q2) = 9%• What about in the next generation?

Practical Applications of HWE1. Genotyping error

– If your genotypes are grossly off of the expected from HWE calculations

2. Artificial Selection3. Population Genetics

– Determining genetic risk in different populations

4. Disease risk5. Forensic Biology

Genotyping ErrorAre genotypes present in expected proportions with allele frequencies?

1/1 = p2 * total # genotypes1/2 = 2pq * total # genotypes2/2 = q2 * total # genotypes

Expected:

Genotype Observed Expectedp(1) = 0.62 1/1 123 116.9 CHI2: 2.207q(2) = 0.38 1/2 131 143.2 df: 1

2/2 50 43.9 p-value: 0.137304 304

HWE Calculations:Allele freq.

Artificial SelectionThis 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? (q2) • What are problems here?

– Multifactorial, more than two alleles, etc

Disease RiskCouple 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 RiskCouple wants to know their risk of being

carriers for disease (2pq)• Population genetics tells us how frequent

phenotype is in population• That’s q2

• Square root – calculate q• Calculate p• Calculate 2 pq – That’s carrier frequency

Carrier Frequencies:

X-linked is different…• Females follow same HWE formula:p2 + 2pq + q2 = 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 q2

Forensic BiologyUsing 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 BiologyIdentifying individuals with DNA:1. Genotype a few polymorphisms (~10)2. All on different chromosomes3. All highly polymorphic

– More than two alleles– More alleles, more information per

polymorphism4. Match to crime scene

– Or body, or baby to father in paternity

Forensic BiologyIdentifying 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 = q2 = (.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:

Het = 2pq = 2(.6)(.3) = .36

Genotype Five Bi-allelicPolymorphism

Het = 2pq = 2(.5)(.3) = .30Het = 2pq = 2(.15)(.8) = .24

Het = 2pq = 2(.80)(.18)= .29Homo = q2 = (.2)(.2) = .04

HWE and Product Rule:

= .36(.36)(.3)(.24)(.04)(.29) = 0.00031

Or 1/3,226

Therefore, the chance of this matchingthe wrong person is 1/3,226

= .30= .24

= .29= .04

Summary• Hardy-Weinberg Equilibrium(HWE) states:p + q = 1p2 + 2pq + q2 = 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!