03 Population Genetics

8/10/2019 03 Population Genetics

1/11

Bi2030 / Lecture #3 / Slide 1 Bi2030 / Lecture #3 / Slide 2


2/11

Bi2030 / Lecture #3 / Slide 3

Natural populations are reservoirs of great genetic variability:

For example, all humans contain the same numbers and kinds of genes, but no twoindividuals look alike (except twins).

The human population must contain multiple alleles of many, many genes to explainthis natural variability.

Hardy & Weinberg showed that the human population could be treated as a gene pool,

a reservoir of gametes that combine at random, to produce progeny whose phenotypesand proportions can be predicted if the frequencies of different alleles in the populationare known. If the allele frequencies are not known, but the population is in Hardy-Weinberg equilibrium, then the gene frequencies can be calculated from thepopulations phenotypic frequencies.


All Hardy-Weinberg arithmetic involves relative gene and phenotype FREQUENCIESexpressed as fractions (i.e, the frequencies of all the alternative alleles of a particulargene in the population must sum to 1). These fractions are also probabilities, i.e., thechances of finding a particular allele in a gamete or an individual with a particularphenotype.

A note on genotype notation:A/A

(orA/a

ora/A

ora/a

) indicates the two copies of aparticular gene in a diploid individual. The slash separating the two alleles indicatesthat they reside on homologous chromosomes it is NOT a division sign for anarithmetic operation!


3/11


Note that the fractional frequencies (or probabilities) of the three possible phenotypesmust sum to 1.



4/11


(a) Let q = the frequency of the recessive cf allele. Assuming HW equilibrium, thenq2= 0.0003 and q = (0.0003)1/2= 0.017. Then p, the frequency of the wild-typeallele, = (1 - q) = 0.983.

(b) We know that the woman is heterozygous, so she has a 50% chance of passingthe cf allele to any of her offspring. The chance that her unaffected mate is also

a heterozygous carrier (assuming no family history of CF) is [2pq / (p2+ 2pq)] =

0.033 / (1 - 0.0003) !3.3%. The chance that this couple will have an affectedchild is therefore (0.5)(0.033)(0.5) !0.8%.

(c) No! We cannot be certain that a population is at HW equilibrium unless we

empirically measure (not calculate) ALL relevant genotype frequencies.Fortunately, in the case of cystic fibrosis, there's now a genetic screen for

carriers, so families at risk can get objective estimates of the risk factors. (With ascreening test, it would also be possible, in principle, to determine whether cysticfibrosis is at HW equilibrium.)

Bi2030 / Lecture #3 / in-class slide


5/11


The MN blood group antigens are specified by CODOMINANT alleles (each makes aproduct that contributes to the organisms phenotype). In this case, the Mand Nallelesmake antigenically different proteins that are displayed on the surface of red bloodcells. Heterozygotes produce both antigens, so they display a combined phenotype.Because there is no complete dominance in this system, the organisms MN phenotypereflects his/her genotype, which allows for simple tests of whether or not a population isat Hardy-Weinberg equilibrium, as summarized in the next slide.



6/11



7/11


The ABO blood group antigens are specified by a gene with three different alleles. TheIAandIBalleles both make distinctive antigens on the red cell surface and so they arecodominant to each other. The iallele makes no antigen and so is recessive to boththe IAandIBalleles.

Note that, unlike the simpler MN system, it is not possible to calculate gene frequenciesin the ABO system directly from the ABO blood group frequencies. The data in thetable must have been generated by assuming that each population was at Hardy-Weinberg equilibrium, then using the various phenotype frequencies and the HWformula to calculate allele frequencies.

(See the next slide for details of extending the HW analysis to a three-allele system.)



8/11


If a population is NOT at HW equilibrium, it can reach equilibrium in one generation ofrandom mating (assuming that the population is large enough to avoid sampling errorsin individual matings). The matrix shows the proportion of matings involving each of thethree genotypes in the population. For example, if 0.4 of the individuals have EEgenotype and matings are at random, then (0.4)(0.4) or (0.16) of the matings should bebetween two EEindividuals. (This is directly analogous to how gene combinations arecalculated in HW.) Those matings will yield all EEoffspring, so 0.16 of the EE offspring

in the next generation will come from EE x EEmatings. To check this calculation,simply count the dark gray squares at the junction of the EE parents. Similarly, EE x Eematings will contribute to EE individuals in the next generation. Although there are twoways to get such a mating (father EE and mother Ee or father Eeand mother EE), only0.5 of the gametes from Eewill carry the Eallele, so the EE x Eematings will contribute(2)(0.5)(0.4)(0.2) = 0.08 of the EEindividuals to the next generation. Finally, Ee x Eematings can generate EEindividuals, but only (0.25) of the matings will yield EEoffspring, so this contribution is (0.25)(0.2)(0.2) = 0.01. Thus, the total EEindividualsafter one generation of random mating will be 0.16 + 0.08 + 0.01 = 0.25. Similarcalculations will yield the other phenotype frequencies. The three resulting phenotypesare shown as different gray shades in the matrix. Count the squares of each type tocheck your calculation.


Matings in populations may not be random with respect to all phenotypic traits. Forexample, humans choose mates of a similar skin color more often than chance.

Assortative mating patterns cause deviations from the idealized HW phenotypefrequencies: preferential matings between like phenotypes (positive assortative mating)tends to reduce the proportion of heterozygotes in a population and increase theproportions of homozygotes; preferential matings between unlike phenotypes (negativeassortative mating) tends to increase heterozygosity and decrease homozygosity.

Note that nonrandom mating alters both genotype and phenotype frequencies, but doesNOT change the allele frequencies in the population.


9/11


Inbreeding definitely reflects a nonrandom pattern of mating, although in this case thefactors influencing mate selection are probably based more on proximity and socialcustom than on phenotypic appearance. Again, genotype and phenotype frequenciesare changed, but allele frequencies are not (unless there is selection for or againstcertain phenotypes in the population).



10/11



11/11


Documents

03 Population Genetics