Introduction to Genetics. Topics Darwin and Mendel Probability Mendelian genetics –Mendel's...

Introduction to Genetics

Topics• Darwin and Mendel• Probability• Mendelian genetics

– Mendel's experiments– Mendel's laws

• Introduction to Population Genetics• Introduction to Quantitative Genetics

Darwin & Mendel• Darwin (1859) Origin of Species

– Instant Classic, major immediate impact– Problem: Model of Inheritance

• Darwin assumed Blending inheritance• Offspring = average of both parents• zo = (zm + zf)/2• Fleming Jenkin (1867) pointed out problem

– Var(zo) = Var[(zm + zf)/2] = (1/2) Var(parents)– Hence, under blending inheritance, half the

variation is removed each generation and this must somehow be replenished by mutation.

Mendel• Mendel (1865), Experiments in Plant

Hybridization• No impact, paper essentially ignored

– Ironically, Darwin had an apparently unread copy in his library

– Why ignored? Perhaps too mathematical for 19th century biologists

• The rediscovery in 1900 (by three independent groups)

• Mendel’s key idea: Genes are discrete particles passed on intact from parent to offspring

Probability & Genetics

• Pr(A) falls between 0 and 1

• Pr(not A) = 1-Pr(A)

Since genes are passed on at random, an understandingof probability is critical to understanding genetics

Let A denote an event of interest (getting a head onthe flip of a coil, rolling a 5 on a dice, getting a QQgenotype)

Let Pr(A) denote the probability that event A occurs

• The sum of the probabilities for all (non-overlapping) events is one --- Probabilities sum to one

ExampleConsider the offspring in a cross of two Qq parents

What is the probability that an offspring is Anything EXCEPT qq?

Pr(not qq) = 1- Pr(qq) = 1-1/4 = 3/4

The AND Rule Suppose the events A and B are independent ---knowing that A has occurred does not change the probability that B occurs.

The Pr(A AND B) = Pr(A)*Pr(B)

“AND Rule” -- if see “AND”, multiply probabilities

Pr(A AND B AND C) = Pr(A)*Pr(B)*Pr©

The OR Rule Suppose the events A and B are mutually exclusive ---Non-overlapping

For example, A = roll even number of dice, B = rollA six are NOT mutually exclusive, but if B = roll 5 theyare

Pr(A OR B) = Pr(A) + Pr(B)

“OR Rule” --- see OR = add probabilities

Genetics examplesAgain consider offspring from Qq x Qq cross

Prob(Not qq) = Pr(QQ or Qq) = Pr(QQ) + Pr(Qq) = 3/4

Pr(QQ) = Pr(Q from father AND Q from mother) = Pr(Q from father)*Pr(Q from mother) = (1/2)*(1/2) = 1/4

Pr(Qq) = Pr([f = Q AND m = q] OR [f = q AND m = q]) = Pr(f = Q AND m = q) + Pr(f = q AND m = q) = Pr(f = Q)*Pr(m = q) + Pr(f = q )*Pr( m = q) = (1/2)*(1/2) + (1/2)*(1/2) = 1/2

Conditional probabilityLet Pr(A | B) = Pr(A) given that we observe event B

Pr(A | B) = Pr(A and B) / Pr(B) = Pr(A,B)/Pr(B)Pr(A,B) is called the joint probability of A & BExample: Suppose QQ and Qq give purple offspring,While qq = green offspring. What is the probabilityAt a purple offspring from a Qq x Qq cross is QQ?

Pr(QQ | F1 Purple) = Pr(QQ and Purple)/Pr(Purple) = (1/4)/(3/4) = 1/3

Mendel’s experiments with the Garden Pea7 traits examined

Mendel crossed a pure-breeding yellow pea linewith a pure-breeding green line.

Let P1 denote the pure-breeding yellow (parental line 1)P2 the pure-breed green (parental line 2)

The F1, or first filial, generation is the cross ofP1 x P2 (yellow x green).

All resulting F1 were yellow

The F2, or second filial, generation is a cross of two F1’sIn F2, 1/4 are green, 3/4 are yellow

This outbreak of variation blows the theory of blending inheritance right out of the water.

Mendel also observed that the P1, F1 and F2 Yellow lines behaved differently when crossed to pure green

P1 yellow x P2 (pure green) --> all yellow

F1 yellow x P2 (pure green) --> 1/2 yellow, 1/2 green

F2 yellow x P2 (pure green) --> 2/3 yellow, 1/3 green

Mendel’s explanationGenes are discrete particles, with each parent passingone copy to its offspring.Let an allele be a particular copy of a gene. In Diploids,each parent carries two alleles for every gene

Pure Yellow parents have two Y (or yellow) allelesWe can thus write their genotype as YY

Likewise, pure green parents have two g (or green) allelesTheir genotype is thus gg

Since there are lots of genes, we refer to a particular geneby given names, say the pea-color gene (or locus)

Each parent contributes one of its two alleles (atrandom) to its offspring

Hence, a YY parent always contributes a Y, whilea gg parent always contributes a g

In the F1, YY x gg --> all individuals are Yg

An individual carrying only one type of an allele(e.g. yy or gg) is said to be a homozygote

An individual carrying two types of alleles issaid to be a heterozygote.

The phenotype of an individual is the trait value weobserve

For this particular gene, the map from genotype tophenotype is as follows:

YY --> yellowYg --> yellowgg --> green

Since the Yg heterozygote has the same phenotypicvalue as the YY homozygote, we say (equivalently)

Y is dominant to g, org is recessive to Y

Explaining the crossesF1 x F1 -> Yg x Yg

Prob(YY) = yellow(dad)*yellow(mom) = (1/2)*(1/2)

Prob(gg) = green(dad)*green(mom) = (1/2)*(1/2)

Prob(Yg) = 1-Pr(YY) - Pr(gg) = 1/2

Prob(Yg) = yellow(dad)*green(mom) + green(dad)*yellow(mom)

Hence, Prob(Yellow phenotype) = Pr(YY) + Pr(Yg) = 3/4Prob(green phenotype) = Pr(gg) = 1/4

Dealing with two (or more) genes

For his 7 traits, Mendel observed Independent Assortment

The genotype at one locus is independent of the second

RR, Rr - round seeds, rr - wrinkled seedsPure round, green (RRgg) x pure wrinkled yellow (rrYY)

F1 --> RrYg = round, yellowWhat about the F2?

Let R- denote RR and Rr. R- are round. Note in F2,Pr(R-) = 1/2 + 1/4 = 3/4

Likewise, Y- are YY or Yg, and are yellow

Phenotype Genotype FrequencyYellow, round Y-R- (3/4)*(3/4) = 9/16

Yellow, wrinkled Y-rr (3/4)*(1/4) = 3/16

Green, round ggR- (1/4)*(3/4) = 3/16

Green, wrinkled ggrr (1/4)*(1/4) = 1/16

Or a 9:3:3:1 ratio

Probabilities for more complex genotypes

Cross AaBBCcDD X aaBbCcDdWhat is Pr(aaBBCCDD)?

Under independent assortment, = Pr(aa)*Pr(BB)*Pr(CC)*Pr(DD) = (1/2*1)*(1*1/2)*(1/2*1/2)*(1*1/2) = 1/25

What is Pr(AaBbCc)?= Pr(Aa)*Pr(Bb)*Pr(Cc) = (1/2)*(1/2)*(1/2) = 1/8

Mendel was wrong: Linkage

Phenotype

Genotype Observed Expected

Purple long P-L- 284 215Purple round

P-ll 21 71

Red long ppL- 21 71Red round ppll 55 24

Bateson and Punnet looked at flower color: P (purple) dominant over p (red )

pollen shape: L (long) dominant over l (round)

Excess of PL, pl gametes over Pl, pLDeparture from independent assortment

LinkageIf genes are located on different chromosomes they(with very few exceptions) show independent assortment.

Indeed, peas have only 7 chromosomes, so was Mendel luckyin choosing seven traits at random that happen to allbe on different chromosomes? Problem: compute this probability.

However, genes on the same chromosome, especially ifthey are close to each other, tend to be passed ontotheir offspring in the same configuation as on theparental chromosomes.

Consider the Bateson-Punnet pea data

Let PL / pl denote that in the parent, one chromosomecarries the P and L alleles (at the flower color andpollen shape loci, respectively), while the other chromosome carries the p and l alleles.

Unless there is a recombination event, one of the twoparental chromosome types (PL or pl) are passed ontothe offspring. These are called the parental gametes.

However, if a recombination event occurs, a PL/pl parent can generate Pl and pL recombinant chromosomesto pass onto its offspring.

Let c denote the recombination frequency --- theprobability that a randomly-chosen gamete from theparent is of the recombinant type (i.e., it is not aparental gamete).

For a PL/pl parent, the gamete frequencies are

Gamete type Frequency Expectation under independent assortment

PL (1-c)/2 1/4pl (1-c)/2 1/4pL c/2 1/4Pl c/2 1/4

Parental gametes in excess, as (1-c)/2 > 1/4 for c < 1/2Recombinant gametes in deficiency, as c/2 < 1/4 for c < 1/2

Expected genotype frequencies under linkageSuppose we cross PL/pl X PL/pl parents

What are the expected frequencies in their offspring?

Pr(PPLL) = Pr(PL|father)*Pr(PL|mother) = [(1-c)/2]*[(1-c)/2] = (1-c)2/4

Recall from previous data that freq(ppll) = 55/381 =0.144Hence, (1-c)2/4 = 0.144, or c = 0.24

Likewise, Pr(ppll) = (1-c)2/4

Generally, to compute the expected genotypeprobabilities, need to consider the frequenciesof gametes produced by both parents.

Suppose dad = Pl/pL, mom = PL/pl

Pr(PPLL) = Pr(PL|dad)*Pr(PL|mom) = [c/2]*[(1-c)/2]

Notation: when PL/pl, we say that alleles P and Lare in coupling

When parent is Pl/pL, we say that P and L are in repulsion

Allele and Genotype Frequencies

62pi = freq(A i) = freq(AiAi) + 1 Xi=j

freq(AiA j )

Given genotype frequencies, we can always compute allelefrequencies, e.g.,

The converse is not true: given allele frequencies we cannot uniquely determine the genotype frequencies

For n alleles, there are n(n+1)/2 genotypes

If we are willing to assume random mating,

freq(AiA j ) =Ωp2i for i = j

2pipj for i6= jHardy-Weinberg proportions

Hardy-Weinberg• Prediction of genotype frequencies from allele freqs• Allele frequencies remain unchanged over generations, provided:

• Infinite population size (no genetic drift)• No mutation• No selection• No migration

• Under HW conditions, a single generation of randommating gives genotype frequencies in Hardy-Weinbergproportions, and they remain forever in these proportions

Gametes and Gamete Frequencies

freq(AABB) = freq(ABjfather) freq(ABjmother)

freq(AaBB) =freq(ABjfather)freq(aBjmother)+freq(aBjfather)freq(ABjmother)

When we consider two (or more) loci, we follow gametes

Under random mating, gametes combine at random, e.g.

Major complication: Even under HW conditions, gametefrequencies can change over time

ABAB

abab

abAB

ABab

In the F1, 50% AB gametes50 % ab gametes

If A and B are unlinked, the F2 gamete frequencies are

AB 25% ab 25% Ab 25% aB 25%

Thus, even under HW conditions, gamete frequencies change

Linkage disequilibrium

freq(AB) = freq(A) freq(B)freq(ABC) = freq(A)freq(B)freq(C)

Random mating and recombination eventually changesgamete frequencies so that they are in linkage equilibrium (LE).Once in LE, gamete frequencies do not change (unless actedon by other forces)

At LE, alleles in gametes are independent of each other:When linkage disequilibrium (LD) present, alleles are nolonger independent --- knowing that one allele is in the gamete provides information on alleles at other loci

freq(AB)6= freq(A) freq(B)The disequilibrium between alleles A and B is given byDA B = freq(AB) ° freq(A)freq(B)

freq(AB) = freq(A) freq(B) + DA B

D(t) = D(0)(1 c)t°

The Decay of Linkage DisequilibriumThe frequency of the AB gamete is given by

LE valueDeparture from LEIf recombination frequency between the A and B loci

is c, the disequilibrium in generation t is

Initial LD valueNote that D(t) -> zero, although the approach can beslow when c is very small

Quantitative GeneticsThe analysis of traits whose variation is determined by

both a number of genes and environmental factors

Phenotype is highly uninformative as tounderlying genotype

Complex (or Quantitative) trait

• No (apparent) simple Mendelian basis for variation in the trait

• May be a single gene strongly influenced by environmental factors

• May be the result of a number of genes of equal (or differing) effect

• Most likely, a combination of both multiple genes and environmental factors.

• Example: Blood pressure, cholesterol levels– Known genetic and environmental risk factors

Phenotypic distribution of a traitConsider a specific locus influencing the trait

For this locus, mean phenotype = 0.15, whileoverall mean phenotype = 0

Goals of Quantitative Genetics

• Partition total trait variation into genetic (nature) vs. environmental (nurture) components

• Predict resemblance between relatives– If a sib has a disease/trait, what are your odds?

• Find the underlying loci contributing to genetic variation – QTL -- quantitative trait loci

• Deduce molecular basis for genetic trait variation• Prediction of selection response• Prediction of the effects of selfing & assortative

mating

Dichotomous (binary) traitsPresence/absence traits (such as a disease) can (and usually do) have a complex genetic basisConsider a disease susceptibility (DS) locus underlying a disease, with alleles D and d, where allele D significantly increases your disease risk

In particular, Pr(disease | DD) = 0.5, so that thePenetrance of genotype DD is 50%

Suppose Pr(disease | Dd ) = 0.2, Pr(disease | dd) = 0.05dd individuals can rarely display the disease, largelybecause of exposure to adverse environmental conditions

If freq(d) = 0.9, what is Prob (DD | show disease) ?freq(disease) = 0.12*0.5 + 2*0.1*0.9*0.2 + 0.92*0.05 = 0.0815From Bayes’ theorem, Pr(DD | disease) = Pr(disease |DD)*Pr(DD)/Prob(disease) = 0.12*0.5 / 0.0815 = 0.06 (6 %)

dd individuals can give rise to phenocopies 5% of the time,showing the disease but not as a result of carrying therisk allele

Pr(Dd | disease) = 0.442, Pr(dd | disease) = 0.497

Thus about 50% of the diseased individuals are phenocopies

Introduction to Genetics. Topics Darwin and Mendel Probability Mendelian genetics –Mendel's...

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