Upload File

genotype frequencies without random matingrogers/ant5221/lecture/inbreed-2x3.pdf · inbreeding alan...

  • Home
  • Documents
  • Genotype frequencies without random matingrogers/ant5221/lecture/inbreed-2x3.pdf · Inbreeding Alan R. Rogers March 24, 2020 1/23 Inbreeding depression in corn Genotype frequencies
4
Inbreeding Alan R. Rogers March 24, 2020 1 / 23 Inbreeding depression in corn (Jones 1924) Two plants on left are from inbred homozygous strains. Next: the F 1 offspring of these strains Then offspring (F 2 ) of two F 1 s. Then F 3 And so on. 2 / 23 Inbreeding depression in humans ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) (Bittles and Neel, 1994) 0 10 20 30 40 50 0 10 20 30 40 50 Outbred Cousin marriage Offspring mortality: inbred versus outbred Offspring of cousin marriages are less likely to survive. (Bittles and Neel 1994) 3 / 23 Genotype frequencies without random mating Genotype Frequency A 1 A 1 p 2 + pqF A 1 A 2 2pq(1 - F ) A 2 A 2 q 2 + pqF Describes any bi-allelic locus. F = 0 under random mating. Reduces to Hardy-Weinberg. F > 0 under inbreeding. Gives excess of homozygotes. F is the coefficient of inbreeding. 4 / 23 Example Assume p =1/2 Frequency Genotype F =0 F =0.1 A 1 A 1 0.25 0.275 A 1 A 2 0.50 0.450 A 2 A 2 0.25 0.275 Inbred population has more homozygotes. Suffers if either Heterozygotes tend to have high fitness Deleterious alleles tend to be recessive. 5 / 23 Outline of theory Inbreeding increases F , which increases homozygosity, which decreases fitness. 6 / 23

Upload: others

Post on 11-Nov-2020

3 views

Category:

Documents


4 download

Report

TRANSCRIPT

Page 1: Genotype frequencies without random matingrogers/ant5221/lecture/inbreed-2x3.pdf · Inbreeding Alan R. Rogers March 24, 2020 1/23 Inbreeding depression in corn Genotype frequencies

Inbreeding

Alan R. Rogers

March 24, 2020

1 / 23

Inbreeding depression in corn

(Jones 1924)

I Two plants on left arefrom inbredhomozygous strains.

I Next: the F1 offspringof these strains

I Then offspring (F2)of two F1s.

I Then F3I And so on.

2 / 23

Inbreeding depression in humans

●

●

●

●●

●

●

●●

●

●

●●

●●

●

●

●

●

●●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)(Bittles and Neel, 1994)0

10

20

30

40

50

0 10 20 30 40 50Outbred

Cou

sin

mar

riage

Offspring mortality: inbred versus outbred

Offspring of cousinmarriages are less likely tosurvive.(Bittles and Neel 1994)

3 / 23

Genotype frequencies without random mating

Genotype Frequency

A1A1 p2 + pqFA1A2 2pq(1− F )A2A2 q2 + pqF

Describes any bi-allelic locus.

F = 0 under random mating.Reduces to Hardy-Weinberg.

F > 0 under inbreeding. Givesexcess of homozygotes.

F is the coefficient of inbreeding.

4 / 23

Example

Assume p = 1/2Frequency

Genotype F = 0 F = 0.1

A1A1 0.25 0.275A1A2 0.50 0.450A2A2 0.25 0.275

Inbred population has morehomozygotes.

Suffers if either

I Heterozygotes tend to havehigh fitness

I Deleterious alleles tend tobe recessive.

5 / 23

Outline of theory

Inbreeding increases F , which increases homozygosity, whichdecreases fitness.

6 / 23

Page 2: Genotype frequencies without random matingrogers/ant5221/lecture/inbreed-2x3.pdf · Inbreeding Alan R. Rogers March 24, 2020 1/23 Inbreeding depression in corn Genotype frequencies

Decay of heterozygosity under selfing

Gen. A1A1 A1A2 A2A2

0 N

. .. ...

. . .14

12

14

. .. ...

. . .

1 14N 1

2N 1

4N

... . .. ...

. . ....

1 14

12

14

1... . .

. .... . .

...2 3

8N 1

4N 3

8N

7 / 23

Decay of heterozygosity under selfing

Half of heterozygosity is lost each generation.

8 / 23

What about cousin mating, or mating between sibs?

It is extremely difficult to work this out, using the method we justused.

Between 1903 and 1915, no one could get it right.

Pearl (1913): Only for brother-sister matings does inbreedingreduce heterozygosity. [Wrong!]

Solution: build theory looking backwards in time, not forwards.

9 / 23

Kinds of gene identity

There are two senses in which a pair of gene copies may be“identical:”

identity in state : copies of same alleleidentity by descent : copies of same gene copy in an ancestor

Abbreviation: IBD = Identity by Descent

10 / 23

Gametes a and b are identical by descent

◦ •.............................

.......................................................................................................

............................................................................................................................................................................................................

◦ •.............................

.......................................................................................................

............................................................................................................................................................................................................

◦ •.............................

.......................................................................................................

............................................................................................................................................................................................................

...........................................................................................................................................................................................................................

...........................

......................................................................................................................................................................................................................................................

a b

············

············

11 / 23

Identical in state, not by descent

• •.............................

.......................................................................................................

............................................................................................................................................................................................................

◦ •.............................

.......................................................................................................

............................................................................................................................................................................................................

◦ •.............................

.......................................................................................................

............................................................................................................................................................................................................

........................................................................................................................................................................................................

...........................

......................................................................................................................................................................................................................................................

a b

············

············

May be IBD relativeto an earliergeneration.

Not IBD relative tothe pedigree shown

here.

IBD is alwaysrelative to a

particulargeneration.

12 / 23

Page 3: Genotype frequencies without random matingrogers/ant5221/lecture/inbreed-2x3.pdf · Inbreeding Alan R. Rogers March 24, 2020 1/23 Inbreeding depression in corn Genotype frequencies

Identical neither in state nor by descent

◦ •.............................

.......................................................................................................

............................................................................................................................................................................................................

• ◦.............................

.......................................................................................................

............................................................................................................................................................................................................

◦ •.............................

.......................................................................................................

............................................................................................................................................................................................................

........................................................................................................................................................................................................

...........................

......................................................................................................................................................................................................................................................

a b

············

············

13 / 23

Uniting gametes

Consider the two gametes that unite to form an individual.

I F , is the probability that they are IBD

I What is the probability that they both carry A1?

Event Probability

IBD from A1-bearing ancestor FpDescend from two random ancestors who bothcarry A1

(1− F )p2

P11 = Fp + (1− F )p2

= p2 + pqF

14 / 23

All three genotypes

Genotype Frequency

A1A1 p2 + pqFA1A2 2pq(1− F )A2A2 q2 + pqF

Same formulas as before.

F is no longer arbitrary.

F is probability that unitinggametes are IBD.

15 / 23

Calculating F from a pedigree

F is probability that b and e are IBD,which I abbreviate as Pr[b = e].

F = Pr[b = c]× Pr[c = a]× Pr[a =a′]× Pr[a′ = d ]× Pr[d = e].

Pr[b = c] = 1/2, because b has anequal chance of coming from C or from

B’s other parent.

Ditto Pr[c = a], Pr[a′ = d ], &Pr[d = e].

16 / 23

Pr[a = a′] is also 1/2

Let ⊕ and ⊗ represent the two genecopies in A. There are four equally

likely possibilities for gametes a and a′:(⊕,⊕), (⊕,⊗), (⊗,⊕), or (⊗,⊗).

a = a′ in half of these possibilities, soPr[a = a′] = 1/2.

Regardless of which gene copy Acontributes to C , there is a 50% chancethat he contributes the same one to D.

17 / 23

Calculating F from a pedigree (conclusion)

F = Pr[b = c]× Pr[c = a]× Pr[a =a′]× Pr[a′ = d ]× Pr[d = e].

Each of the probabilities in this productequals 1/2, so F = 1/25.

In general, each loop in the pedigreecontributes 1/2n, where n is the number

of ancestors in the loop.

18 / 23

Page 4: Genotype frequencies without random matingrogers/ant5221/lecture/inbreed-2x3.pdf · Inbreeding Alan R. Rogers March 24, 2020 1/23 Inbreeding depression in corn Genotype frequencies

Mating between full siblings

..............................................................................................................................

.............................................................................................................................................................................................................................................................................................. ...........................

..............................................................................................................................................................................................................................................................................................

...........................

..............................................................................................................................

........................................................................................................................................................................... ...........................

...........................................................................................................................................................................

...........................

A B

C D

E

The two loops are mutuallyexclusive: the two gene copies inE cannot be IBD from A andalso from B.

F is the sum of the probabilitiesof the 2 loops.

Each loop contributes 1/23.

F = 2× 123

= 1/22

19 / 23

A complex pedigree

..............................................................................................................................

.............................................................................................................................................................................................................................................................................................. ...........................

..............................................................................................................................................................................................................................................................................................

...........................

..............................................................................................................................

..............................................................................................................................

.............................................................................................................................................................................................................................................................................................. ...........................

..............................................................................................................................

........................................................................................................................................................................... ...........................

...........................................................................................................................................................................

...........................

A B

C D

E F

G

Three loops with probabilities1/23, 1/25, and 1/25.

F = 1/23 + 2× 125

= 3/24

20 / 23

Darwin-Wedgewood Genealogy

..............................................................................................................................

...................................................................................................................................................................................................................................................................................................................................................................................................... ...........................

......................................................................................................................................................................................................................................................................................................................................................................................................

...........................

..............................................................................................................................

..............................................................................................................................

..............................................................................................................................

........................................................................................................................................................................................................................... ...........................

...........................................................................................................................................................................................................................

...........................

Josiah Wedgewood Sarah Wedgewood

Susannah Wedgewood Josiah Wedgewood II

Charles Darwin Emma Wedgewood

George Darwin

21 / 23 22 / 23

Summary

I The inbreeding coefficient, F , is the probability that the twogene copies in an individual are IBD from some givengeneration in the past.

I Inbreeding subtracts 2pqF from heterozygosity and adds pqFto the frequency of each of the two homozygous genotypes.

I Each loop in a pedigree contributes 1/2n to F , where n is thenumber of ancestors in the loop.

I Sum across loops to calculate F .

23 / 23


Inbreeding and inbreeding depression in endangered red ... · Inbreeding and inbreeding depression in endangered red wolves (Canis rufus) KRISTIN E. BRZESKI,* DAVID R. RABON JR,†

polysat version 1.3 Tutorial Manual - Amazon S3 · PDF file6.4.1 Allele diversity and frequencies . . . . . . . . . . . . .43 6.4.2 Genotype frequencies ... convert genotype data

24 RANDOM MATING INBREEDING, OUTBREEDING, AND · INBREEDING, OUTBREEDING, AND 24 RANDOM MATING Objectives •Determine how nonrandom breeding affects allele and genotype frequencies

Inbreeding Trends in Pedigree Beef Cattle - ICBF€¦ · Inbreeding Trends in Pedigree Beef Cattle. 2 Inbreeding What is it? •Inbreeding is defined as the probability that two alleles

1 Weak forces in Evolution Dan Graur. 2 Changes in allele frequencies are important. Changes in genotype frequencies are not so important. 1. 2

General analysis of inbreeding

Summary of previous lesson Dominant vs. codominant genetic markers Concept of “genotype” Alternatively fixed allele vs.difference in frequencies PLANT

Academic Inbreeding: An Evaluation

-mating population, the genotype and allele frequencies

Forming groups to calculate inbreeding coefficients · Inbreeding coefficients – The level of inbreeding is calculated as the probability of two alleles being identical by descent

A Genomewide Assessment of Inbreeding Depression… · A Genomewide Assessment of Inbreeding Depression: Gene Number, Function ... Assessment of Inbreeding Depression: Gene ... recesivos

Estimating Allele Frequencies and Inbreeding Coefficients ...pg2113/index_files/Gorroochurn & Hodge-MPS.pdf · Some of the methods of estimation of allele frequencies and inbreeding

Population Genetics Genetic variation Genotype frequencies Allele frequencies Hardy-Weinberg Equilibrium Inferring Evolution Applying HWE p 2 + 2pq + q

Lecture 5 Inbreeding and Crossbreeding · Genotype frequencies under inbreeding •The inbreeding coefficient, F •F = Prob(the two alleles within an individual are IBD)--identical

Variation in inbreeding rates across the range of Northern ...inbreeding associated with F¼0.125 (Table 1), where F is the inbreeding coefficient that reflects the probability that

The dangers of inbreeding in the Arabian horse · 2019-02-12 · The danger of inbreeding in the Arabian horse • Inbreeding results in: - inbreeding depression (fitness and reproduction)

Lecture 7: Inbreeding and Crossbreeding. Inbreeding Inbreeding = mating of related individuals Often results in a change in the mean of a trait Inbreeding

Lecture 3: Allele Frequencies and Hardy-Weinberg …sdifazio/popgen_12/lectures/aug27_HW...Hardy-Weinberg Equilibrium After one generation of random mating, genotype frequencies remain

1997.INBREEDING IN AMMOTRAGUS - Digital CSICdigital.csic.es/bitstream/10261/118245/1/High levels of inbreeding in... · inbreeding coefficient. Previous considerations and the strong

15.1 SELECTIVE BREEDING (Hybridization vs. Inbreeding; biotechnology) 15.1 SELECTIVE BREEDING (Hybridization vs. Inbreeding; biotechnology)

BIOL 432 - Evolution - Western Washington Universityfire.biol.wwu.edu/.../biol432_S12/BIOL_432_F11_58_ho.pdf1 BIOL 432 - Evolution Lecture 8 Expected Genotype Frequencies in the Absence

KEY CONCEPT Hardy-Weinberg equilibrium provides a framework … · 2020. 3. 1. · Hardy-Weinberg equilibrium describes populations that are not evolving. • Genotype frequencies

TLR4 genotype and environmental LPS mediate RSV ...dm5migu4zj3pb.cloudfront.net/manuscripts/75000/75183/JCI75183.… · in Hardy-Weinberg equilibrium, and the frequencies were sim-ilar

Last day… examined basic equation of population genetics, Hardy-Weinberg equilibrium (H-W eq.), describes expected genotype frequencies in stable pop

Maximum Likelihood Estimation for Allele Frequenciescsg.sph.umich.edu/abecasis/class/666.07.pdf · Next Series of Lectures zEstimating allele and haplotype frequencies from genotype

Inbreeding - jvanderw.une.edu.au

Equilibrium - NIST1 Duncan Taylor [email protected] Hardy Weinberg equilibrium Allele frequencies Genotype probabilities Confidence intervals Databases Linkage When a DNA profile

Chapter 23: Population Genetics (Microevolution) · Chapter 23: Population Genetics (Microevolution) Microevolution is a change in allele frequencies or genotype frequencies in a

What is Population Genetics? About microevolution (evolution within species) The study of the change of allele frequencies, genotype frequencies, and

Genotype Frequencies - Rice Universitynakhleh/COMP571/Slides-Fall2010/GenotypeFrequencies.pdfGenotype Frequencies In 1908, Hardy and Weinberg formulated the relationship that can be

Inbreeding Depression

Homework problem set 3 - New Mexico State Universitylithornis.nmsu.edu/~phoude/homework_set_3_2017_solutions.pdf · Homework problem set 3 . 14a) Which of the following genotype frequencies

Chapter 23: Population Genetics (Microevolution)Microevolution is a change in allele frequencies or genotype frequencies in a population over time allele frequency – proportion of

Analysis of Coyote Mitochondrial DNA Genotype Frequencies ... · Analysis of Coyote Mitochondrial DNA Genotype Frequencies: ... which equilibrium gene frequencies can be studied

Heterosis And Inbreeding Depression in Sorghum - …shodhganga.inflibnet.ac.in/bitstream/10603/19531/5/chapter 3.pdf · TCSV 930'28 104B Genotype Important feature Nan-tan, semi-loose