In a species of grass, height is determined by 5 genes. Each gene has two allele
ID: 177031 • Letter: I
Question
In a species of grass, height is determined by 5 genes. Each gene has two alleles, one allele contributes 2cm to plant height and the other is non-contributing (adds 0 cm to height). Plants with the genotype aabbccddee are 12cm tall (all alleles noncontributing), while plants with the genotype AABBCCDDEE are 32cm tall (all alleles contributing).
a) If you cross aabbccddee x AABBCCDDEE, what will be the height of the progeny?
b) If you cross the F1 from above to itself, what proportion of the F2 progeny will be 20cm tall?
c) What proportion of the F2 progeny will be 12cm?
Explanation / Answer
a) crossing aabbccddee x AABBCCDDEE will yield all heterozygous individuals, i.e. AaBbCcDdEe. The five contributing alleles will add 2 cm each to the height. Therefore, height will be 12 + (2×5) = 22 cms
b) To be 20 cm in height, the grass needs to be heterozygous for any four genes [(20-12)/2 = 4] and homozygous recessive for the fifth gene.
The probability that an offspring will be heterozygous for any one gene is 1/2. Similarly, the probability that an offspring will be homozygous recessive for any one gene is 1/4.
Therefore, the probability of an offspring being heterozygous for four genes and homozygous recessive for the fifth gene = (1/2)×(1/2)×(1/2)×(1/2)×(1/4) = 1/64
The combined probability of an offspring being heterozygous for any four genes and homozygous recessive for the fifth gene = 5×1/64 = 5/64
c) To be 12 cm in height, the plant needs to be homozygous recessive for all genes, i.e. aabbccddee.
Again, the probability that an offspring will be homozygous recessive for any one gene is 1/4. Therefore, the probability that it will be homozygous recessive for all genes = (1/4)×(1/4)×(1/4)×(1/4)×(1/4) = 1/1024