please help with these 4 questions 1. In wild type fruit flies, gray body G and
ID: 217564 • Letter: P
Question
please help with these 4 questions
1. In wild type fruit flies, gray body G and regular wings R are dominant to black body g and vestigial wings r. A homozygous wild type fly is crossed with a homozygous reces A. Show this cross using the letter symbols. If the resulting heterozygous offspring were test crossed with a known homozygous recessive and the two traits were not linked, what would be the expected phenotypic ratio? C. However, the actual results from the above test cross yielded the following phenotypes: gray-regular black-vestigial gray-vestigial black-regular 110 90 30 20 Which of the above is a recombinant phenotype and which is a parental phenotype? Has linkage occurred? How do you know? Calculate how many map units the body color locus is from the wing length locus?Explanation / Answer
Answer:
1).
A). GGRR (wild) x ggrr (recessive)
B). GgRr (wild) is F1 progeny
GgRr x ggrr----Test cross
gr
GR
GgRr-25%
Gr
Ggrr-25%
gR
ggRr-25%
gr
ggrr-25%
Expected phenotypic ratio = 1:1:1:1
C). Hint: Recombinant phenotypes are always less in number than the parental combinations in the progeny.
So the two genes are linked and the parental phenotypes are gray-regular and black-vestigial. Recombinant phenotypes are gray-vestigial and black-regular.
Recombination frequency = (no. of recombinants / Total progeny) *100
= (50/250)100
= 20%
Recombination frequency (%) = Distance between the genes (map units)
Therefore the distance between the two loci are 20 map units.
----------------------------------------
2).
A). Parental phenotypes are slate-red-long and blue-green-short.
Double crossover phenotypes are slate-green-short and blue-red-long.
B). Middle gene is “S”
C). R----------29.43cM--------S-----7.8cM------L
C. Coefficient of coincidence = 0.78
Expected double cross over = 0.023
D). Interference = 0.22
It means that the there is a difference in expected and observed recombinant frequency.
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is SRL/srl.
1).
If single crossover occurs between S & R..
Normal combination: SR / sr
After crossover: Sr/sR
Sr progeny= 402+27=429
sR progeny = 427+27= 454
Total this progeny = 883
The recombination frequency between S&R = (number of recombinants/Total progeny) 100
RF = (883/3000)100 = 29.43%
2).
If single crossover occurs between R&L..
Normal combination: RL / rl
After crossover: Rl/rL
Rl progeny= 427+95= 522
rL progeny = 402+85=487
Total this progeny = 1009
The recombination frequency between R&L = (number of recombinants/Total progeny) 100
RF = (1009/3000)100 = 33.63%
3).
If single crossover occurs between S&L..
Normal combination: SL / sl
After crossover: Sl/sL
Sl progeny= 27+95= 122
sL progeny = 27+85=112
Total this progeny = 234
The recombination frequency between S&L = (number of recombinants/Total progeny) 100
RF = (234/3000)100 = 7.8%
Recombination frequency (%) = Distance between the genes (cM)
R----------29.43cM--------S-----7.8cM------L
Expected double crossover frequency = (RF between R & S) * (RF between S & L)
= 29.43% * 7.8% = 0.023
The observed double crossover frequency = 27+27 / 3000 = 0.018
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.018 / 0.023
= 0.78
Interference = 1-COC
= 1-0.78 = 0.22
-------------------------------------------------------------------------------------------
3).
A). w----------35cM-------m-----20cM------f
B). Coefficient of Coincidence (COC) = 0.057
C). Interference = 1-0.057 = 0.943
It means that the there is a difference in expected and observed recombinant frequency.
Explanation:
If the genes are not linked and they are assorted independently, the trihybrid testcross ratio is “1:1:1:1:1:1:1:1”
The number of gametes produced is not in the ratio mentioned above. Therefore the genes are linked.
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is +m+ / w+f.
1).
If single crossover occurs between + & m.
Normal combination: +m/w+
After crossover: ++/wm
++ progeny= 168+3=171
wm progeny = 178+1=179
Total this progeny = 350
The recombination frequency between +&m = (number of recombinants/Total progeny) 100
RF = (350/1000)100 = 35%
2).
If single crossover occurs between m&+.
Normal combination: m+/+f
After crossover: mf/++
mf progeny= 95+1=96
rL progeny = 101+3+104
Total this progeny = 200
The recombination frequency between m&+ = (number of recombinants/Total progeny) 100
RF = (200/1000)100 = 20%
3).
If single crossover occurs between +&+.
Normal combination: ++/wf
After crossover: +f/w+
+f progeny= 168+95=263
w+ progeny = 178+101= 279
Total this progeny = 542
The recombination frequency between +&+ = (number of recombinants/Total progeny) 100
RF = (542/1000)100 = 54.2%
Recombination frequency (%) = Distance between the genes (cM)
w----------35cM-------m-----20cM------f
Expected double crossover frequency = (RF between w & m) * (RF between m & f)
= 0.35 * 0.2 = 0.07
The observed double crossover frequency = 3+1 / 1000 = 0.004
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.004 / 0.07
= 0.057
Interference = 1-COC
= 1-0.057 = 0.943
----------------------------------------------------------
4).
A).
If the genes are not linked and they are assorted independently, the trihybrid testcross ratio is “1:1:1:1:1:1:1:1”
The number of gametes produced is not in the ratio mentioned above. Therefore the genes are linked.
B). TSR
C). T----------16.29cM-------S-----6.36cM------R
D). Interference = 1-0.5 = 0.5
Explanation:
If the genes are not linked and they are assorted independently, the trihybrid testcross ratio is “1:1:1:1:1:1:1:1”
The number of gametes produced is not in the ratio mentioned above. Therefore the genes are linked.
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is TSR/tsr.
1).
If single crossover occurs between T&S.
Normal combination: TS/ts
After crossover: Ts/tS
Ts progeny= 11+2=13
tS progeny = 12+0=12
Total this progeny = 25
Total progeny = 393
The recombination frequency between T&S = (number of recombinants/Total progeny) 100
RF = (25/393)100 = 6.36%
2).
If single crossover occurs between S&R
Normal combination: SR/sr
After crossover: Sr/sR
Sr progeny= 33+0=33
sR progeny = 2+29=31
Total this progeny = 64
The recombination frequency between S&R = (number of recombinants/Total progeny) 100
RF = (64/393)100 = 16.29%
3).
If single crossover occurs between T&R.
Normal combination: TR/tr
After crossover: Tr/tR
Tr progeny= 33+11=44
tR progeny = 29+12=41
Total this progeny = 85
The recombination frequency between T&R = (number of recombinants/Total progeny) 100
RF = (85/393)100 = 21.63%
Recombination frequency (%) = Distance between the genes (cM)
T----------16.29cM-------S-----6.36cM------R
Expected double crossover frequency = (RF between T & S) * (RF between S & R)
= 0.1629 * 0.0636 = 0.01
The observed double crossover frequency = 0+2 / 393 = 0.005
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.005 / 0.01
= 0.5
Interference = 1-COC
= 1-0.5 = 0.5
gr
GR
GgRr-25%
Gr
Ggrr-25%
gR
ggRr-25%
gr
ggrr-25%
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