1Intro to population genetics

• Population genetics is the study of traits and their inheritance in populations.– Information is important in assessing (genetic)

health of population of organisms• Changes in genetic make-up can be spotted• Diversity of traits usually a good thing

– In humans and others, controlled matings are not possible, or don’t give accurate view

– Provides information on how a population is evolving

• Hardy-Weinberg principle (and equation)– Provides the theoretical and mathematical basis

2Points to remember

• Number of alleles– Most organisms are diploid; maximum 2 alleles!– Two individuals may have completely different

alleles: there can be many alleles in a population.

• Population: a group of individuals of the same species, usually that can interbreed. Local pop.

• Gene pool: sum of all the alleles in a population. (or all gametes (haploid) made)

• Two items of interest: frequency of a genotypes & frequency of alleles in a population.

3Genotype frequency

• Frequency is a decimal fraction (0 to 1)

• Example: the M and N blood groups– M & N are codominant, so phenotype →genotype.– MM = type M; MN = type MN; NN = type N

• The Data: MM = 298 MN = 489 NN = 213

• The calculations: easy. Total is 1000, so– Frequency of MM = 298/1000 = 0.298– Frequency of MN = 0.489; of NN = 0.213– Notice that 0.298+0.489+0.213= 1.0. Has to.

4Allele frequency

• MN blood groups, continued.• What are frequencies of the individual alleles (M&N)?

– There are 1000 people who are diploid, 2000 alleles– If there are 298 people with the MM genotype, and

each person has 2 copies, the # of M alleles is 2 x 298 = 596. Also contributing: 1 M allele in every MN person (489) so 596 + 489 = 1085.

– There are 2000 copies of alleles. So N = 2000 – 1085 = 915.

– 1 N in every MN (489) plus 2 x 213 (426) = 915. Good.

• Frequencies: M = 1085/2000 = 0.5425; N = 0.4575• Frequencies add up to 1.0 (0.5425 + 0.4575)

5Methods

• Different alleles detected as molecular polymorphisms– Enzymes: alternative forms are allozymes

• Proteins vary by 1 or 2 amino acids, differ in charge, move differently in electrophoresis

• Gel typically starch or cellulose acetate sheet• Stain for enzyme activity (or just protein)

– DNA markers: inheritable patterns • RFLPs: mutation changes restriction fragments• VNTRs: repeats vary in number

– Identified using PCR, produces different size bands.

• Some alleles readily apparent– Homozygous recessive: phenotype → genotype

6Study of allele frequencies

• A population has two or more alleles for 15-40% of its genes, depending on organism.

• Having multiple alleles is to be genetically diverse; this is correlated with “fitness”– Multiple alleles means population can adapt to

environmental changes more easily.– Few alleles means organisms are nearly clones;

what affects one (e.g. disease) could affect most.

• Alleles can be lost (freq. = 0) or fixed (freq = 1)

7Calculating allele frequencies

• Hardy and Weinberg came up with the same equation and same mathematical assumptions about the same time.

• Principle: given certain things being true, the allele frequencies in a populations do not change by themselves, but remain constant.

• The principle is mathematical. The assumptions (that are generally violated) are biological.

8Hardy Weinberg equation

• p2 + 2pq + q2 = 1 and p + q = 1.• IMPORTANT: p + q are allele frequencies (in a two

allele system).– p2 , 2pq , and q2 are Genotype frequencies!– p2 and q2 are the frequencies of homozygotes– 2pq is the frequency of heterozygotes.

• This can be seen from thePunnett Square; the frequency of

alleles in the gametes reflects the

frequency of alleles in the pop.

p q

p p*p p*q

q p*q q*q

9Hardy Weinberg sample calculation

• Make believe problem: wild peas.– Two alleles for flower color, W (purple) & w (white)– Can’t tell WW from Ww, but you know that all the

white flowered plants are ww.– Assign freq p to Purple allele, freq q to White.

• You want to know the heterozygote frequency

• You count all the plants: 100. Of these, 4 have white flowers. White means homozygous recessive. This is q2! q2 = 4/100 or 0.04

• Arithmetic. Sq root of 0.04 = 0.2 = q

10Sample calc continued

• If genotype ww freq is 0.04, allele frequency (q) is square root of that or 0.2.

• p + q = 1 always! So p = 1 – 0.2 = 0.8.

• Back to genotype freq: heterozygotes are 2pq so 2 x 0.8 x 0.2 = 0.32.

• Check your work: p2 + 2pq + q2 = 10.64 + 0.32 + 0.04 = 1

11Hardy-Weinberg assumptions

1. Individuals of all genotypes have equal rates of survival and reproduction (no selection).

2. No new alleles created (by mutation)3. No migration (movement of individuals)4. Population is infinitely large (small populations

create sampling errors, statistical anomalies).5. Individuals in the population mate randomly.• If these things hold true, allele frequencies in

a population will not change.– “Dominant” does NOT mean “will take over”!

12Use of Hardy-Weinberg

• Understanding the basis for population change– A population in which allele frequencies are not

changing is said to be in “H-W equilibrium”– Violation of the assumptions of H-W can be

detected; impact on & evolution of the population can be studied.

• Genetic counseling

• Forensic DNA analysis

13Violation of H-W assumptions

• Natural selection– Selection acts on genotypes, but this changes

allele frequencies.– Selection results in survival and reproduction of

organisms with favorable genotypes.– An individual’s genetic contribution to future

generations is measured by “Fitness”. Math.• Alleles making up less fit genotypes decrease in

frequency over time.• Humans: disease selects for some genotypes.

– CCR5 allele: HIV, viral hemorrhagic fever?

14Mutations

• Mutations: the only way new alleles are made.– Mutations have little effect on allele frequencies

because they are rare, but provide genetic variation for selection to act on.

• Harmful recessive mutations– Persist in population because present in

heterozygotes, so seldom subject to selection.– Heterozygote superiority: also persist because

heterozygotes may have a selective advantage.• Sickle cell anemia;

15Migration

• Immigration or emigration changes allele freq.– Extent of change depends on #s of individuals and

difference in allele freq between the 2 populations.– Human history can be mapped.– Equations exist.

http://anthro.palomar.edu/vary/images/map_of_B_blood_clines_in_Europe.gif

16Genetic drift

• This kind of “drift” has nothing to do with physical movement!– Random statistical fluctuations that can occur in

small populations, a “drift” in the allele frequencies.– Genetic drift can result in loss of or fixing of alleles.

• Founder effect: population arises from a few individuals not representative of whole pop.

• Bottleneck: survivors of natural disaster (not selection!) have different gene pool than originals.

http://ro.hallmarkchannel.com/data/images/PG_USA_Gilligan.jpg

17Non-random mating

• Non-random mating assumption often violated– Assortative mating: individuals choose mates who

are similar (positive) or dissimilar (negative).– Extreme example: inbreeding; this too can be

described mathematically.• In plants: self-fertilization

– Inbreeding results in fewer heterozygotes, increased homozygotes.• Allele frequencies do not change, genotype

frequencies change. Still has an effect.

18More effects of inbreeding

• Sometimes positive– Selective inbreeding produces plants, livestock with

fixed positive alleles, lost harmful alleles.

• Often (consanguineous in humans) harmful.– Increased chances for inheritance of 2 harmful

alleles, resulting in failed development, disease.– Many human cultures forbid such matings.

19Calculating heterozygote frequencies

• Asking if a population is in H-W equilibrium– Compare freq of heterozygotes calculated from

allele frequencies to freq of heterozygotes observed– Excess heterozygotes: could be selection is

occurring; Deficiency: could be inbreeding.

20Heterozygotes-more

• Example: a 2 allele system. Separate allozymes in a gel. You can see genotypes. – First, count the number of the different alleles.– Second, count the number of heterozygotes.

What’s their frequency?– Last, use the number of alleles you counted to

calculate 2pq. Is it the same as what you counted?

21Diagram and math

S

F

Let p be the allele frequency of the S allele, q the allele freq of the F allele.

Where there is one band, homozygosity is assumed.

Ten individuals tested; 14 S, 6 F; so p = 14/20 = 0.7; and q = 6/20 = 0.3. The actual frequency of heterozygotes is 2/10 = 0.2. The calculated: 2pq = 1 – 0.49 – 0.09 = 0.42. There are only half as many heterozygotes as expected!

http://www.cf.adfg.state.ak.us/geninfo/research/genetics/techfac/images/allozyme/gelpgm.jpg

22More about Heterozygotes

• Where do we find most recessive alleles?– In heterozygotes. – Using our H-W equation p2 + 2pq + q2 = 1, if q is

0.1, the ratio of heterozygotes/homozgotes is 20; – when q = 0.01 this ratio is 200. – The more rare the recessive allele, the more the

allele is found in heterozygotes.• No eugenics. Can’t get rid of bad alleles.

23Heterozygotes are carriers

• Using H-W to calculate odds of having a child with a genetic disease or being a carrier. – If you know the number of people with recessive

phenotypes, you know the number with that genotype (q2 ) and you can calculate from that.

– Cystic fibrosis: in people of northern European ancestry, incidence of afflicted is 1/2500.• 1/2500 = 0.0004 = q2

• q = SQRT(0.0004) = 0.02, so p = 1-0.02 = 0.98• 2pq = 2(.98)(.02)= 0.04 or 1/25

24Multiple alleles

• In many genetic systems, more than two alleles exist. Good example is human blood groups where A and B are co-dominant and O is recessive. Genotypes are AA, AO, BB, BO, AB, and OO and phenotypes are A, B, O, and AB.

• H-W equation becomes p12 + p2

2 + p32 + 2p1p2 +

2p1p3 + 2p2p3 = 1.

• For more than 3 alleles, calculations depend on polynomial expansion.

25Use of VNTRs

Restriction sites are on either side; fragment length depends on number of repeats in between sites.

26VNTRs

There are various sites or loci throughout human DNA where VNTRs exist. This is a gel showing the results for 9 individuals plus standards; note that lane 1 appears to be homozygous.

Calculations: use the genotype frequency 2pq. The frequency for each allele comes from a table (next); instead of p2 for apparent homozygotes, 2p is used; in forensic work, this gives benefit of the doubt to accused.

genetics.biol.ttu.edu/genetics/ Lecture/red12.html

27

Repeat # Caucasian Hispanic African American

Asian

14 0 0 0 0

15 0 0.001 0 0

16 0.001 0.010 0.002 0.034

17 0.002 0.009 0.028 0.025

18 0.237 0.224 0.073 0.152

19 0.003 0.005 0.003 0.022

20 0.018 0.013 0.032 0.007

21 0.021 0.028 0.115 0.034

22 0.038 0.024 0.081 0.017

Allele frequencies for D1S80 among US population groups

28Working problem

• Looking at the previous 2 slides: look at individual #2. Suppose those 2 bands are for repeat #s 19 and 21and the individual was African American. To find how common that genotype is, look up the frequency of those two bands then multiply: 2pq = 2 x 0.003 x 0.115 = 0.00069. One out of every 1,449 US African Americans would be expected to have that genotype. When you multiply that by the odds for two other VNTRs, you can really narrow it down.