Three linked autosomal loci were studied in squash flowering

Three linked autosomal loci were studied in squash: flowering time (late vs. early), fruit shape (elongate vs. ovate) and stem length (long vs. short). Pure breeding early flowering, elongated squash with long stems were mated to pure breeding late flowering, ovated squash with short stems. The F1 were testcrossed and the testcross progeny are given below.

I found 1 and 2, but need guidance with 3-5. Thanks!

1) How many progeny did you look at for this experiment: 4856

2) Which locus is in the center? Stem Length

3) What is the value for R1?   

4)What is the value for R2?

5)What is the coefficient of coincidence?

Three linked autosomal loci were studied in squash: flowering time (late vs. early), fruit shape (elongate vs. ovate) and stem length (long vs. short). Pure breeding early flowering, elongated squash with long stems were mated to pure breeding late flowering, ovated squash with short stems. The F1 were testcrossed and the testcross progeny are given below. Stem Flowering Number Fruit shape Time Length Elongate Short 577 Late Early Elongate Short 64 Early Elongate Long 1164 Elongate Long 352 Late Ovate Late Long 78 Early Ovate Long 657 Early Short 439 Ovate Short Late Ovate 1525 Use the map to answer the questions below. Fruit Shape Locus B Locus C R1 R2 How many progeny did you look at for this experiment? Answer: 4856

Solution

Answer:

Elngate= A; Ovate=a

Early=B, late=b

Long=C; short=c

Phenotype

Shorthand

Number of progeny

Late, elongate, short

bAc

577

Early, elongate, short

BAc

64

Early, elongate, long

BAC

1164

Late, elongate, long

bAC

352

Late, ovate, long

baC

78

Early, ovate, long

BaC          

657

Early, ovate, short

Bac

439

Late, ovate, short

Bac

1525

Imagine the order of genes is BAC/bac

1.     If single cross over (SCO) occurs between B&A

Normal order= B---------A & b-----a

After cross over= B-----a & b------A

B-----a recombinants are 657+439 = 1096

b------A recombinants are 577+ 352 = 929

Total recombinants = 2025

RF = (2025/4856)*100 =38.2%

2.     If single cross over (SCO) occurs between A & C

Normal order= A---------C & a----c

After cross over= A-----C & a------c

A-----C recombinants are 577+64=641

a------c recombinants are 657+78=735

Total recombinants = 1376

RF = (1376/4856)*100 = 28.33%

3.     If single cross over (SCO) occurs between B & C

Normal order= B---------C & b------c

After cross over= B-----c & b------C

B-----c recombinants are 64+439=503

b------C recombinants are 78+352=430

Total recombinants = 933

RF = (933/4856)*100 =19.21%

% RF = Map unit distance

The order of genes is -----

B------19.21 m.u.------C------28.33m.u.-------A

RF of R1 between B & C = 19.21% = 0.1921

RF of R2 between C & A = 28.33% = 0.2833

Expected double cross overs = 0.1921 * 0.2833= 0.0544

Observed double cross overs = 78 + 64 = 142 /4856 = 0.0292

Coefficent of coincidence (COC) = Obeserved dco / Expected dco

COC = 0.0292/0.0544 = 0.54

Phenotype	Shorthand	Number of progeny
Late, elongate, short	bAc	577
Early, elongate, short	BAc	64
Early, elongate, long	BAC	1164
Late, elongate, long	bAC	352
Late, ovate, long	baC	78
Early, ovate, long	BaC	657
Early, ovate, short	Bac	439
Late, ovate, short	Bac	1525