Question

In Parakeets, Yellow (B) is dominant to Blue (b), long beaks (S) are dominant to short beaks (s), and being quiet (N) is dominant to being noisy (n). You perform a test cross of a trihybrid parakeet (BbSsNn) and a triple recessive parakeet. The following results are obtained 

Phenotype:	Genotype	of progeny
Blue, short, quiet		13
Yellow, long. noisy		13
Yellow, long, quiet		436
Blue, short, noisy		450
Yellow, short, quiet		165
Blue, long, noisy		162
Blue, long, quiet		82
Yellow, short, noisy		79
Total		1400

A. Out of the genotypes, list the two that are parental 

B. List the two genotypes that are double recombinants and identify the middle gene 

C. Calculate the recombination frequencies for each gene pair and draw the map of the three genes on the chromosome. Make sure you clearly indicate the gene pair you are referring to in each case and do not forget to include the units. 

D. What is the interference for this data set? (Round to 4 decimal places while doing calculations, but round your final answer to two decimal places or express as a %. Example: 0.77 or 77%).

Answer 1

According to the given question-

Here we have a condition, where parakeets have three characteristics, such as color, size of the beak, and nature of expression.

The yellow color is dominant over the blue color, long beaks are dominant over the short beaks, and the quiet is dominant over the noisy.

The allele for yellow color = B, the allele for blue color = b

The allele for long beaks = S, the allele for short beaks = s

The allele for quiet = N and the allele for noisy - n

When we make a cross between the parents which are true-breeding for these three characteristics then in the F1 generation we get all the offspring are heterozygous having yellow color, long beaks, and are quiet.

When we make a cross between the Offspring of the F1 generation, with the parents which are true-breeding having recessive allele for these three characteristics then in the F2 generation we get the following offspring.

Blue, short beaks, quiet (b s N)= 13

Yellow, long beaks, noisy (B S n) = 13

Yellow, long beaks, quiet (B S N) = 436

Blue, short beaks, noisy (b s n) = 450

Yellow, short beaks, quiet (B s N) = 165

Blue, long beaks, noisy (b S n) = 162

Blue, long beaks, quiet (b S N) = 82

Yellow, short beaks, noisy (B s n) = 79

First, we have to find the gene present in the middle-

Then arrange the number of offspring in descending order-

Blue, short beaks, noisy (b s n) = 450

Yellow, long beaks, quiet (B S N) = 436

Yellow, short beaks, quiet (B s N) = 165

Blue, long beaks, noisy (b S n) = 162

Blue, long beaks, quiet (b S N) = 82

Yellow, short beaks, noisy (B s n) = 79

Blue, short beaks, quiet (b s N)= 13

Yellow, long beaks, noisy (B S n) = 13

TOtal number of offspring = 1400

Here the parentals are Blue, short beaks, noisy (b s n) = 450 and Yellow, long beaks, quiet (B S N) = 436, while the double recombinant or double crossover is Blue, short beaks, quiet (b s N)= 13 and Yellow, long beaks, noisy (B S n) = 13, single cross over are Yellow, short beaks, quiet (B s N) = 165, Blue, long beaks, noisy (b S n) = 162, Blue, long beaks, quiet (b S N) = 82, and Yellow, short beaks, noisy (B s n) = 79.

parentals B S N = Yellow long quit Blue, short, noisy. Þ 8 2 B b the double crois over N 3 B n S = 13, Yellow, short, quit = 13, Blue, long, nawy 8 b 2

thus the gene present in the middle is N

N S Gene Order B

The distance between gene B and gene N = 165 + 162 + 13 +13 $\div$ 1400 = 353 $\div$ 1400 = 0.2521 or 25.21 cM

The distance between gene N and gene S = 82 + 79 + 13 +13 $\div$ 1400 = 187 $\div$ 1400 = 0.1335 or 13.35 cM

The distance between gene B and gene S = 0.2521 + 13.35 = 38.56 cM

N s B Map distance 25.21413-35

The Coefficient of coincidence = all double cross over offspring present in the population $1596088087399_blob.png$ (distance present between first two genes) (distance present between last two genes) $1596088087403_blob.png$ total number of offspring present in the population

= (13 + 13) $1596088088022_blob.png$ (0.2521) $1596088087447_blob.png$ (0.1335 ) $1596088087419_blob.png$ 1400

= 26 $1596088087450_blob.png$    0.03365535 $1596088087447_blob.png$ 1400 = 26 $1596088087488_blob.png$  47.11749 = 0.5518

Coefficient of coincidence = 0.5518

Interference (I) = 1- Coefficient of coincidence

Interference I = 1 - 0.5518 = 0.45 or 45 %.