Question

30) A homozygous groucho fly (g, bristles about the eye) is crossed with a homozygous rough fly (r, eye abnormality). The F1 females are test crossed to a groucho, rough fly. The following data was obtained.

514 Groucho

466 Rough

11 Groucho, rough

9 Wild type

Are these genes linked? Is the F1 in coupling or repulsion phase. Draw out the heterozygous parents genotype on chromosomes.

33)

Three corn genes, R, D and Y lie on chromosome 9. The map of the three genes, using map units is:
            20        40
R -------D--------------Y

A heterozygous plant (R D Y/r d y) is crossed with a testcross parent (r d y/r d y) :

how many double cross over progeny would be expected in a progeny of 1000 if there were no interference?

What is the interference if there are 50 DCOs?

B= brown body, b = black body, N = non-waxy wings, n = waxy wings, R = red eyes, r = cinnabar eyes

What is the order of the genes?

What is the recombination frequency for each gene pair?

Draw a map with the genes listed and the correct map units.

Is there interference? If so what is the value.

Answer 1

30).

gR/gR (Groucho) x Gr/Gr (rough)—Parents

gR/Gr (Wild type)-----F1

gR/Gr (Wild type) x gr/gr (groucho, rough)----------Test cross

Dihybrid test cross ratio must be in 1:1:1:1, if the genes are not linke. But the given number is not similar, so that they are said to be linked.

F1 progeny are in repulsion phase.

Heterozygous genotype is gR/Gr.

33).

RDY / rdy X rdy / rdy

Class of gametes	Phenotype	Frequency of reciprocal pair	Numbers	Frequency of each class	Number of progeny frequency * 1000
Parental	RDY	1- all recombinant progeny	1-(0.2+0.4+0.08) = 0.32	0.16	160
	Rdy			0.16	160
SCO 1	Rdy	RF in region 1 = SCO1	SCO1 = 0.2	0.1	100
	rDY			0.1	100
SCO 2	RDy	RF in region 2 = SCO2	SCO2 = 0.4	0.2	200
	rdY			0.2	200
DCO	RdY	(RF in region 1) * (RF in region 2)	0.2 * 0.4 = 0.08	0.04	40
	rDy			0.04	40

If there are 50 DCOs, then the interference…..

Class of gametes	Phenotype	Frequency of reciprocal pair	Numbers	Recombination Frequency of each class	Number of progeny frequency * 1000
Parental	RDY	1- all recombinant progeny	1-(0.12+0.32+0.08) = 0.48	0.24	240
	rdy			0.24	240
SCO 1	Rdy	RF in region 1 = SCO1	SCO1 = 0.2 – 0.08 = 1.2	0.06	60
	rDY			0.06	60
SCO 2	RDy	RF in region 2 = SCO2	SCO2 = 0.4 – 0.08 = 3.2	0.16	160
	rdY			0.16	160
DCO	RdY	(RF in region 1) * (RF in region 2)	0.2 * 0.4 = 0.08	0.04	40
	rDy			0.04	40

1). Order of genes = R—D--Y

2)

Expected Recombination frequency (%)

RF between R&D = 12%

RF between D&Y = 32%

RF between R&Y = 12%+32% = 0.44

3).

Recombination frequency (%) = Distance between the genes (mu)

Gene map = R----12 mu-----D------32------Y

4). Interference is there and its value = 0.375

Explanation:

Expected double crossovers = 0.08

Observed double crossovers = 50/1000 = 0.05

Coefficient of coincidence (COC)= Observed double crossovers / Expected double crossovers

= 0.05 / 0.08 = 0.625

Interference = 1-COC

= 1-0.625 = 0.375