Researchers cross a corn plant that is purebreeding for the

Researchers cross a corn plant that is pure-breeding for the dominant traits colored aleurone (C1), full kernel (Sh), and waxy endosperm (Wx) to a pure-breeding plant with the recessive traits colorless aleurone (c1), shrunken kernel (sh), and starchy (wx).

alculate the recombination fraction between the Cl-Sh gene pair.

Enter your answer to three decimal places (example 0.023).

Recombination fraction .035

Part D

Calculate the recombination fraction between the Sh-Wx gene pair.

Enter your answer to three decimal places (example 0.234).

Recombination fraction .184

Part E

Calculate the recombination fraction between the Cl-Wx gene pair

Enter your answer to three decimal places (example 0.234).

Recombination fraction .219

QUESTION: What is the interference value for this data set?

Enter your answer to three decimal places (example 0.234).

Problem 5.13 Kernel Phenotype Number Colored, shrunken, starchy 116 601 Colored, full, starchy Colored, full, waxy 2538 Colored, shrunken, waxy Colorless, shrunken starchy 2708 2 Colorless, full, starchy Colorless, full, waxy 113 Colorless, shrunken, waxy 626 6708 Part E Calculate the recombination fraction between the C-Wx gene pair Enter your answer to three decimal places (example 0.234) Recombination fraction .219 Submit My Answers Give Up Correct Part F What is the interference value for this data set? Enter your answer to three decimal places (example 0.234) Interference value .34 Submit My Answers Give Up incorrect; Try Again; 5 attempts remaining Provide Feedback Continue

In the given cross, the following are the OBSERVED results:

The map distance (recombination fraction, the distance between genes) can be calculated using the formula:

Rf = n° of recombinants/total number of individuals

Total number of individuals = 2538 + 2708 + 113 + 116 + 601 + 626 + 2 + 4 = 6708

The recombination fraction for Cl-Sh: (113+116+4+2)/6708 = 235/6708 = 0.035

The recombination fraction for Sh-Wx: (601+626+4+2)/6708 = 1233/6708 = 0.184

The recombination fraction for Cl-Wx: (113+116+601+626+4+4+2+2)/6708 = 1468/6708 = 0.219

Cl---3.5 m.u.---Sh----------18.4 m.u.----------Wx

0.035 + 0.184 = 0.219 = 21.9 m. u.

You can prove the calculus was done correctly comparing to an example from the book: An Introduction to Genetic Analysis. 7th edition. in NCBI website (just search the topic: Three-point testcross in that book).

QUESTION: What is the interference value for this data set?

To get the interference values first you need to get the coefficient of coincidence (c.o.c.):

c.o.c. = observed double recombinants/expected double recombinants

Expected double recombinants = (Rf of single crossover 1 x Rf of single crossover 2) x total n° of individuals

Expected double recombinants = (0.035 x 0.184) x (6708) = 0.00644 x 6708 = 43.19952

c.o.c. = (4+2)/43.19952 = 6/43.19952 = 0.13889

Interference value = 1 - 0.13889 = 0.861

Please be kind to like the answer if it was helpful for you