application of population genetics
- 1. Applications Of Population Genetics ATHIRA P J 1 MSc BIOCHEMISTRY
- 2. • Population calculations are applied to real-life situations by determining the value of q2 (homozygous recessives) from the incidence of an inherited disease in a population, deducing p and q, and then using the Hardy-Weinberg equation to predict the likelihood that a person is a carrier. • The Hardy-Weinberg equation is applied to population statistics on genetic disease incidence to derive carrier risks. • To determine allele frequencies for autosomal recessively inherited characteristics, we need to know the frequency of one genotype in the population.
- 3. CYCTIC FIBROSIS • Affects 1 in 2,000 people. • How do we know that what % of people are homozygous dominant or heterozygous dominant. • If we assume the population is in hardy – weinberg equillibrium, ie, p2+2pq+q2=1,
- 4. The homozygous recessive frequency (cc if c represents the disease-causing allele) is 1/2,000, or 0.0005 in this population. q2 =0.0005 q=√f(cc)=√0.0005=0.02 We know that p+q=1, therefore, p=1-q =1-0.02 =0.98 F(CC)=p2 =(0.98)2 = 0.960 ie, 1920 in 2000 F(Cc)=2pq =2(0.98)(0.02) =0.0392 ie,79 in 2000 If there is no cystic fibrosis in a family, a person’s risk of having an affected child, derived from population statistics, is relatively low.
- 5. For X-linked traits, different predictions of allele frequencies apply for males and females. • For a female, who can be homozygous recessive, homozygous dominant, or a heterozygote, the standard Hardy- Weinberg equation of p2 + 2pq + q2 applies, as it usually would to an autosomal recessive trait. • In males, the allele frequency is the phenotypic frequency, because a male who inherits an X-linked recessive allele exhibits it in his phenotype.
- 6. HEMOPHILIA • The incidence of X-linked hemophilia (XhY), for example, is 1 in 10,000 male births. • Therefore, q (the frequency of the“h” allele) equals 0.0001 • We know that p+q=1, therefore, p=1-q =1- 0.0001 =0.9999.
- 7. The incidence of carriers (XhXH), who are all female= 2pq =(2)(0.0001)(0.9999) =0.00019; Rounding…, = 0.0002 = 0.02 %, Ie, about 1 in 5,000. The incidence of a female having hemophilia (XhXh) = q2 =(0.0001)2 = 0.00000001 ie about 1 in 100 million.
- 8. •Mendelian disorders are usually very, very rare. • Therefore, the q component of the Hardy- Weinberg equation does not contribute much. Because this means that the value of p approaches 1, the carrier frequency, 2pq, is very close to just 2q. •Therefore, the carrier frequency is approximately twice the frequency of the rare, disease-causing allele.
- 9. TAY-SACHS DISEASE • Occurs in 1 in 3,600 people. q2 = 1/3,600 = 0.0003. q=√ 0.0003= 0.017. • The frequency of the dominant allele, p= 1 – 0.017=0.983. 2pq= (2)(0.983)(0.017)=0.033. • This is very close to double the frequency of the mutant allele, 0.017. While these examples all trace a gene with just two alleles, the Hardy-Weinberg equation can also be modified to analyze genes that have more than two alleles.
- 10. DNA Fingerprinting—A Practical Test of Hardy-Weinberg Assumptions
- 11. •The Hardy-Weinberg equation and the product rule are used to derive the statistics that back up a DNA fingerprint. •First, the DNA fingerprint pattern shows whether an individual is a homozygote or a heterozygote for each repeat, because a homozygote only has one band representing that gene. •Genotype frequencies are then calculated using parts of the Hardy-Weinberg equation.
- 12. •Logic then enters the equation. If an allele combination is exceedingly rare in the population the suspect comes from, and if it is found both in the suspect’s DNA and in crime scene evidence, the suspect’s guilt appears to be highly likely. •Applying the product rule, the genotype frequencies are multiplied. The result is the probability that this particular combination of DNA sequences would occur in the population.
- 13. Molecular clock • In paleonotological studies ,estimation of divergence of species from common ancestry is studied using the applicaton of population genetics. • It is done by comparing homologous proteins in different species. • The difference in aminoacid sequence between two species with an unknown common ancestor may serve as a molecuar clock,indicating the time since the two species diverged.
- 14. For example, This difference would state that a mutaion occurred in one of these species at these positions and has Subsequently become fixed by genetic drift. If the aminoacid sequences are closely related ,we can say that they have common ancestor.
- 16. References • Lewis-human genetics-concepts and application- edition 5;page no:267-274 • Population Genetics ;A Concise Guide ; John H Gillespie • Genetics ;Robert F Weaver,philip W Hedrick
Editor's Notes
- #3: Wht is an autosomal resesive trait????
- #4: There can be 3 types of genotypes, homozygous dominant or heterozygous dominant. And heterozygous recessive (diseased)
- #5: Consider a Caucasian couple with no family history of cystic fibrosis asking a genetic counselor to calculate the risk that they could conceive a child with this illness. The genetic counselor tells them that the chance of each potential parent being a carrier is about 4.3 percent, or 1 in 23. But this is only part of the picture. The chance that both of these people are carriers is 1/23 multiplied by 1/23—or 1 in 529—because the probability that two independent events will occur equals the product of the probability that each event will happen. However, if they are both carriers, each of their children would face a 1 in 4 chance of inheriting the illness, based on Mendel’s first law, of gene segregation. Therefore, the risk that two unrelated Caucasian individuals with no family history of cystic fibrosis will have an affected child is 1/4 × 1/23 × 1/23, or 1 in 2,116. This couple has learned their chance of producing an affected child from disease incidence statistics.