Genetica per Scienze Naturali a.a. 03-04 prof S. Presciuttini Estimating nucleotide mutation rate in human The approach by Nachman and Crowell to measuring

Genetica per Scienze Naturalia.a. 03-04 prof S. Presciuttini

Estimating nucleotide mutation rate in human The approach by Nachman and Crowell to measuring the The approach by Nachman and Crowell to measuring the

human mutation rate takes advantage of the well-known human mutation rate takes advantage of the well-known result that for neutral mutations, the mutation rate is equal result that for neutral mutations, the mutation rate is equal to the rate of mutation substitutionto the rate of mutation substitution

Knowing Knowing divergence time and generation lengthdivergence time and generation length of two of two species allowed themspecies allowed them to to estimate of the rate and pattern ofestimate of the rate and pattern of mutation mutation in in stretches of DNA without functionstretches of DNA without function

They They sequenced 18 pseudogenes in humans and sequenced 18 pseudogenes in humans and chimpanzees, including 12 on autosomes and 6 on the X chimpanzees, including 12 on autosomes and 6 on the X chromosomechromosome. In this way, they. In this way, they1.1. estimated the average mutation rate per nucleotide siteestimated the average mutation rate per nucleotide site2.2. compared mutation rates for different sites and for different compared mutation rates for different sites and for different

classes of mutation to evaluate heterogeneity of mutation rateclasses of mutation to evaluate heterogeneity of mutation rate3.3. compared rates of divergence on the X chromosome and on compared rates of divergence on the X chromosome and on

autosomes to evaluate the hypothesis that the X chromosome autosomes to evaluate the hypothesis that the X chromosome has a lower mutation rate than the autosomeshas a lower mutation rate than the autosomes

4.4. provided an approximation of the genomic deleterious provided an approximation of the genomic deleterious mutation rate by considering the total mutation rate and the mutation rate by considering the total mutation rate and the fraction of the genome that is subject to constraintfraction of the genome that is subject to constraint



Nachman and Crowell main results


Nachman and Crowell results (1)








Mutation rates per gene in humans

A list of traits specified by dominant mutant alleles whose mutation rates have been estimated is given below

Mutation rates may be estimated by a direct method (counting affected individuals born from normal parents among a large number of births) or by an indirect method (using the frequency of affected individuals in a population and estimating their average selective disadvantage)


Summary of mutation rates in human

Mutation rates per generationMutation rates per generation• Per base pair ~10Per base pair ~10-8 -8

• Per gene ~10Per gene ~10-6 -6 - 10- 10-5-5

• Per genomePer genomeAll point mutations: All point mutations: ~~100 per gamete100 per gameteDeleterious mutations: 1 Deleterious mutations: 1 ~~ 2 per gamete 2 per gamete


Selection Changes in a species in response to a changing environment occur because the Changes in a species in response to a changing environment occur because the

different genotypes produced by mutation and recombination have different abilities different genotypes produced by mutation and recombination have different abilities to survive and reproduce.to survive and reproduce.

The The differential rates of survival and reproduction are what is meant by differential rates of survival and reproduction are what is meant by selectionselection, and the process of selection results in changes in the frequencies of the , and the process of selection results in changes in the frequencies of the various genotypes in the population.various genotypes in the population.

Darwin called the process of differential survival and reproduction of different types Darwin called the process of differential survival and reproduction of different types natural selectionnatural selection by analogy with the artificial selection carried out by animal and by analogy with the artificial selection carried out by animal and plant breeders when they deliberately select some individuals of a preferred type.plant breeders when they deliberately select some individuals of a preferred type.

The relative probability of survival and rate of reproduction of a phenotype or The relative probability of survival and rate of reproduction of a phenotype or genotype is now called its genotype is now called its Darwinian fitnessDarwinian fitness..

Although geneticists sometimes speak loosely of the fitness of an individual, the concept Although geneticists sometimes speak loosely of the fitness of an individual, the concept of fitness really applies to the average survival and reproduction of individuals in a of fitness really applies to the average survival and reproduction of individuals in a phenotypic or genotypic class. Because of chance events in the life histories of phenotypic or genotypic class. Because of chance events in the life histories of individuals, even two organisms with identical genotypes and identical environments individuals, even two organisms with identical genotypes and identical environments will differ in their survival and reproduction rates. It is the will differ in their survival and reproduction rates. It is the fitness of a genotype on fitness of a genotype on averageaverage over all its possessors that matters over all its possessors that matters


How selection works The simplest way to see the effect of selection is to consider an allele, The simplest way to see the effect of selection is to consider an allele,

aa, that is completely lethal before reproductive age in homozygous , that is completely lethal before reproductive age in homozygous condition.condition. Suppose that, in some generation, the allele frequency of this gene is 0.10. Suppose that, in some generation, the allele frequency of this gene is 0.10.

Then, in a randomThen, in a random mating population, the proportions of the three genotypes mating population, the proportions of the three genotypes after fertilization are.after fertilization are.

How are these genotype frequencies connected to the allele frequency How are these genotype frequencies connected to the allele frequency of 0.1?of 0.1?

The answer is provided by the The answer is provided by the law of Hardy-Weinberglaw of Hardy-Weinberg, which is the way an , which is the way an elementary statistical principle is named in population geneticselementary statistical principle is named in population genetics


Hardy-Weinberg Equilibrium If the frequency of allele If the frequency of allele AA is is pp in both the sperm and the eggs and the frequency of in both the sperm and the eggs and the frequency of

allele allele aa is is qq = = 1-1-pp, then the consequences of random unions of sperm and eggs , then the consequences of random unions of sperm and eggs may may be worked out be worked out in in a a Punnett square. Punnett square.

The probability that both the sperm and the egg will carry The probability that both the sperm and the egg will carry AA is is pp × × pp = = pp22, so , so pp22 will will be the frequency of be the frequency of AA//AA homozygotes in the next generation. In like manner, the homozygotes in the next generation. In like manner, the chance of heterozygotes chance of heterozygotes AA//aa will be ( will be (pp × × qq) + () + (qq × × pp) = 2) = 2pqpq, and the chance of , and the chance of homozygotes homozygotes aa//aa will be will be qq × × qq = = qq22. The three genotypes, after a generation of . The three genotypes, after a generation of random mating, will be in the frequencies random mating, will be in the frequencies pp22:2:2pqpq::qq22..

IIn the n the nextnext generation, the frequencies of the three genotypes will again be generation, the frequencies of the three genotypes will again be pp22:2:2pqpq::qq22, and so forth, forever., and so forth, forever.

The Hardy-Weinberg equilibrium frequencies that result from random mating. The frequencies of A and a among both eggs and sperm are p and q ( = 1-p), respectively. The total frequencies of the zygote genotypes are p2 for A/A, 2pq for A/a, and q2 for a/a. The frequency of the allele A in the zygotes is the frequency of A/A plus half the frequency of A/a, or p2 + pq = p(p + q) = p.


How selection works Going back to our example of a recessive letal gene with frequency of 0.1, we have Going back to our example of a recessive letal gene with frequency of 0.1, we have

the genotype frequencies among newborns ofthe genotype frequencies among newborns of

At reproductive age, however, the homozygotes At reproductive age, however, the homozygotes aa//aa will have already died, leaving will have already died, leaving the genotypes at this stage asthe genotypes at this stage as

But these proportions add up to only 0.99 because only 99 percent of the population But these proportions add up to only 0.99 because only 99 percent of the population is still surviving. Among the actual surviving reproducing population, the is still surviving. Among the actual surviving reproducing population, the proportions must be recalculated by dividing by 0.99 so that the total proportions proportions must be recalculated by dividing by 0.99 so that the total proportions add up to 1.00. After this readjustment, we haveadd up to 1.00. After this readjustment, we have

The frequency of the lethal The frequency of the lethal aa allele among the gametes produced by these survivors allele among the gametes produced by these survivors is thenis then

and the change in allelic frequency in one generation, expressed as the new value and the change in allelic frequency in one generation, expressed as the new value minus the old one, has been 0.091minus the old one, has been 0.091- - 0.100 = 0.019. We can repeat this calculation in 0.100 = 0.019. We can repeat this calculation in each successive generation to obtain the predicted frequencies of the lethal and each successive generation to obtain the predicted frequencies of the lethal and normal alleles in a succession of future generationsnormal alleles in a succession of future generations


Fitness and selection We may take the phenotype with highest absolute fitness, whose We may take the phenotype with highest absolute fitness, whose

overall expectation of progeny that reach reproductive age is overall expectation of progeny that reach reproductive age is maximum),maximum),WWmaxmax, as the standard phenotype, with which the fitnesses , as the standard phenotype, with which the fitnesses

WWjj of other phenotypes can be compared.of other phenotypes can be compared.

Thus, the Thus, the relative fitness relative fitness wwjj of phenotypeof phenotype j j is is wwjj = = WWjj/W/Wmaxmax and it can and it can

assume any value included between 0 and 1.assume any value included between 0 and 1. We define the We define the selection coefficient selection coefficient or selective disadvantage of or selective disadvantage of

phenotypephenotype j j to beto be ssjj = 1 – w = 1 – wjj. Sometimes it is easier to reason in terms . Sometimes it is easier to reason in terms

of fitness (of fitness (ww), sometimes it is easier in terms of selection coefficient ), sometimes it is easier in terms of selection coefficient ((ss).). A letal phenotype has fitness 0 and the selection coefficient against it is 1.A letal phenotype has fitness 0 and the selection coefficient against it is 1.


Change of allele frequency in one generation• We can now consider the example of a deleterious recessive gene in general terms. We can now consider the example of a deleterious recessive gene in general terms.

Multiplying the initial frequency by the fitness of each genotype we obtain the Multiplying the initial frequency by the fitness of each genotype we obtain the contribution of each genotype to the gametes that will form the next generation:contribution of each genotype to the gametes that will form the next generation:

Genotypes AA Aa aa TotalGenotypes AA Aa aa Total Frequency before selection pFrequency before selection p22 2pq q 2pq q22 1,00 1,00 Fitness 1 1 1-s Fitness 1 1 1-s Frequency after selection pFrequency after selection p22 2pq q 2pq q22 (1-s) 1-sq (1-s) 1-sq22

• To find the frequency of a genes in the progeny, we can apply the calculation To find the frequency of a genes in the progeny, we can apply the calculation already seen, from which we arrive at the following general formula for the change already seen, from which we arrive at the following general formula for the change of the allele frequency in one generation :of the allele frequency in one generation :

sqsq22(1-q)(1-q)q = - q = - 1 – sq1 – sq22 • We will use this equation to obtain the number of generations of selection needed to We will use this equation to obtain the number of generations of selection needed to

reach a specific change of gene frequency. An aswer to this question is sometimes reach a specific change of gene frequency. An aswer to this question is sometimes required in connexion with breeding programmes or proposed eugenic measures. required in connexion with breeding programmes or proposed eugenic measures. We will consider, for simplicity, the special case of s = 1 (complete selection We will consider, for simplicity, the special case of s = 1 (complete selection against aa).against aa).


Number of generations required By working with the recursive equation for By working with the recursive equation for q, we may easily arrive at the result q, we may easily arrive at the result

qqtt = q = q00/(1+tq/(1+tq00), from which we obtain t = 1/q), from which we obtain t = 1/qtt – 1/q – 1/q00..

Consider albinism as an example, and ask the question: how long it take to reduce Consider albinism as an example, and ask the question: how long it take to reduce its frequency to half its present value if albinos are prevented from reproduction ? its frequency to half its present value if albinos are prevented from reproduction ? The present frequency is about 1/20,000, and this makes qThe present frequency is about 1/20,000, and this makes q00 = 1/141 (q = 1/141 (q22 = 1/20,000, = 1/20,000,

so q = so q = qq22, from the Hardy-Weinberg law). The objective is q, from the Hardy-Weinberg law). The objective is q22 = 1/40,000, which = 1/40,000, which makes qmakes qtt = 1/200. So, from the above equation t = 200 – 141 = 59 generations. With = 1/200. So, from the above equation t = 200 – 141 = 59 generations. With

25 years to a generation it would take nearly 1,500 years to achieve this modest 25 years to a generation it would take nearly 1,500 years to achieve this modest result.result.

A general conclusion from the above exampleA general conclusion from the above example is that it is extremely difficult to is that it is extremely difficult to significantly reduce the frequency of an allele that is already rare in a population. significantly reduce the frequency of an allele that is already rare in a population. Thus, eugenic programs designed to eliminate deleterious recessive genes from Thus, eugenic programs designed to eliminate deleterious recessive genes from human populations by preventing the reproduction of affected persons do not work. human populations by preventing the reproduction of affected persons do not work. Of course, if all heterozygotes could be prevented from reproducing, the gene could Of course, if all heterozygotes could be prevented from reproducing, the gene could be eliminated (except for new mutations) in a single generation. Because every be eliminated (except for new mutations) in a single generation. Because every human being is heterozygous for a number of different deleterious genes, however, human being is heterozygous for a number of different deleterious genes, however, no one would be allowed to reproduce.no one would be allowed to reproduce.


Selection in complex situations A case that illustrates A case that illustrates a more complexa more complex relation of fitness to relation of fitness to

environment is sickle-cell anemia. An allelic substitution at the environment is sickle-cell anemia. An allelic substitution at the structural-gene locus for the structural-gene locus for the chain of hemoglobin results in chain of hemoglobin results in substitution of valine for the normal glutamic acid at chain position 6. substitution of valine for the normal glutamic acid at chain position 6. The abnormal hemoglobin crystallizes at low oxygen pressure, and The abnormal hemoglobin crystallizes at low oxygen pressure, and the red cells deform and hemolyze.the red cells deform and hemolyze.

Homozygotes Homozygotes HbHbSS//HbHbSS have a severe anemia, and survivorship is low. have a severe anemia, and survivorship is low. Heterozygotes have a mild anemia and under ordinary circumstances Heterozygotes have a mild anemia and under ordinary circumstances

exhibit the same or only slightly lower fitness than normal exhibit the same or only slightly lower fitness than normal homozygotes homozygotes HbHbAA//HbHbAA..

However, in regions of Africa with a high incidence of malaria However, in regions of Africa with a high incidence of malaria ((Plasmodium Plasmodium falciparumfalciparum)), heterozygotes (, heterozygotes (HbHbAA//HbHbSS) have a ) have a higherhigher fitness than normal homozygotes because the presence of some fitness than normal homozygotes because the presence of some sickling hemoglobin apparently protects them from the malaria. sickling hemoglobin apparently protects them from the malaria. Where malaria is absent, the fitness advantage of heterozygosity is Where malaria is absent, the fitness advantage of heterozygosity is lostlost


Testing Hardy-Weiberg equilibrium for HbA/S

In the course of a survey conducted in an African hospital, 1000 newborn children In the course of a survey conducted in an African hospital, 1000 newborn children were typed for hemoglobin A and S, and the following genotype counts were were typed for hemoglobin A and S, and the following genotype counts were obtained: obtained: 778 778 AA, AA, 205 205 AS, and AS, and 1717 SS. Are these frequencies in H-W equilibrium? SS. Are these frequencies in H-W equilibrium?

Genotypes Observed Expected Chisq Genotypes Frequency

AA 761 763.0 0.005 AA 0.761AS 225 221.0 0.073 AB 0.225SS 14 16.0 0.251 BB 0.014Total 1000 1000.0 0.328 Total 1

P = 0.56664Allele count Allele Frequency

A 1747 A 0.874B 253 B 0.127Tot 2000 Tot 1.000

The answer is: “Yes, they are; there is no evidence of significant deviation of the The answer is: “Yes, they are; there is no evidence of significant deviation of the observed frequecies from those expected under the H-W law”observed frequecies from those expected under the H-W law”


HWE in an adult population

Deviation from HWE is now striking. What is going on? Genotype frequencies are Deviation from HWE is now striking. What is going on? Genotype frequencies are in HW proportion at birth, whereas a large excess of heterozygotes is observed in HW proportion at birth, whereas a large excess of heterozygotes is observed among adults.among adults.

We may hypothesize that homozygote deficiency is due to selection. Increased We may hypothesize that homozygote deficiency is due to selection. Increased mortality of SS homozygotes from severe anemia is obvious; less obvious is an mortality of SS homozygotes from severe anemia is obvious; less obvious is an increased mortality of AA homozygotes.increased mortality of AA homozygotes.

It has been proven both by direct observation and by experiments that mortality It has been proven both by direct observation and by experiments that mortality from malaria is higher among AA than among AS.from malaria is higher among AA than among AS.

Genotypes Observed Expected Chisq Genotypes Frequency

AA 9365 9523.9 2.7 AA 0.756AS 2993 2675.3 37.7 AB 0.242SS 29 187.9 134.3 BB 0.002Total 12387 12387.0 174.7 Total 1

P = 0.00000Allele count Allele Frequency

A 21723 A 0.877B 3051 B 0.123Tot 24774 Tot 1.000

In another survey (Edington, 1957) a large sample of the Yoruba adult population In another survey (Edington, 1957) a large sample of the Yoruba adult population from Ibadan (Nigeria) was typed for hemoglobin A and S, with the following from Ibadan (Nigeria) was typed for hemoglobin A and S, with the following results:results:


Hetrozygote advantage We now proceed to estimate the selection coefficients against the two homozygotesWe now proceed to estimate the selection coefficients against the two homozygotes

Genotypes Observed Expected Obs/Exp Relative fitness Selection coefficients

AA 9365 9523.9 0.983 wAA = 1-t = 0.983/1.119 = 0.879 t = 0.121

AS 2993 2675.3 1.119 wmax = 1 = 1.119/1.119 = 1.000

SS 29 187.9 0.154 wSS = 1-s = 0.154/1.119 = 0.138 s = 0.862Total 12387 12387.0

These estimates are based on higly simplified assumptions; nevertheless, they are These estimates are based on higly simplified assumptions; nevertheless, they are remarkably close to the values required to explain this polymorphism.remarkably close to the values required to explain this polymorphism.

In fact, if the fitness of the heterozygote is higher than that of both homozygotes, In fact, if the fitness of the heterozygote is higher than that of both homozygotes, the polymorphism is the polymorphism is stablestable. Selection cause the relative frequencies of alleles to . Selection cause the relative frequencies of alleles to rise or fall until they reach equilibrium with both alleles represented at high rise or fall until they reach equilibrium with both alleles represented at high frequency in the population.frequency in the population.

A polymorphism such as HbA polymorphism such as HbAA//HbHbSS, that is maintained by selection in favor of , that is maintained by selection in favor of heterozygoteheterozygotess, is called a , is called a balanced polymorphismbalanced polymorphism. .


Equilibrium allele frequencies for balanced polymorphisms• We can now work out the equilibrium allele frequency of a balanced We can now work out the equilibrium allele frequency of a balanced

polymorphism:polymorphism: Genotypes AA AS SS TotalGenotypes AA AS SS Total Frequency before selection pFrequency before selection p22 2pq q 2pq q22 1,00 1,00 Fitness 1-t 1 1-s Fitness 1-t 1 1-s Frequency after selection pFrequency after selection p22(1-t) 2pq q(1-t) 2pq q22(1-s) 1 - tp(1-s) 1 - tp22 - sq - sq22

pq(tq-sp)pq(tq-sp)q = q = 1 – tp1 – tp22 -sq -sq22

qq22 (1-s) + pq (1-s) + pqq’ = q’ = 1-tp1-tp22 - sq - sq22

• After selection, the new frequency of allele HbAfter selection, the new frequency of allele HbSS is therefore: is therefore:

• Equilibrium occurs at Equilibrium occurs at q = 0, or (tp-sq)=0, from which we find:q = 0, or (tp-sq)=0, from which we find:

, from which, from which

ssppeqeq==

(t + s)(t + s) andand

ttqqeqeq==

(t + s)(t + s)

• The fact that the values estimated for theseselections coefficients are roughly The fact that the values estimated for theseselections coefficients are roughly similar to those measured for the real genotypes in areas where malaria is endemic similar to those measured for the real genotypes in areas where malaria is endemic means that the observed allele frequencies may be near their equilibrium values.means that the observed allele frequencies may be near their equilibrium values.

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Genetica per Scienze Naturali a.a. 03-04 prof S. Presciuttini Estimating nucleotide mutation rate in human The approach by Nachman and Crowell to measuring