Biology 442 - Human Genetics

 Quantitative Genetics

Probability, Hardy-Weinberg, risk assessment Bayes Theorem

Probability Theory

Definition of the probability of an event A (event of interest)

= # event of interest = p(A)
    # of all possible events = total # of possible events

"p" ranges from 0 to 1 and is usually expressed as a fraction.
An example is: probability of having a girl

= 1 (girl)
   2 (# of sexes)

=    1/2

Law of addition

The probability of event A OR event B occurring, if the events are mutually exclusive, is:
p(A) + p(B)
for example: probability of having a boy OR a girl is ½ + ½ = 1

Law of multiplication

The probability of event A AND event B occurring, if the event s are not mutually exclusive, is: p(A) X p(B)
for example: the probability of first having of boy AND THEN having a girl

= ½ X ½ = 1/4

Binomial Distribution

However, if the question is the probability of have a girl and a boy in a two child family then there are two ways that can happen if one does not specify the order: one could have girl first AND boy second OR boy first AND girl second, the probability is as follows:

½ (girl first) X ½ (boy second) + ½ (boy first) X ½ (girl second) = 1/4 + 1/4 = 1/2

When there are two alternative events with probability p and q (where p + q) = 1, the frequencies of the possible combination in a series of n trials are given by the expansion of (p + q)ⁿ.
The general term can be written as:
n!          p^m q^n-m
m! (n - m)!

where n = the total number in the series
m = number of times the specified event (p) occurs
n - m = number of times the alternative event (q) occurs
p = the probability of a specified event
q = the probability of the alternative event

For example a questions could be: In a sib ship of three, what is the probability of two girls and one boy?
n = # in sib ship = 3 m = # "affected" = # of girls = 2
p = probability of being affected = probability of being a girl = ½
q = probability of not being affected = probability of being a boy = ½
therefore:

3!/2! 1! X (½ )² X (½ )¹ = 3/8

You can use the same equation to determine the probability in a sib ship of three, of having all girls, all boys, two boys and a girls, or two girls and a boy.

Hardy Weinberg Equilibrium

The H-W equilibrium (a binomial distribution) used in population genetics describes the distribution of genotypes in a population at steady-state (random matings with regard to the genes, no selection, etc.). It says that the proportions of genotypes in a population will not change from generation to generation if these conditions hold. It allows you to determine the various genotype frequencies using the gene frequencies found in the gene pool of specific populations. The gene frequencies of the various (non selective) polymorphic markers such as the VNTRs and STRs alleles used in forensic genetics and paternity testing are determined by examining members of specific ethnic groups.

(p + q)² = p² (AA) + 2pq (AA') + q ²(A'A')

The frequency of AA homozygotes = p²
The frequency of AA' heterozygotes = 2pq
The frequency of A'A' homozygotes = q²

In clinical genetics this equation is useful in calculating heterozygote (carrier) frequencies for specific disorders. One need only know the incidence of the disorder in the population (q²) = incidence of those affected with the autosomal recessive (AR) disorder to calculate the carrier frequency. If q² = incidence of A'A' individuals in the population, the square root of q² = q = gene frequency of A'; and the gene frequency of the "normal" allele, A = 1 - q = p. Because q is usually a very small number, p = 1. The heterozygote frequency is 2pq .With X linked genes the incidence of affected males is equal to q, however, the incidence of affected females is q².

Using Tay-Sachs Disease as an example of an AR trait:
The incidence among Ashkenazi Jews is 1/3900 which equals q².
Therefore, q = 1/62 = 0.016 and the carrier frequency (2pq) = 1/31 or 0.032.
In non Ashkenazi Jews the incidence is less, 1/112000 = q².
Therefore, q = 1/335 or 0/003 and the carrier frequency (2pq) = 1/168 or 0.006

Using colorblindness as an example of an X linked trait:
The incidence among males of colorblindness if 1/12 = 8%
Therefore q = 0.08 and p = .92
The incidence of colorblindness among females would be expected to be q² = 0.0064 and the incidence of carrier females would be 2pq = 0.15.

Bayesian Calculations

Bayes Theorem allows one to calculate risks incorporating additional information which is useful when the person's genotype cannot be determined. To determine the probability of a person being a carrier when additional information is available one does the following steps:

Person a carrier or Person is NOT a carrier

1. Prior probability P(A) 1 - P(A)

2. Conditional Probability (based on new information) P(B) P(B')

3. Joint Probability  P(A)•P(B)  [1-P(A)]•P(B)

4. Posterior Probability P(A)•P(B)

[P(A)•P(B)} + {[1-P(A)]•P(B)]} [1 - P(A)] • P(B)

[P(A)•P(B) + {[1-P(A)]•P(B)]}

We can use carrier testing for cystic fibrosis (CF) as an example of how Bayes theorem is used. If there is NO family history and the individual is of northern european caucasian ancestry, after testing negative for the thirteen most common mutations which cause CF, we can calculate the probability that the person is a carrier despite the negative results.

Person is a carrier or Person is NOT a carrier

Prior probability 0.04 (heterozygote frequency, 2pq) in that population) 0.96

Conditional probability (based on testing for 13 mutations) 0.15 (assuming that 15% are not detected) 0.85

Joint probability 0.04 X 0.15 = 0.006 0.96 X 0.85 = 0.816

Posterior probability 0.006/(0.006 + 0.816) =

0.006/0.822 = 0.007 0.816/(0.006 + 0.816) =

0.816/0.822 = 0.993