User:Rounak Ray/sandbox

 Theorem 1 : If 𝚡 = number of hybrid pairs of contrasting traits, number of one genotypic combination = 2𝚡

Proof:             In a genotypic combination, let, there are x hybrid pairs of contrasting traits, like-

Aa, Bb, Cc, Dd, Ee, ………………( n pairs)

The  combination  ‘Aa’ can come in the cheakerboard in two ways:

(i)’A’ as male gamate,  ‘a’ as female gamate

(ii) ’A’ as female gamate,  ‘a’ as male gamate

For each of these 2 cases there are other 2 cases of any other pair.

So, the required number of  genotypic combination will be = 2 x 2 x 2 x…….( 𝚡 times) = 2𝚡

 Theorem 2 : The phenotypic ratio of a polyhybrid cross having n pairs of contrasting traits will be [3n-r] for nCr times.

Proof:           Let there are n pairs in a gene combination.Like,

Aa, Bb, Cc, Dd,………..………( n number of pairs)

i)                    During determining phenotypic ratio, at first we take all dominant characters.

So, dominant character can occur in two ways:

i)AA, BB, CC, DD,……………..( 𝚗 pairs)

ii)Aa, Bb, Cc, Dd, ,…………….( 𝚗 pairs)

All the (i)-cases can occur only in one way. (i.e. ’A’ as male gamate,  ‘A’ as female gamate)

But all the (ii)-cases can occur in two ways. (i.e. ’A’ as male gamate, ‘a’ as female gamate

& ’A’ as female gamate, ‘a’ as male gamete)

So, for each pair of contrasting traits, we get, (2+1) or 3 ways.

So, number of dominant (all characters) combination = 3n

ii) In second step, we take all dominant but one recessive character. And that is:

in 1st  case - 1st character

in 2nd case – 2nd character  and so on….

This phenotypic combination occurs for nC1 times

And likewise, on third step we take all dominant but 2 recessive.

So this phenotypic combination occurs for nC2 times

And in each case number of dominant pairs of trait decreases by 1.

So, in the required phenotypic ratio we have to put the number

3(n-r) for nCr times with ‘:’ sign in between, where 𝚛 = 0, 1, 2, 3....….. 𝚗.

We have to continue till 𝚗 = 𝚛.

Application of theorem 1: Let in an experiment, where seven pairs of contrasting traits are considered, genotype combination AaBBCCDdEeFFGg comes. How many such combinations will be found in the checkerboard?

Sol. Number of hybrid pairs of contrasting traits in the given combination is: 4.

Therefore according to above theorem, number of given genotype combination in the checker board equals to 24 = 16.

Application of theorem 2: What will be the phenotypic ratio when 5 pairs of contrasting traits are considered in a hybridization?

Sol. Let the number of contrasting traits be n.

Therefore n = 5  So in the ratio 3n will occur for nC0 times that is for once. So in the ratio 3n-1 will occur for nC1 times that is for 5 times and so on.

Therefore the ratio will be: 35:34:34:34:34:34:33:33:33:33:33:33:33:33:33:33:32:32:32:32:32:32:32:32:32:32:31:31:31:31:31:30

=729:81:81:81:81:81:27:27:27:27:27:27:27:27:27:27:9:9:9:9:9:9:9:9:9:9:3:3:3:3:3:1