Download B 1 = B 2

Selection and Mutation

If either of the following occurs then the
population is responding to selection.
1. Some phenotypes allow greater survival to
reproductive age.
-or2. All individuals reach reproductive age but some
individuals are able to produce more viable
(reproductively successful) offspring.
If these differences are heritable then
evolution may occur over time.
Caution




It needs to be mentioned that most
phenotypes are not strictly the result of
their genotypes.
Environmental plasticity and
interaction with other genes may also be
involved.
In other words it is not as simple as we
are making it here but we have to start
somewhere.
1.
2.
Selection may alter allele frequencies or
violate conclusion #1
Selection may upset the relationship
between allele frequencies and genotype
frequencies.
Conclusion #1 is not violated but conclusion #2 is
violated.
In other words the allele frequencies remain stable but
genotype frequencies change and can no longer be
predicted accurately from allele frequencies.

After random mating which produces 1000
zygotes we get:
Initial frequencies
B1= 0.6; B2 = 0.4
B1B1
B1B2
B2B2
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
B1B2
B2B2
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
B2B2
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
360 B1B1
480 B1B2
160 B2B2
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
360 B1B1
100%
survive
480 B1B2
160 B2B2
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
360 B1B1
480 B1B2
100%
survive
75 %
survive
160 B2B2
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
number surviving
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
number surviving
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
360
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
number surviving
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
360
360
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
number surviving
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
360
360
80
1000 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
number surviving
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
360
360
80
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
differential survival of
offspring leads to reduced
numbers of some
genotypes
number surviving
The genotype frequencies
of mating individuals
which survive is
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
360
360
80
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
differential survival of
offspring leads to reduced
numbers of some
genotypes
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
differential survival of
offspring leads to reduced
numbers of some
genotypes
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
.10
differential survival of
offspring leads to reduced
numbers of some
genotypes
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
.10
B1 =
B2 =
differential survival of
offspring leads to reduced
numbers of some
genotypes
The resulting allelic
frequencies in the new
reproducing population is
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
.10
differential survival of
offspring leads to reduced
numbers of some
genotypes
The resulting allelic
frequencies in the new
reproducing population is
B1 =
.45+1/2(.45)
= 0.675
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
.10
B1 =
.45+1/2(.45)
= 0.675
B2 =
differential survival of
offspring leads to reduced
numbers of some
genotypes
The resulting allelic
frequencies in the new
reproducing population is
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
.10
B1 =
.45+1/2(.45)
= 0.675
B2 =
1/2(.45)+0.10
= 0.325
differential survival of
offspring leads to reduced
numbers of some
genotypes
The resulting allelic
frequencies in the new
reproducing population is
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
.10
B1 =
.45+1/2(.45)
= 0.675
B2 =
1/2(.45)+0.10
= 0.325
an increase of
.075
a decrease of
.075
differential survival of
offspring leads to reduced
numbers of some
genotypes
The resulting allelic
frequencies in the new
reproducing population is
1000 total
800 total
Initial frequencies
B1= 0.6; B2 = 0.4
360 B1B1
480 B1B2
160 B2B2
100%
survive
75 %
survive
50 %
survive
number surviving
360
360
80
The genotype frequencies
of mating individuals
which survive is
.45
.45
.10
B1 =
.45+1/2(.45)
= 0.675
B2 =
1/2(.45)+0.10
= 0.325
an increase of
.075
a decrease of
.075
differential survival of
offspring leads to reduced
numbers of some
genotypes
The resulting allelic
frequencies in the new
reproducing population is
1000 total
800 total
Thus, conclusion #1 is violated and the allele frequencies are
changing; we are not in equilibrium. The population is evolving!



analyze the population on the basis of the
fitness of the offspring produced.
The fittest individuals will survive the
selection process and leave offspring of their
own.
We are going to define fitness as the survival
rates of individuals which survive to
reproduce.
MEAN FITNESS

If :
w11 = fitness of allele #1 homozygote (exp
B1B1)
w12 = fitness of the heterozygote (exp B1B2)
w22 = fitness of allele #2 homozygote exp
(B2B2)
mean fitness of the population will be described by the
formula: ŵ = p2w11 + 2pqw12 + q2w22
CAUTION! Use ONLY allele frequencies in these
formulas NOT genotype frequencies!


B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
Figure the mean fitness now.



B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
Figure the mean fitness now.
ŵ= (.6)2(1)+



B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
Figure the mean fitness now.
ŵ= (.6)2(1)+(2(.6)(.4)(.75)) +
B1= 0.6 and B2 = 0.4 and
fitness of B1B1 = 1.0 (100% survived)
fitness of B1B2 = .75 ( 75% survived)
fitness of B2B2 = .50 (50% survived)
 Figure the mean fitness now.
ŵ= (.6)2(1)+(2(.6)(.4)(.75)) + (.4)2 (.5) = .80

B1B1 = P2w11
ŵ
B1B2 = 2pqw12
ŵ
B2B2 = q2w22
ŵ
We can use these formulas which
can calculate the new expected
genotype frequencies based on the
fitness of each genotype and the
allele frequencies in the current
generation.
B1 = p2w11+pqw12
ŵ
B2 = pqw12+q2w22
ŵ
Δ B1 = Δp =
p (pw11+qw12 – ŵ)
ŵ
Δ B2 = Δq = q (pw12+qw22 – ŵ)
ŵ


Go back to the problem we did in class last
time. Taking this current population that you
have already analyzed, figure out what the
new genotype and allele frequencies will be if
the fitness of these individuals is actually as
follows:
SS individuals 0.8 ; Ss individuals 1.0 and the
ss individuals 0.6.
Last time we calculated S = .82 and s = .18
Now we set the fitnesses at w11(SS)=.8;w12(Ss)=1;w22(ss)=.6
Calculate the ŵ and B1B1; B1B2; and B2B2 values for the next
generation now
ŵ = p2w11 + 2pqw12 + q2w22
ŵ= (.82) 2 (.8) + 2(.82)(.18)(1.0) + (.18)2 (.6)
ŵ = .537 + .295 + .019 = .85
B1B1 = P2w11
ŵ
; SS = (.82)2(.8) / .85 = .633
B1B2 = 2pqw12
ŵ
;Ss = 2(.82)(.18)(1.0) / .85 = .347
B2B2 = q2w22
ŵ
;ss = (.18)2(.6) / .85 = 0.023
Hint: Do they add up to 1.0?
We an also calculate the new allele frequencies as well
B1 =
B2 =
p2w11+pqw12
ŵ
pqw12+q2w22
ŵ
S = (.82)2(.8) + (.82)(.18)(1.0) = .806
.85
s = (.82)(.18)(1.0) + (.18)2(.6) = .196
.85
So…… B1B1 = .63
B1B2 = .35
B2B2 = .02
and
B1 =
.80
B2 = .20
Is this population in equilibrium?
Have the allele frequencies changed?
Can we predict the genotype frequencies from the
allelic frequencies?
Fruit fly experiments of
Cavener and Clegg



Worked with fruit flies having two versions
of the ADH (alcohol dehydrogenase)
enzyme, F and S. (for fast and slow moving
through an electrophoresis gel)
Grew two experimental populations on
food spiked with ethanol and two control
populations on normal, non-spiked food.
Breeders for each generation were picked
at random.
Took random samples of flies every few
generations and calculated the allele
frequencies for AdhF and AdhS
Figure 6.13 pg 185




only difference is ethanol in food
no migration
assured random mating
population size, drift?
no mutation.
Must be selection for the fast form of gene.
Indeed studies show that AdhF form breaks
down alcohol at twice the rate as the AdhS
form.
Therefore offspring carrying this allele are
more fit and leave more offspring and the
make-up of gene pool changes.


Selection may upset the relationship
between allele frequencies and genotype
frequencies.
Conclusion #1 ( allele frequencies do not
change) is not violated but conclusion #2
(that we can predict genotype frequencies
from allele frequencies) is violated
.
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
Number of survivors to
reproductive age
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
125 / 750
0.167
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
125 / 750
0.167
500 / 750
0.667
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
125 / 750
0.167
500 / 750
0.667
125 / 750
0.167
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
The resulting allelic
frequencies in new mating
population
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
125 / 750
0.167
500 / 750
0.667
125 / 750
0.167
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
The resulting allelic
frequencies in new mating
population
B1 =
.167+1/2 (0.667)
= 0.5
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
125 / 750
0.167
500 / 750
0.667
125 / 750
0.167
B1 =
.167+1/2 (0.667)
= 0.5
B2 =
½ (.667) + .167
= 0.5
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
The resulting allelic
frequencies in new mating
population
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
125 / 750
0.167
500 / 750
0.667
125 / 750
0.167
B1 =
.167+1/2 (0.667)
= 0.5
B2 =
½ (.667) + .167
= 0.5
No change
No change
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
The resulting allelic
frequencies in new mating
population
1000 Total
750 total
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
125 / 750
0.167
500 / 750
0.667
125 / 750
0.167
B1 =
.167+1/2 (0.667)
= 0.5
B2 =
½ (.667) + .167
= 0.5
No change
No change
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
The resulting allelic
frequencies in new mating
population
1000 Total
750 total
Thus conclusion #1 is not violated therefore this population has not
evolved…..but…..
Initial B1 = 0.5
Initial B2= 0.5
250 B1B1
500 B1B2
250 B2B2
Differential fitness of the
genotypes
Fitness .50
Fitness 1.0
Fitness .50
125
500
125
125 / 750
0.167
500 / 750
0.667
125 / 750
0.167
B1 =
.167+1/2 (0.667)
= 0.5
B2 =
½ (.667) + .167
= 0.5
No change
No change
Number of survivors to
reproductive age
The genotype frequencies of
mating individuals which
survive
The resulting allelic
frequencies in new mating
population
1000 Total
750 total
Change in allele frequency
Thus conclusion #1 is not violated therefore this population has not
evolved….but…..
Conclusion #2 is violated. We are not in equilibrium.
Frequency of B1B1 =.167 which is not equal to (.5)2
Kuru example among the Foré in New Guinea
◦ Pg 188-191
◦ Wanted to determine if there was a genetic basis to
the resistance of kuru infection.
 Ritualistic mortuary feasts, only young women ate
the contaminated nervous system tissue leading to
CJD (similar to mad cow disease)
 Among young women who never participated he
Met allele = 0.48 and the Val allele 0.52;
Genotypes were: Met/Met 0.22; Met/Val 0.51 and
Val/ Val 0.26 very close to the values expected for
H-W.






Met = 0.52 and Val = 0.48
The expected genotypes are
Met/Met 0.27 ; Met/ Val 0.5 and Val/Val 0.23
The actual were:
Met/Met 0.13 ; Met/ Val 0.77 and Val/Val
0.10
Appears homozygotes are susceptible but
heterozygotes are protected.


HIV example in book. pg 191
Two conditions must be met
1. Need a high enough frequency of the
beneficial allele in the population gene
pool
2. There must be high selection
pressure for the allele in the same
area. In this case a high incidence of
HIV infection.

If selection is acting, does the rate of
evolution of a particular allele depend on
whether it is….
dominant or recessive?
heterozygote or homozygote?
Tribolium Beetle example






Dawson’s Flour beetle example
Studied a gene locus that had a wild type
(+) allele and a lethal allele.
+/+ or +/L are normal L/L is lethal.
Two experimental populations composed
of all heterozygotes +/L
Therefore started with + = 0.5 and L =0.5.
Expected populations to evolve toward
lower frequency of the L allele.
Results showed that the
recessive lethal did
drop rapidly at first but
slowed down over
successive generations.
WHY?
In each succeeding generation all LL
are lost and ++ makes up a greater
proportion of the survivors.
As you go on there are less and less
homozygous lethals for selection to act on
and the lethal allele hides in the
heterozygotes




Dawson showed that dominance and allele
frequency interact to determine the rate of
evolution when acted on by selection
If a recessive allele is common evolution is
rapid because there are a lot of homozygotes
that express the phenotype for selection to
act on.
If recessive allele is rare, evolution is slow
because the rare allele is hidden in the
heterozygotes where selection cannot act.
His experiments also demonstrated that
◦ controlled lab situations can accurately predict the
course of evolution
◦ populations do what you would expect if selection is
occurring as predicted by the evolutionary theory.



Normally in a recessive/ dominant gene, the
fitness of the heterozygote will be equal to
one of the homozygotes
Also, it is possible for the heterozygotes to
have a fitness intermediate to the two
homozygotes.
Thirdly we may find Heterozygote Superiority
or Inferiority




Studied a gene in which
Homozygotes for one allele are viable
Homozygotes for the other allele are
not viable and are lethal.
Heterozygotes have a higher fitness
than either homozygotes

Started with all heterozygotes to
establish a new population (each allele
=.5)
After several generations equilibrium was
reached at .79 frequency for the viable allele
This means that the lethal allele was maintained
at frequency of 0.21! How could this be?
Started more populations beginning with
frequency of .975 of viable allele. Expect the
population to eliminate all lethal alleles and fix
the viable allele at 1.0.
But .....
The viable allele dropped in frequency and the same
equilibrium around a frequency of .79 was reached for the
viable allele!
Figure 6.18 pg 200



There is some advantage to the heterozygote
condition and the heterozygote actually has a
superior fitness to either homozygote.
Example in humans is sickle cell anemia
Leads to the maintenance of genetic diversity
= balanced polymorphism




Where the heterozygote condition is
inferior to either of the homozygotes
What do you predict would happen to the
allele frequencies here?
Leads to fixation of one allele in the
population, while the other is lost.
Either allele may be fixed depending on
conditions and beginning frequencies of
each allele in the gene pool.


Leads to a loss of genetic diversity
Although if different alleles are fixed in
different populations can help maintain
genetic diversity among populations


When one allele is consistently favored it will
be driven to fixation
When heterozygote is favored both alleles are
maintained and at a stable equilibrium
(balanced polymorphism) even though one of
the alleles may be lethal in the homozygous
state.




The Elderflower orchid
example in book
Population’s allele
frequencies remain at or
near an equilibrium but
it is due to the direction
of selection fluctuating.
First one allele is favored
and then the other.
The population
fluctuates around an
equilibrium point.





Bumblebees visit yellow and purple flowers
alternately
The least frequent phenotype is visited more often
and receives more pollination events.
In subsequent generations this color becomes
more and more frequent until it becomes the
dominant color.
Once this happens then the same color becomes
less frequently visited and the other color
becomes favored.
Oscillation between the two colors continues and
the favored allele alternates over time around
some mean equilibrium value.


Mutation is the source of all new alleles
Mutation provides the raw material on
which selection can act




Mutation alone is a weak or nonexistent
evolutionary force
If all mutations that happened, occurred in
gametes so that they would be immediately
passed on to their offspring and ….
the rate of mutation were high, say Aa at
a rate of 1 in 10,000 per generation.
then the rates are very slow as shown in
figure 6.23
Figure 6.23 pg. 211
In concert with selection, mutation becomes a
potent evolutionary force.
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Used a strain of E. coli that cannot exchange
DNA (conjugation) so the only possible source
of genetic variation is mutation.
Showed steady increases in fitness and size
over 10,000 generations in response to a
demanding environment. (little over 4 years)
However, increases in fitness occurred in jumps
when a beneficial mutation occurred and then
spread rapidly through the population
Figure 6.25 pg 213
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When mutations are deleterious
Selection acts to eliminate them
Deleterious Mutations persist because they
are created anew over and over again
When the rate at which deleterious mutations
are formed exactly equals the rate at which
they are eliminated by selection the allele is
in equilibrium. = mutation-selection balance
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If the mutation is only mildly deleterious and
therefore selection against it is weak; and
Mutation rate is high then
◦ The equilibrium frequency of the mutated
allele will be relatively high in the
population.
If, on the other hand, there is strong
selection against a mutation (the mutation is
highly deleterious) and the mutation rate is
low then
◦ Equilibrium ratio of the mutated allele will
be low in the population
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Spinal muscular atrophy, second most
common lethal autosomal recessive disease
in humans. Selection coefficient is .9 against
the disease mutations.
However, among Caucasians 1 in 100 people
carry the disease causing allele.
Research shows that the mutation rate for
this disease is quite high
Mutation selection balance is proposed
explanation for persistence of mutant
alleles.
http://www.smafoundation.org
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Cystic fibrosis is the most common lethal
autosomal recessive disease in Caucasians
Mutation-selection balance alone cannot
account for the high frequency of the allele
= .02
Appears to also be some heterozygote
superiority involved
Heterozygotes are resistant to typhoid
fever bacteria and have superior fitness
during typhoid fever epidemic.
At the current time it is believed that CF is
an example of heterosis and not mutationselection balance
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An autosomal dominant allele
Is actually increasing in the human
population.
Any ideas why?
May be because it increases the tumor
supressor activity in cells dramatically
lowering the incidence of cancer in those with
the defective allele.
They survive through the reproductive years
and leave more offspring than their
unaffected siblings.