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.
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.
15.
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
Wht is an autosomal resesive trait????
There can be 3 types of genotypes, homozygous dominant or heterozygous dominant. And heterozygous recessive (diseased)
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.
