Question

Answer the next five questions based on the following information. You are studying a particular type of spider and have identified four traits that follow simple mendelian genetics. The B gene governs body color and black body color is dominant to brown. The D gene governs whether or not the spider is poisonous, and poisonous is dominant to non-poisonous. The E gene governs eye color and black eyes are dominant to red eyes. The H gene governs hair type and hairy is dominant to hairless. Indicate your answer numerically not alphabetically. For instance, answer "2" not "two." For each answer that is not a whole number, give your answer as a fraction reduced to the lowest possible whole numbers. For instance, 3/9 would be reduced to 1/3.

How many different types of gametes can be formed by a spider that has the genotype BbDdEeHH?

If you chose a single gamete at random from a spider that has the genotype BbDdEeHh, what is the probability that the gamete would have the genotype BdEh?

You mate a BbDDEeHh spider with a BbDdEEHh spider. What is the probability that one randomly chosen offspring will have the genotype BbDdEEHH?

Answer 1

1)The number of possible gametes from a given genotype can be found by using the fromula 2ⁿ, where 'n' is the number of heterozygous locus. In the above question, the genotype is BbDdEeHH. Here, there are 3 heterozygous locus (BbDdEe).

Hence number of possible gametes = 2ⁿ = 2³ = 2*2*2 = 8.

2) The number of gametes possible from the genotype BbDdEeHh are 2⁴ (Since there are 4 hetrozygous locus).ie., 16. So, the probability of getting a gamete BbEh from BbDdEeHh is 1/16.

3) The genotype of parents is BbDDEeHh and BbDdEEHH.

BbDDEeHh X BbDdEEHH

BbDdEEHH

The probability is = (1/2*1/2)+(1/2*1/2) *1*1/2*1/2*1*1/2*1/2

   = (1/4+1/4)*1*1/2*1/2*1*1/2*1/2

= 1/32

{ the first locus of offspring is heterozygous and that of both parent is heterozygous. So, (1/2*1/2) + (1/2*1/2). Because it can be either Bb OR bB}.

AND Rule- Multiplication

OR Rule - Addition