Question

2. Suppose an individual has the genotype B/b:D/D:Ele:G/g. a. What are the odds that the individual will produce a gamete with the genotype be? Ignore the D and G genes for now.) b. Now considering all four genes, how many different unique kinds of gametes could this individual produce? c. What are the odds that the individual will produce a gamete with the genotype B:D:e G?

Answer 1

Given that the genotype is B/b;D/D;E/e;G/g

a) Only considering B/b;E/e genotype (ignoring D/D;E/e genes), all possile games that can be formed are So, the gametes with genotype b;e will constitute 25% of total gametes formed.

b) Different kind of gametes that will be formed is given by the formula 2ⁿ (where 'n' denotes number of heterozygous genes)

Here there are 3 heterozygous genes i.e B/b, E/e, G/g.

So types of gemetes are -( 2³=8) . These are:-

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c) So from the above answer (b), it is seen that gamete with genotype B;D;e;G is having a 1/8 chance of occurance i.e 12.5%.