30) A homozygous groucho fly (g, bristles about the eye) is crossed with a homoz
ID: 301938 • Letter: 3
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.
Explanation / Answer
Answer:
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
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
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