Population Genetics Dr Mohammed Hussein M B Ch

Population Genetics Dr. Mohammed Hussein M. B. Ch. B, MSC, Ph. D, DCH (UK), MRCPCH

Definition • Population genetics is the study of genetic variation in populations. • Allow us to understand how and why the prevalence of various genetic diseases differs among populations.

: British Mathematician : German Physician

Hardy-Weinberg Equilibrium

• If a population is large and if individuals mate at random with respect to their genotypes at a locus, the population should be in H-W equilibrium.

• This means that there is a constant and predictable relationship between genotype frequencies and allele frequencies. • The Hardy Weinberg equation, allows one to estimate genotype frequencies if one knows allele frequencies, and vice versa.

2 P + 2 pq + p = a normal allele q = a disease producing allele 2 q =1

2 P + 2 pq + 2 q =1 • p = a normal allele • q = a disease-producing allele • pp = the genotype pp (homozygous normal) = P 2 • qq = the genotype qq (homozygous affected) = q 2 • pq = heterozygous ( affected or carrier) = 2 pq

Heterozygous male pq Heterozygous female pq

p p q p p p q Homozygous for the normal allele Heterozygous q 1/4 p q q q pp pq 1/4 1/4 pp + pq + qq = 1 p 2 + 2 pq + q 2 pq qq = 1 Heterozygous Homozygous for the diseased allele

2 P + 2 pq + 2 q =1 p = frequency of the normal allele q = frequency of the disease-producing allele 2 p = frequency of the homozygous normal 2 q= frequency of the homozygous affected 2 pq = frequency of heterozygous

Simplification • Generally p, the normal allele frequency in the population, is very close to 1 (e. g. , most of the alleles of this gene are normal). • In this case, we may assume that p ~ 1, and the equation simplifies to: 2 P + 2 pq + 1 + 2 q + 2 q =1 =1

1 + 2 q + 2 q =1 q= frequency of the diseased-producing allele 2 q = frequency of heterozygous for diseased allele 2 q = frequency of homozygous for diseased allele

• In Autosomal Dominant diseases • The diseased allele (q) is dominant over the normal allele (p), so pp = normal qq = homozygous affected pq = heterozygous affected

2 P 2 q + 2 pq + = 1 2 1 + 2 q + q = 1 • In Autosomal Dominant diseases q = frequency of the diseased-producing allele 2 q = heterozygous affected = prevalence of the disease 2 q = homozygous affected (sever disease)

• In Autosomal Recessive diseases • The normal allele (p) is dominant over the diseased allele (q), so pp is normal pq is heterozygous carrier qq is homozygous affected

2 P 2 q + 2 pq + = 1 2 1 + 2 q + q = 1 • In Autosomal Recessive diseases q = frequency of the diseased-producing allele 2 q = heterozygous carrier 2 q = homozygous affected (prevalence of the disease)

• In X-linked Recessive diseases • The normal allele (p) is dominant over the diseased allele (q) • As the diseased allele presented only on the X chromosome p X Y = normal male q X Y = affected male p p X X = normal female p q X X = female carrier q q X X = affected female

2 P 2 q + 2 pq + = 1 2 1 + 2 q + q = 1 q = frequency of the diseased-producing allele q = frequency of the hemizygous affected males (Xq Y ) (prevalence of the disease) 2 q = heterozygous female carrier (Xp Xq ) 2 q = frequency of homozygous affected females (Xq Xq )

q 2 q q 2 Diseasedproducing allele Heterozygous Homozygous X-L R Disease prevalence AD Disease prevalence AR Carrier X-L R Female carrier AD Sever affected AR Disease prevalence X-L R Affected females

q AD diseases: Øq = frequency of the disease-producing allele Ø 2 q = heterozygous affected = prevalence of the disease Øq 2 = homozygous affected (sever disease) q. AR diseases: Øq = frequency of the disease-producing allele Ø 2 q = heterozygous carriers Øq 2 = homozygous affected = prevalence of the disease q. X-L - R diseases: Øq = frequency of the disease-producing allele Øq = prevalence of the disease Ø 2 q = heterozygous female carriers Øq 2 = homozygous affected females (sever disease)

A Practical Application of the H-W Principle • If the prevalence of homocystinuria (AR genetic disease) is 1/10, 000, find out 1. The frequency of the disease-producing allele 2. The carrier frequency

Answer • q 2 =Disease prevalence q = 1/100 • q 2 = 1/10, 000 Frequency of the disease-producing allele = 1/100 • 2 q = Carrier frequency • 2 x 1/100 = 2/100 = 1/50 The carrier frequency = 1/50 Disease prevalence 1/10, 000 Disease-producing allele ? The carrier frequency ? AR Disease q = disease-producing allele 2 q = carrier frequency q 2 = disease prevalence

Example 2 • The prevalence of hyperlipidemia (AD disease) is 1/500, calculate 1. 2. The frequency of the disease-producing allele The frequency of the homozygous affected

Answer • 2 q =Prevalence 2 q= 1/500 q= 1/1000 Frequency of the disease-producing allele = 1/1000 • q 2 = homozygous affected • q 2 = 1/1000 x 1/1000 = 1/1000000 = 1/106 Frequency of the homozygous affected = 1/106 Disease prevalence 1/500 Disease-producing allele ? The homozygous affected? AD Disease q = disease-producing allele 2 q = disease prevalence q 2 = homozygous affected

Example 3 • The prevalence of hemophilia in males is 1/10, 000, calculate 1. The frequency of the disease-producing allele 2. The prevalence (frequency) of female carriers 3. The frequency of the females affected.

Answer Disease prevalence 1/10, 000 Disease-producing allele ? The female carrier? The female affected? X-linked R Disease q = disease-producing allele (disease prevalence) 2 q = heterozygous female carrier q 2 = homozygous affected female (sever disease) • q = The prevalence of the disease • q = 1/10, 000 as q = frequency of the disease-causing allele Frequency of the disease-producing allele = 1/10, 000 • 2 q = heterozygous female carrier • 2 q = 2 x 1/10, 000 = 2/10, 000 = 1/5, 000 Frequency of the female carrier= 1/5, 000 • q 2 = affected female • q 2 = 1/10, 000 x 1/10, 000 = 1/100, 000 = 1/108 Frequency of the females affected= 1/108

How to calculate the probability for a family to have a child with an autosomal recessive disease?

• The probability is based on the joint occurrence of 4 events: 1. 2. 3. 4. The probability that the male carry the abnormal allele The probability that the he will pass the allele to the child The probability that the female carry the abnormal allele The probability that the she will pass the allele to the child 1 * 2 * 3 * 4 = the risk of having a child with the disease

The probability that the parent carry the abnormal allele If the parent is known affected by the disease or known carrier Then the probability that he/she carries the abnormal gene is 100% which equal to 1

The probability that the parent carry the abnormal allele If the parent is don’t know whether carrier or not Then we have to find the carrier risk or the carrier frequency If non of his/her family members is affected or carrier then we need to calculated the carrier frequency of the population However if one of his/her brothers is affected then the probability is 67% or 2/3

The probability that the parent will pass the abnormal allele If the parent is known affected by the disease This means that the probability to pass the gene is 100% or 1 Otherwise This means that the probability to pass the gene is 50% or 1/2

A Practical Application of the H-W Principle • A 20 -year-old female has an AR genetic disease called Phenylketonuria (PKU). She asks you 2 questions: 1. What is the chance that she had to marry a man with the disease-producing allele. 2. What is the probability that she will have a child with PKU? • Note: the prevalence of PKU in the population is 1/10, 000.

Answer Disease prevalence 1/10, 000 The carrier frequency ? • q 2 =Disease prevalence • q 2 = 1/10, 000 q = 1/100 Frequency of the disease-causing allele = 1/100 AR Disease q = disease-causing allele 2 q = carrier frequency q 2 = disease prevalence • 2 q = Carrier frequency • 2 x 1/100 = 2/100 = 1/50 The carrier frequency = 1/50 The chance that she had to marry a man with disease-producing allele is 1/50

• The answer to the second question is based on the joint occurrence of 4 events: 1. The probability that the female carries the abnormal allele is (1) as she is known affected 2. The probability that the she will passes her abnormal allele to the child is 1 as the disease is AR as she is affected 3. The probability that the male carries the abnormal allele is 1/50 as we calculate it before 4. The probability that the he will passes his abnormal allele to the child is 1/2 as the disease is AR as he is carrier 1 * 1/50*1/2 = 1/100

• A couple come for genetic counseling about the risk that their child will have cystic fibrosis. • The husband's sister has the disease, but he is not affected and there is no history in the wife's family of the disease. • Note: the prevalence of CF is 4/10000.

Answer Disease prevalence 4/10, 000 The carrier frequency ? • q 2 =Disease prevalence • q 2 = 4/10, 000 q = 2/100 Frequency of the disease-causing allele = 2/100 AR Disease q = disease-causing allele 2 q = carrier frequency q 2 = disease prevalence • 2 q = Carrier frequency • 2 x 2/100 = 4/100 = 1/25 The carrier frequency = 1/25 The carrier risk of CF in population, which is the risk for the wife = 1/25

Answer • Regarding the husband, as his sister is affected, so both parents must be carriers. • Since he is not affected therefore there is a 2/3 chance that he is a carrier. pp pq pq qq

1. 2. 3. 4. The probability that the female carry the abnormal allele The probability that the she will pass her abnormal allele to the child The probability that the male carry the abnormal allele The probability that the he will pass his abnormal allele to the child 1/25 * 1/2 * 2/3 * 1/2 * = 1/150

• Although human populations are typically in H-W equilibrium for most loci, deviations from equilibrium can be produced by: 1. Mutation 2. Natural Selection 3. Genetic Drift 4. Gene Flow 5. Consanguinity

• Mutation is ultimately the source of all new genetic variation in populations. • Founder effect

• Natural selection acts upon genetic variation, increasing the frequencies of alleles that promote survival or fertility (referred to as fitness) and decreasing the frequencies of alleles that reduce fitness.

• There is now evidence for heterozygote advantages for several other recessive diseases that are relatively common in some populations. Examples include: q Cystic fibrosis (heterozygote resistance to typhoid fever) q Hemochromatosis (heterozygote advantage in iron-poor environments) q Glucose-6 -phosphate dehydrogenase deficiency (heterozygote resistance to malaria)

• Mutation rates do not vary significantly from population to population, although they can result in significant differences in allele frequencies when they occur in small populations

• Gene flow refers to the exchange of genes among populations. • Because of gene flow, populations located close to one another often tend to have similar gene frequencies

• Consanguinity refers to the mating of individuals who are related to one another (typically, a union is considered to be consanguineous if it occurs between individuals related at the second-cousin level or closer).

• Dozens of empirical studies have examined the health consequences of consanguinity, particularly first-cousin matings. • These studies show that the offspring of first-cousin matings are approximately twice as likely to present with a genetic disease as are the offspring of unrelated matings.

Thank you for Your Attention Dr. Mohammed Hussein M. B. Ch. B, MSC, Ph. D, DCH (UK), MRCPCH