14. In Drosophila, a cross was made between females-all express- ing the three X

Question

14. In Drosophila, a cross was made between females-all express- ing the three X-linked recessive traits scute 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 F2 generation, and 1000 offspring were counted, with the results shown in the fol- lowing table Offspring 314 280 150 156 46 Phenotype sC S sc+ scV 30 10 SC No determination of sex was made in the data (a) Using proper nomenclature, determine the genotypes of the P, and Fi parents. (b) Determine the sequence of the three genes and the map distances between them (c) Are there more or fewer double crossovers than expected? (d) Calculate the coefficient of coincidence. Does it represent positive or negative interference?

Explanation / Answer

Analyses: Drosophila, females will have two (2) X chromosomes and males do have one (1) X chromosome and one (1) Y chromosomes. Given female has all three (3) X-linked recessive trait, so those locus should be in homozygous conditions for the recessive allele. Thus if 'sc' stands for scute bristles, 's' stands for sable body and 'v' stands for vermilion eyes and + stands for all respective wild alleles, we suppose genotypes for P (parental) cross as below:

female : sc s v / sc s v and male : + + + / Y . the croos between them and the genotypes of F1 progeny will supposedly look like as given below:

P: sc s v / sc s v X + + + / Y

F1 : sc s v / +++ (females) and sc s v / Y (males)

Genotypes of F1 progeny is showing all females having respective wild type allele, making all three locus heterozygous and outcome phenotype will be wild type. Whereas, all the males are having only one X chromosmes with all the three recessive allele (hemizygous condition), resulting all of them to express recessive phenotypes.

Now, according to the question F1 cross will supposedly look like this, as given below :

F1 : sc s v / +++ X sc s v / Y

Now being in hemizygous condition, males wont show any recombination (Drosophila males do not showlack recombination for other locuses also). Males here will contribute to the offspring either sc s v or Y chromosmes. In both cases, offspring of phenotypes will be regulated by F1 females chromosme. F1 Females for these three traits have different patterns of cross-over (CO) as given below:

1) CO between sc and s locus : resulting phenotype of offspring will be as : sc + + , + s v, sc + v (from Double CO between sc X s and s X v, simultaneously) and + s + (from Double CO between sc X s and s X v, simultaneously), according to the question the numbers are 156, 150, 46 and 30, respectively, among 1000 offsprings. Thus frequency of CO between sc ands s locus is = {(156+150+46+30)/1000} = (382/1000) = 0.382. This frequency also denotes the relative distance between these two locus as centiMorgan (cM) unit. Thus the distnace between sc and s locus is approx 38.2 cM.

2) CO between s and v locus : resulting phenotype of offsprings : sc s +, + + v, sc + v (from Double CO between sc X s and s X v, simultaneously) and + s + (from Double CO between sc X s and s X v, simultaneously), according to the question the numbers are 10, 14, 46, 30. Thus the frequency of CO = {(10+14+46+30)/1000} = (100/1000) = 0.1. Thus the relative distance between s and v locus is 10.0 cM.

3) CO between sc and v locus : resulting phenotypes of offsprings : sc + +, + s v, sc s + and + + v, according to the question the respective numbers are 156, 150, 10, 14. Thus the frequency of CO = {9156+150+10+14)/1000} = (330/1000) = 0.330. Thus the relative distance between sc and v locus is 33.0 cM.

4) Double Cross-Over between sc X s and s x V, simultaneously : resulting phenotypes of offsprings : sc + v and + s +, according to the question the respective numbers are 46 and 30. Thus the observed Double Cross over frequency is = {(46+30)/1000} = (76/1000) = 0.076.

According to the results from 1), 2) and 3) we can say the distance between sc and s is highest (38.2 cM) among all the three and locus v is lying betwen these two. Thus the speculated relative distances and arrangement of these three given locus viz. sc, s and v, on X chromosme, is going to be as shown below:

sc.................................v..........s

33 cM 10cM

Now according to the arrangements, the expected double croos over frequency is = (cross over frequency between sc and v * croos over frequency between v and s) = (0.330 * 0.1) = 0.033.

So, the coeffcient of coincidence (COC) = (observed cross over frequency / expected cross over frequency) = (0.076/0.033) = 2.303030

Interference = (1 - COC) = (1 - 2.303030) = - 1.303030

Answers: Now depending on the above analyses the answers of the given question will be :

a) Genotypes of P and F1 parents:

P : sc v s / sc v s X + + + / Y

F1: sc v s / + + + X sc v s / Y

b) The sequence of the genes are as follows : sc v s and the map distance are i) between sc and v = 33 cM, ii) between v and s = 10 cM and iii) between sc and s = 38.2 cM.

c) As expected cross over frequency is 0.033 and observed ferquency is 0.076; we could conclude that there are more double cross over phenomeona than expected.

d) The COC is 2.303030, which indicates a negative interfernce, implying cross over at one region is influencing positively cross over at the other position to occur simultaneously.

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