Question

1. Populations of a fish species are studied in four different adjacent lakes. A microsatellite genetic locus is studied in fifty fish from each lake, revealing the frequencies of three microsatellite alleles Al, A2, and A3: ssig frequencies of: A2 A3 Lake 1: Lake 2: Lake 3: Lake 4: 0.64 0.36 0.58 0.74 0.89 0.18 0.17 0.11 a) what is the expected H in each lake? (Note: H-, (1-homozygotes) is easiest.) Lake 1: H,= Lake 2 H,- Lake 3: H, Lake 4: H,-, di 10 1C H in suhnopulations? (i.e, what is the average H of the four lakes?)

Answer 1

Lake 1

Let A2 be p allele

A1 be q allele

According to Hardy-Weinberg equation

p² + 2pq + q² = 1

(0.64)² + 2 x 0.64 x 0.36 + (0.36)² =1

0.4096 + 0.4608 + 0.1296 = 1

Expected H in Lake 1 = 1 – homozygotes

                                    = 1 – (0.4096 + 0.1296)

                                    = 1 – 0.5392 = 0.4608

Lake 2

Let A2 be p allele

A1 be q allele

A3 be r allele

According to Hardy-Weinberg equation

p² + 2pq + q² + 2pr + 2qr + r² = 1

(0.24)² + 2 x 0.24 x 0.58 + (0.58)² + 2 x 0.24 x 0.18 + 2 x 0.58 x 0.18 + (0.18)² = 1

0.0576 + 0.2784 + 0.3364 + 0.0864 + 0.2088 + 0.0324 = 1

Expected H in Lake 2 = 1 – homozygotes

                                    = 1 – (0.0576 + 0.3364 + 0.0324)

                                    = 1 – 0.4264 = 0.5736

Lake 3

Let A2 be p allele

A1 be q allele

A3 be r allele

According to Hardy-Weinberg equation

p² + 2pq + q² + 2pr + 2qr + r² = 1

(0.17)² + 2 x 0.17 x 0.74 + (0.74)² + 2 x 0.17 x 0.09 + 2 x 0.74 x 0.09 + (0.09)² = 1

0.0289 + 0.2516 + 0.5476 + 0.0306 + 0.1332+ 0.0081 = 1

Expected H in Lake 3 = 1 – homozygotes

                                    = 1 – (0.0289 + 0.5476 + 0.0081)

                                    = 1 – 0.5846 = 0.4154

Lake 4

Let A2 be p allele

A1 be q allele

According to Hardy-Weinberg equation

p² + 2pq + q² = 1

(0.11)² + 2 x 0.11 x 0.89 + (0.89)² =1

0.0121 + 0.1958 + 0.7921 = 1

Expected H in Lake 4 = 1 – homozygotes

                                    = 1 – (0.0121 + 0.7921)

                                    = 1 – 0.8042 = 0.1958