Predict the results of two-point mapping between d and c, d and vg, and d and ad
ID: 184448 • Letter: P
Question
Predict the results of two-point mapping between d and c, d and vg, and d and adp. 13. Two different female Drosophila were isolated, each heterozy- gous for the autosomally linked genes b (Black body), d (dachs tarsus), and c(curved wings). These genes are in the order d-b- with b being closer to d than to c. Shown here is the genotypic arrangement for each female along with the various gametes formed by both Fenale A d b+ Female B Gamete formation (1) d b c (5)d +(1)db(5) d b (3)++c (7)d +(3)d+7d++ Identify which categories are noncrossovers (NCOs), single crossovers (SCOs), and double crossovers (DCOs) in each case. Then, indicate the relative frequency in which each will be produced. 14. In Drosophila, a cross was made between females-all express- ing the three X-linked recessive traits soute bristles (sc), sable body (s), and vermilion eyes (v)-and wild-type males. In the Fi, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F: generation, and 1000 offspring were counted, with the results shown in the fol- lowing table. Offspring 314 280 150 156 46 30 10 Phenotype No determination of sex was made in the data (a) Using proper nomenclature, dedermine the genotypes of the P, and F parents { b) Determine the sequence of the nee genes and the map distances between them (c) Are th e more or fever doubk aosovertán expecte d Cakulate the codfficient of coincidence. Doa it represent positine or negative interfa anceExplanation / Answer
Answer:
13).
Female A:
Single crossover between second locus and third locus: 1). dbc & 2). +++
Noncrossovers: 3). ++c & 4). db+
Double crossovers: 5). d++ & 6) +bc
Single crossover between first locus and second locus: 7). d+c & 2). +b+
Female B:
Double crossovers: 1). db+ & 2) ++c
Single crossover between second locus and third locus: 3). d+c & 4). +b+
Single crossover between first locus and second locus: 5). dbc & 6). +++
Noncrossovers: 7). d++ & 8). +bc
14).
a).
P1 genotypes:
Femlae genotype--- sc s v / sc s v
Male genotype--- + + + / Y
F1 genotypes:
Femlae genotype--- + + + / sc s v
Male genotype--- sc s v / Y
b). sc----------38.2cM--------s-----------9.4cM--------------v
c).
Expected double crossover frequency = (RF between h & w) * (RF between w & b)
= 38.2% * 9.4% = 0.036
Observed double crossovers = 0.036*1000 = 36
The observed double crossovers sc + v & + s + = 46+30 =76
So observed double crossovers are more than the expected double crossovers
d).
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.076 / 0.036
= 2.11
Interference = 1-COC
= 1- 2.11 = -1.11
Interference is negative interference
Explanation:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotypes is sc s v / + + +.
1).
If single crossover occurs between sc & s..
Normal combination: sc s / ++
After crossover: sc +/+ s
sc + progeny = 156+46 = 202
+ s progeny = 150+30=180
Total this progeny = 382
The recombination frequency between sc&s = (number of recombinants/Total progeny) 100
RF = (382/1000)100 = 38.2%
2).
If single crossover occurs between s & v..
Normal combination: s v / ++
After crossover: s+/+v
s+ progeny= 30+10=40
+v progeny = 46+14=54
Total this progeny = 94
The recombination frequency between s&v = (number of recombinants/Total progeny) 100
RF = (94/1000)100 = 9.4%
3).
If single crossover occurs between sc & v..
Normal combination: sc v / ++
After crossover: sc +/+ v
sc + progeny= 156+10=166
+ v progeny = 150+14=164
Total this progeny = 330
The recombination frequency between sc&v = (number of recombinants/Total progeny) 100
RF = (330/1000)100 = 33%
Recombination frequency (%) = Distance between the genes (cM)
sc----------38.2cM--------s-----------9.4cM--------------v
Expected double crossover frequency = (RF between h & w) * (RF between w & b)
= 38.2% * 9.4% = 0.036
The observed double crossover frequency = 30+46 / 1000 = 0.076
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.076 / 0.036
= 2.11
Interference = 1-COC
= 1- 2.11 = -1.11