Problem of quantitative genetics

Problem: Fruit colour of wild Solanum nigrum is controlled by two alleles of a gene (A and a). The frequency of A, p=0.8 and a, q=0.2. In a neighbouring field a tetraploid genotype of S. nigrum was found. After critical examination five distinct genotypes were found; which are AAAA, AAAa, AAaa, Aaaa and aaaa. Following Hardy Weinberg principle and assuming the same allele frequency as that of diploid population, the numbers of phenotypes calculated within a population of 1000 plants are close to one of the following: AAAA : AAAa : AAaa : Aaaa : aaaa

1. 409 : 409 : 154 : 26 : 2

2. 420 : 420 : 140 : 18 : 2

3. 409 : 409 : 144 : 36 : 2

4. 409 : 420 : 144 : 25 : 2                                                      (CSIR June 2016)

Solution:

Frequency of allele A = 0.8 and a = 0.2

	AA	Aa	Aa	aa
AA	AAAA	AAAa	AAAa	AAaa
Aa	AAAa	AAaa	AAaa	Aaaa
Aa	AAAa	AAaa	AAaa	Aaaa
aa	AAaa	Aaaa	Aaaa	aaaa

From above chart phenotypic ratio is,

AAAA : AAAa : AAaa : Aaaa : aaaa = 1 : 4 : 6 : 4 : 1

From Hardy-Weinberg principle:

p⁴ : 4p³q : 6p²q² : 4pq³ :q⁴

for AAAA= p⁴ x 1000 = (0.8)⁴ x 1000

= 409.6 » 409

for AAAa = 4p³q x 1000 = 4x(0.8)³x0.2x1000

= 409.6  » 409

for AAaa = 6p²q² x 1000 = 6x(0.8)²x(0.2)² x1000

= 153.6  » 154

for Aaaa = 4pq³ x 1000 = 4x0.8x(0.2)³ x1000

= 25.6  » 26

for aaaa = q⁴ x 1000 = (0.2)⁴ x1000

= 1.6 » 2

Number of phenotypes are 

AAAA : AAAa : AAaa : Aaaa : aaaa = 409 : 409 : 154 : 26 : 2

Option 1 is correct