hardy Weinberg in human population and forensics genetics Flashcards
1
Q
HWE in human population
A
- No human population can meet all the requirements of HW law
2
Q
can we still use it to calculate genotype proportions
A
yes, - Various test have been carried out on population data comparing predicted genotype frequencies under HWE and measuring if there is significant deviation
- Significant deviations have not been detected in the vast majority of populations
3
Q
recessive inherited alleles
A
- HW law can be used to calculate the frequencies of harmful recessive alleles
- Homozygote recessive disorders such as cystic fibrosis
- AA= ‘normal’ phenotype, Aa= carrier, aa= disease phenotype
- Alleles A(p) and a (q) p+q=1
- Genotypes AA (p2), Aa(2pq) and aa(q2)
- If the aa genotype expresses the disease phenotype the proportion of affected individuals in a population would be q2
- The frequency of heterozygous carriers is 2pq
4
Q
e.g 1 in 1000 children have cystic fibrosis (1/1000)
A
- q2 = 0.001
- q = 0.03
- p =0.97
- 2pq = 0.0582
5
Q
forensic dna analysis
A
- the HWE is used along with allele frequency databases to calculate genotype frequencies
6
Q
allele frequency databases
A
- an allele frequency database is constructed by measuring the occurrence of alleles within a defined population
- it has been recommended that a sample of at least 200 alleles per locus (100 individuals) is used
7
Q
frequency of STR profiles
A
- the genotype proportions for each locus are calculated using p2 for homozygotes and 2pq for heterozygotes
- the overall profile frequency is calculated by multiplying the genotype frequency at each locus
- this is called the product rule
8
Q
the product rule
A
- multiplication of individual genotype frequencies at different loci to give the overall profile frequency
- population must be in linkage equilibrium
- this is true for the standard STRs
- all on different chromosomes and are inherited independently
9
Q
calculation of STR frequency
A
look at ppt
