15. A cross in Drosphila involved the recessive x-linked genes yellow body (y),
ID: 57038 • Letter: 1
Question
15. A cross in Drosphila involved the recessive x-linked genes yellow body (y), white eyes (w), and cut wings (ct). A yellow bodied, white-eyes female with normal wings was crossed to a males whose eyes and body were normal, but whose wings were cut. The f1 females were wild type for all three traits, while the F1 males expressed the yellow-body, white-eyes traits. The cross was carried to F2 progeny, and only male offspring were tailed. On the basis of the data shown here, a genetic map was constructed. (a) Diagram the genotypes of the F1 parents. (b) Construct a map, assuming that w is at locus 1.5 on the X chromosome. (c) Were any double-crossover offspring expected?(d) could the F2 female offspring be used to construct the map? Why or why not?
*given:
phenotype Male offspring
y + ct 9
+ w + 6
y w ct 90
+ + + 95
+ + ct 424
y w + 376
y + + 0
+ w ct 0
(a)F1 parents:
16. D is a mutation on chromosome III with a dominant effect of wing shape. It is lethal when homozygous. The genes ebony (e) and pink eye (p) are recessive mutations on chromosome III. Flies from the D stock were crossed to homozygous ebony, pink flies, and the F1 progeny with a D phenotype were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the following table (a) diagram the cross, showing the genotypes of the parents and the offspring of both crosses. (b) What is the sequence and interlocus distance between these three genes?
*given:
phenotype Number
D 401
Ebony, pink 389
D, ebony 84
Pink 96
D, pink 2
Ebony 3
D, ebony, pink 12
Wild type 13
Explanation / Answer
Question 15
a)
Parents were ywCT/ywCT x YWct/- (- is Y chromosome with no genes on it)
X-linked gametes produced by female: all ywCT
X-linked games produced by male: YWct
F1 females all ywCT/YWct (they get one X chromosome from each parent). Males all ywCT/- (they get X from their mother and Y from their father).
Now perform a cross between ywCT/YWct (females) X ywCT/- (males)
X-linked gametes produced by female: YWCT, YWct, YwCT, Ywct, ywCT, yWct, ywCT, ywct,
X-linked gametes produced by male: ywCT
ywCT
-
YWCT
YyWwCTCT (females)
-YWCT (males)
YWct
YyWwCTct(females)
- YWct(males)
YwCT
YywwCTCT(females)
- YwCT(males)
Ywct
YywwctCT(females)
- Ywct(males)
yWCT
yyWwCTCT(females)
- yWCT(males)
yWct
yyWwCTct(females)
- yWct(males)
ywCT
yywwCTCT(females)
- ywCT(males)
ywc
yywwCTct(females)
- ywc(males)
b)
To make the map follow these steps
1. Identify parental classes (these will be the most common and in high frequency i.e. ywCT and YWct)
2. Identify the double crossover classes (these will be the least frequent i.e. yWCT and Ywct.
3. You can now compare these to the parental classes to work out the gene order. The gene that has moved in the double recombinant class compared to the parental class is in the middle.
Look at ywCT and double recombinant yWCT - in the parental class y is found with w and CT, but in the double recombinant it is still with CT but instead of w there is another CT, therefore w is in the middle.
4. Now look for reciprocal single crossovers, they should be similar in numbers and use to calculate recombination frequency = (no. recombinant progeny/total progeny)
Recombination between w and c will give ywc and YWCT, there are no double recombinants so we can ignore these although they need to be included in the calculation where they exist.
RF between w and ct is therefore ((90 +95+0+0)/1000) = 0.185
and RF between w and y using the same logic as above is (9 +6+0+0)/1000 = 0.015
RF between y and ct is (9 + 6 + 90 + 95)/1000 = 0.2
Finally convert RFs into percentages (x100). %recombination = no. map units.
Your map is y.1.5..w.......18.5...............ct
c)
Expected double cross over = (interval yw distance/100) x (interval wct distance/100) x (total offspring) = 2.8
d)
Calculate interferance to answer this part of question
Interference (I) = 1 - (observed double cross overs / expected double cross overs)
= 1 - 0 / 2.9
= 1
In some regions, there are never any observed double recombinants. In these cases, c.o.c.=0, so I=1 and interferenceis complete which holds true for our cross also. Thus, F2 female offsprings can be used to construct genetic map.
ywCT
-
YWCT
YyWwCTCT (females)
-YWCT (males)
YWct
YyWwCTct(females)
- YWct(males)
YwCT
YywwCTCT(females)
- YwCT(males)
Ywct
YywwctCT(females)
- Ywct(males)
yWCT
yyWwCTCT(females)
- yWCT(males)
yWct
yyWwCTct(females)
- yWct(males)
ywCT
yywwCTCT(females)
- ywCT(males)
ywc
yywwCTct(females)
- ywc(males)
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