Inbreeding

Inbreeding

inbreeding coefficient F – probability that given alleles are identical by descent

- note: homozygotes may arise in population from unrelated parents

but: generally will have less overall homozygosity by random chance than from inbreeding

Inbreeding

Result of inbreeding is inbreeding depression:

- loss of fitness due to deficient heterozygosity

- recessive traits are expressed

Inbreeding

inbreeding coefficient F

F = 1 – H F is a function of the ratio

2pq of observed over expected H (# heterozygotes)

H = observed frequency of heterozygotes in the population p = frequency of one allele in the population q = frequency of alternate allele, or 1-p 2pq = expected frequency of heterozygotes in the population

Inbreeding

inbreeding coefficient F

F = 1 – H F is a function of the ratio

2pq of observed over expected H

example: p = 0.5, therefore q = 0.5 2pq = 0.5

Inbreeding

inbreeding coefficient F

F = 1 – H F is a function of the ratio

2pq of observed over expected H

example: p = 0.5, therefore q = 0.5 2pq = 0.5

if in H-W equilibrium, H = 0.5 so F = 0 = random matingif no heterozygotes, H = 0 F = 1 = complete inbreeding

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.5G

ener

atio

n

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.5

0.250 0.250

Gen

erat

ion

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.5

0.250 0.250

0.125 0.250 0.125

Gen

erat

ion

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.50.375 0.250 0.375 0.5 0.5

Gen

erat

ion

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.50.375 0.250 0.375 0.5 0.5

= ½ of het.frequency

= 1 x hom. frequency+ ¼ of het. frequency

Gen

erat

ion

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.50.375 0.250 0.375 0.5 0.50.438 0.125 0.438 0.5 0.50.469 0.063 0.469 0.5 0.50.484 0.031 0.484 0.5 0.50.492 0.016 0.492 0.5 0.50.496 0.008 0.496 0.5 0.50.498 0.004 0.498 0.5 0.50.499 0.002 0.499 0.5 0.50.500 0.001 0.500 0.5 0.50.500 0.000 0.500 0.5 0.5

Gen

erat

ion

= ½ of het.frequency

= 1 x hom. frequency+ ¼ of het. frequency

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.50.375 0.250 0.375 0.5 0.50.438 0.125 0.438 0.5 0.50.469 0.063 0.469 0.5 0.50.484 0.031 0.484 0.5 0.50.492 0.016 0.492 0.5 0.50.496 0.008 0.496 0.5 0.50.498 0.004 0.498 0.5 0.50.499 0.002 0.499 0.5 0.50.500 0.001 0.500 0.5 0.50.500 0.000 0.500 0.5 0.5

Gen

erat

ion

= ½ of het.frequency

= 1 x hom. frequency+ ¼ of het. frequency

If there is a recessive deleterious allele, the population is in trouble….

Inbreeding part 2 – or, how to predict the changes in genotype frequencies

F = 1 – H 2pq

H = observed frequency of heterozygotes in the population2pq = expected frequency of heterozygotes

Inbreeding

F = 1 – H 2pq

H = observed frequency of heterozygotes in the population2pq = expected frequency of heterozygotes

Define H = # heterozygotes (freq Aa) = 2pq –F2pq D = # homozygotes (freq AA)

R = # alternate homozygotes (freq aa)

Inbreedingselfing: F = 0.5 (loss of 50% of total variation per generation)

AA Aa aa p q 1.0 0.5 0.5

0.250 0.500 0.250 0.5 0.50.375 0.250 0.375 0.5 0.50.438 0.125 0.438 0.5 0.50.469 0.063 0.469 0.5 0.50.484 0.031 0.484 0.5 0.50.492 0.016 0.492 0.5 0.50.496 0.008 0.496 0.5 0.50.498 0.004 0.498 0.5 0.50.499 0.002 0.499 0.5 0.50.500 0.001 0.500 0.5 0.50.500 0.000 0.500 0.5 0.5

Gen

erat

ion

heterozygote freq. decreased by F

homozygote freq.increased by ½ F

Inbreeding

Thus, the impact of inbreeding on each genotype is:

freq (AA) = D = p2 + Fpq

freq (Aa) = H = 2pq – 2Fpq

freq (aa) = R = q2 + Fpq

(remember F = 1 is the result of complete inbreeding, no heterozygotes)

Inbreeding

Example of the impact of inbreeding:

p = 0.6, q = 0.4, 2pq = 0.48

F = 0

D = p2 + Fpq 0.36

H = 2pq – 2Fpq 0.48

R = q2 + Fpq 0.16

i.e., no inbreeding, population is in Hardy-Weinberg equilibrium, genotypes are predictable based H-W equation, p2 + 2pq + q2

Inbreeding

Example of the impact of inbreeding:

p = 0.6, q = 0.4, 2pq = 0.48

F = 0 F = 0.5

D = p2 + Fpq 0.36 0.48

H = 2pq – 2Fpq 0.48 0.24

R = q2 + Fpq 0.16 0.28

heterozygotes have decreased by ½ of 0.24; homozygotes have increased by ¼ of 0.24

Inbreeding

Example of the impact of inbreeding:

p = 0.6, q = 0.4, 2pq = 0.48

F = 0 F = 0.5 F = 1

D = p2 + Fpq 0.36 0.48 0.60

H = 2pq – 2Fpq 0.48 0.24 0.0

R = q2 + Fpq 0.16 0.28 0.40

heterozygotes have decreased by all of 0.24; homozygotes have increased by ½ of 0.24

Inbreeding

How much inbreeding is acceptable?Slow increase in inbreeding results in less inbreeding depression than rapid inbreeding– slow purging of deleterious alleles

Inbreeding

How much inbreeding is acceptable?Slow increase in inbreeding results in less inbreeding depression than rapid inbreeding

– slow purging of deleterious alleles

Low genetic variability is much less important than loss of variability

Inbreeding

Deliberate use of inbreeding

- breed out deleterious alleles

- temporary reduction in fitness, then stabilizes

- increase fitness by crossing inbred lines

Inbreeding

Examples?

Beware of file drawer effect/publication bias!- In stable populations, low variation is uninteresting

- In small populations, high variation is uninteresting

Small populations do not evolve

Forces that change neutral genes among sub-populationsfounder effect > reduced diversitygenetic drift > changes in allelic frequency

“Smallness and randomness are inseparable.”M. Soulé (1985)

How big does a population need to be to avoid loss of genetic variation?

What is a ‘small’ population?

Effective population size (Ne)

Ne reflects the probability that genetic variation will not be lost by random chance

Effective population size (Ne)

Ne reflects the probability that genetic variation will not be lost by random chance

Ideal:• 1:1 sex ratio

• all individuals live to maturity and breed

• all adults have equal chances of mating with each other

• all individuals or pairs contribute equal numbers of offspring

• all of the offspring live

Effective population size (Ne) = N only if all of these are true

Effective population size - evidence

Not all individuals can mate with each other:

• 1,000 grizzly bears left in US. (1980s)

• less than 1% of range still occupied

• species now present in 6 isolated subpopulations

• estimated effective population size ~ 25% of census size

(Allendorf et al. 1991)

Effective population size - evidence

Not all parents contribute equally to next generation:

Social structure with mate competition• harem-polygynous species, e.g., lions, some bats• 5.4% of spawning male smallmouth bass

produce 54.7% of progeny (Gross & Kapuscinski 1997)

Sperm competition• mass spawning of 2,000 rainbow trout genetically equiv. to 88.5 spawners• mass spawning of 10,000 Chinook salmon equiv. to 132.5 spawners

(K. Scribner and others)

Effect of sex ratio on Ne

Ne – taking sex ratio (only) into account

Ne = 4Nm Nf

Nm + Nf

Effect of sex ratio on Ne

Ne = 4Nm Nf for 25 of each sex, 4*25*25 = 2,500 = 50Nm + Nf 25 + 25 50

Effect of sex ratio on Ne

Ne = 4Nm Nf for 25 of each sex, 4*25*25 = 2,500 = 50Nm + Nf 25 + 25 50

BUT for 40 males, 10 females

4*40*10 = 1600 =

32 40 + 10 50

Effect of sex ratio on Ne

25:25 Ne = 50

40:10 Ne = 32

49:1 Ne = 3.9

Effect of fluctuating populations on Ne

Calculate Ne as harmonic mean* over several generations

Ne = t (1/Nt)

t = generation (population sample)Nt = number of individuals in generation t

* gives greater weight to small numbers

Effect of fluctuating populations on Ne

Example:Gen. (t)  Pop. Size (Nt)  1/Nt 

1  500  0.002  2  500  0.002  3  500  0.002  4  500  0.002  5  500  0.002  6  50  0.02  7  500  0.002  8  500  0.002  9  500  0.002  10  500  0.002        

Arithmetic   Ne   Mean

455  263

Effect of fluctuating populations on Ne

Example:Gen. (t)  Pop. Size (Nt)  1/Nt 

1  500  0.002  2  50  0.02  3  500  0.002  4  50  0.02  5  500  0.002  6  50  0.02  7  50  0.02  8  50  0.02  9  500  0.002  10  500  0.002        

Arithmetic   Ne   Mean

275  91

Effect of family size on NeEffect of family size on Ne

Neur = k(Nk – 1) Vk + k(k -1)

ur = unequal reproductive outputk = mean number of surviving progenyVk = variance in family sizeN = total progeny

Effect of family size on NeEffect of family size on Ne

Neur = k(Nk – 1) Vk + k(k -1)

ur = unequal reproductive outputk = mean number of surviving progenyVk = variance in family sizeN = total progeny

40 34 40 51 40 62 40 26 40 3 40 99 40 5

Av 40 40N 280 280Ne 287 165

Inbreeding and Ne

rate of inbreeding F = rate at which heterozygosity is lost (or fixation occurs)

1

F = 2Ne

Inbreeding and Ne

effect of changes in sex ratio:

25:25 Ne=50 F = 1/100 = 1%

40:10 Ne=32 F = 1/64 = 1.6%

49:1 Ne=3.9 F = 1/7.8 = 12.8%

1:1 Ne=2 F = 1/4 = 25%

Retention of genetic variation in a small population

# generations Ne 1 5 10 1002 75 24 6 <<16 91.7 65 42 <<110 95 77 60 <120 97.5 88 78 850 99 95 90 36

100 99.5 97.5 95 60

N = 50, M=25, F=25

Assume a population of N = 50As sex ratio changes, equivalent Ne changes

Retention of genetic variation in a small population

# generations Ne 1 5 10 1002 75 24 6 <<16 91.7 65 42 <<110 95 77 60 <120 97.5 88 78 850 99 95 90 36

100 99.5 97.5 95 60

N = 50, M=5, F=45

Retention of genetic variation in a small population

# generations Ne 1 5 10 1002 75 24 6 <<16 91.7 65 42 <<110 95 77 60 <120 97.5 88 78 850 99 95 90 36

100 99.5 97.5 95 60

N = 50, M=3, F=47

Inbreeding

How much inbreeding is acceptable?• 1-3% per generation – 1% preferred• recommended Ne = 50, to maintain inbreeding at <1%

Inbreeding

How much inbreeding is acceptable?• 1-3% per generation – 1% preferred• recommended Ne = 50, to maintain inbreeding at <1%

BUTgenerally only 1/3 to ¼ of popn contributes to next generation

- so N should be 150-200

Small populations: founder effect/bottlenecks, drift, inbreeding

Minimal founder population for captive breeding: 50(M. Soulé 1980)

For long-term breeding, minimal population: 500(J. Franklin 1980 )

the “50/500 rule”

Franklin, I.R. 1980. Evolutionary change in small populations, in Conservation Biology: An evolutionary-ecological perspective. M.E. Soulé and B.A. Wilcox (eds). Sinauer, MA Soulé, M.E. 1980. Thresholds for survival: maintaining fitness and evolutionary potential, in Conservation Biology: An evolutionary-ecological perspective. M.E. Soulé and B.A. Wilcox (eds). Sinauer, MA

Small populations: founder effect/bottlenecks, drift, inbreeding

Minimal founder population for captive breeding: 50(M. Soulé 1980)

For long-term breeding, minimal population: 500(J. Franklin 1980 )

To balance mutation and drift: 5,000(R. Lande 1995)

Franklin, I.R. 1980. Evolutionary change in small populations, in Conservation Biology: An evolutionary-ecological perspective. M.E. Soulé and B.A. Wilcox (eds). Sinauer, MA Soulé, M.E. 1980. Thresholds for survival: maintaining fitness and evolutionary potential, in Conservation Biology: An evolutionary-ecological perspective. M.E. Soulé and B.A. Wilcox (eds). Sinauer, MALande, R. 1995. Mutation and conservation. Conservation Biology 9:782-791.

Inbreeding

When is population size too small (hopeless)?Przewalski’s horse 13 Guam rail 10 black-footed ferret 6 European bison 6 Speke’s gazelle 4 dusky seaside sparrow 2…1..…0

note: these are all captive (regulated) populations….