The F1 progeny of cross 1 were allowed to interbreed and the resulting F2 progeny are shown in the t

Cross 2

The F1 progeny of cross 1 were allowed to interbreed and the resulting F2 progeny are shown in the table below.
Phenotype Number

Red-eyed male  362
Red-eyed female  384
Scarlet-eyed male  141
Scarlet-eyed female  122

1. Calculate the following phenotypic ratios to 1 decimal place (eg 4.9 : 1). (not sure if my ratio's are correct)

F2 male wild type : male scarlet-eyed = 2.6:1
F2 female wild type : female scarlet-eyed = 3.1:1
F2 combined sexes wild type : combined sexes sepia-eyed = 2.8:1

2. What does this result tell you about the behaviour of the alleles of the scarlet gene with respect to Mendel’s Laws?

Cross 3

The table below shows the phenotypes of the F2 generation of a cross between a true breeding ebony-bodied female and a true breeding vestigial-winged male. Note that vestigial-winged flies have very small shrivelled wings.

Phenotype Number
Wild type body and wings
579
Wild type body and vestigial wings
171
Ebony body and wild type wings
201
Ebony body and vestigial wings
68

3. What is the phenotypic ratio for the F2 generation in the table above? (calculate to 1 decimal place eg 7.2 : 4.9 : 3.7 : 1)

4. This result is very close to the expected 9:3:3:1 ratio. What does it tell you about the behaviour of the alleles of both of these genes with respect to Mendel's Laws?

Cross 4

5. What phenotypes and phenotypic ratio(s) would you expect from crossing an ebony-bodied, normal-winged female with an ebony-bodied vestigial-winged male, both from the F2 generation described above? Clue - there are two alternative results for this testcross depending on the genotype of the female.

Either:
Or:

Cross 5

The table below shows the phenotypes of the progeny of a cross between two flies that are both heterozygous for the aristapedia mutation.

Phenotype Number
Wild type antennae
375
Aristapedia (leg-like antennae)
689

6. What phenotypic ratio would you normally expect and what is the actual phenotypic ratio?

Expected ratio:
Actual ratio:

7. The mutant aristapedia allele exerts two separate phenotypic effects, one is dominant to wild type and one is recessive, what are these effects? Note that the recessive effect relates to the unusual phenotypic ratio.

Cross 6

A true breeding vestigial-winged female (vg vg, ri+ri+) was crossed with a true breeding male fly with incomplete wing veination (vg+vg+,ri ri). All the F1 progeny were wild type. The F1 flies were allowed to interbreed and the table overleaf shows the phenotypes of the F2 generation

Phenotype Number
Wild type
552
Vestigial wing
284
Incomplete wing vein
173

8. Explain why there are only three visible phenotypes rather than four. What is this phenomenon called?

Cross 7

9. Starting from the parental cross shown below, write out the genotypes, phenotypes and sex of the F1 progeny and fill in the Punnett square to show the genotypes, phenotypes and sex of the F2 progeny. Note that X = the X chromosome and Y = the Y chromosome.

Parental vg vg, Xw+Xw+ x vg+vg+, XwY
Vestigial-winged female White eyed male

F1 Genotypes =
F1 Phenotypes =
F1 Sex =

F2 Genotypes, phenotypes and sexes displayed in a Punnett square

10. What is unusual about the phenotypic ratios shown in this F2 and what is the cause of this effect? Do the alleles of the white gene segregate independently of the vestigial gene?

Cross 8

The ebony (body) gene locus and the spineless (bristles) gene locus are both on the third chromosome. A true-breeding ebony-bodied female with wild type bristles was crossed with a true-breeding spineless-bristled male with wild type body colour, yielding F1 progeny that were all wild type. An F1 female was then testcrossed with an ebony-bodied spineless-bristled male to produce the progeny recorded in the table below.

Phenotype Number
Wild type body and bristles
66
Wild type body and spineless bristles
449
Ebony body and wild type bristles
442
Ebony body and spineless bristles
49

11. How many map units (centiMorgans) separate these two genes?

The testcross was repeated using an F1 male with an ebony-bodied spineless-bristled female and gave the result shown in the table below.

Phenotype Number
Wild type body and spineless bristles
491
Ebony body and wild type bristles
470

12. What does this result demonstrate about meiosis in male Drosophila?

Cross 9

Both brown (eyes) and dumpy (wing shape) are autosomal recessive mutations. The progeny of a cross between a female fly, heterozygous for both brown and dumpy, with a male fly homozygous for both mutations are shown in the table below.

Phenotype Number
Wild type
269
Brown-eyed
217
Dumpy-winged
215
Brown-eyed, dumpy-winged
283

13. Based on the data given, perform a Chi2 test to determine whether it supports the following Null Hypothesis.

'There is no significant difference between the observed data and the 1:1:1:1 phenotypic ratio expected if the alleles of both genes are assorting independently according to Mendel's Second Law’.

Please show your working and present a sensible conclusion based upon your results.

Phenotype       +             BW                DP                BW + DP
Observed
Expected
(O-E)2
(O-E)2/E

Chi2 =                              Degrees of freedom =

Corresponding value of P = between ___________and___________

Conclusion: