Question

In the fruit fly D. virilis, the mutations dusky body color (dy), cut wings (ct), and white eyes (w) are all recessive alleles located on the X chromosome. Females heterozygous for all three mutations were crossed to dy ct w males, and the following phenotypes were observed among the 500 offspring.

wildtype 18

dusky 153

cut 6

white 76

cut, white 150

dusky, cut 72

dusky, cut, white22

dusky, white 3

14) What are the genotypes of the two maternal X chromosomes, with the genes in the correct order?

a) dy + w / + ct

b) dy + + / + w ct

c) dy w + / + ct

d) dy w ct / + + +

e) + ct + / dy + w

15) What is the recombination frequency between dusky and white?

a) 148/500

b) 46/500

c) 43/500

d) 49/500

e) 40/500

PLEASE EXPLAIN!!!! CLEARLY!!!

Answer 1

The genotype of wild type is + + + (as dominant or wild type charecter is denoted by + and all the three charecter is wild type)

the genotype of dusky (color) is dy + + ( except dusky color other two alleles will be wild type i.e + type )

the genotype of cut (wing) is ct + + ( i.e other two alleles are + type )

the genotype of white(eye) is w + +   

the genotype of cut and white combination is    ct w + ( i.e only one charecter is wild type )

the genotype of dusky and cut combination is dy ct +

the genotype of dusky cut and white is dy ct w ( no charecter is of wild type )

the genotype of dusky and white combination is dy w +

genotype	number of offsprings
+ + +	18
dy + +	153
ct + +	6
w + +	76
ct w +	150
dy ct +	72
dy ct w	22
dy w +	3

the most frequent pair of phenotypic classes ( external charecter) tell us the genotype of original parental combination . here dusky color( dy + + ) and cut white combination ( ct w + ) are the most frequent ( number of offspring is 153 and 150 of these two genotype ). hence these two are the parental combination .

these happens due to linkage i.e the tendency of genes ( present on same loci of chromosome ) to inheritate together .

the second most frequent phenotypic pair are the result of single cross over( S.C.O) between first two gene among the 3 genes and it is known as singel cross over 1 . here ( w + + ) and (dy ct +) genotypic combinations  are the result single cross over 1

the third most frequent phenotypic pair tell us genotypic combination of the result of another single crossing over ( single cross over 2 ) this S.C.O happens between last two genes among three genes . therefore the genotypic combination of S.C.O 2 is (+ + +) and (dy ct w )

and the least frequent combination are the result of double cross over ( D.C.O ) here (dy w +) and (ct + +) are the D.C.O combination .

14) option b is right option and details is explanned.

15) option a is the right option i.e 148/ 500

Order To determine the correct order of gene first we have to determine the middle gene. To determine middle gene we have to compare parental and D.C. O Combination, Parental Comlaination are (dy + +). and (et w t ) and D.C.O Contenation are = (dy wt) and (ct + +). First we assume the order is + dy + and (t + W (assuming dy as middle gene and it first gene) + dy t (Parental) . Gross will be a ct + W From this the double cross over frequency will be + + + and ct dy ow which are not resembles with the given D.O.O There for I can't be middle gene.. dy car = ct dy wettt B cttW

Lets assume 'ct as middle. gene and dy as first one. Therefore cross will be dy + + ct w. dy tt . = + + w, dy ct + + ct w These combination also do not resembels with the given Dco. Lets assume we as middle gene and dy as first in 0 dy + + Cross will be + W ct dy tt = + + ct, dy w t at w ct These & Combination is same as given DCO combination, . w is the middle gene. Next we have to decide first gene from Parental combination and S.C.O Combination

Lets assume dy as first gene : Cross will be e dy t dyy t + w at S.C.O frequence will be day w ct and ttt . which nesembels with the given S.C.O There fore the commect order of gene dyt + + w ct i.e. option (b) of Question number 14. 2 be cause the combination besembler th with the parental combination and we determine the orden. as dy w ct (dy first and w' middle ) Hence nest of the options are not applicable.