Drosophila Practical Referral Work 2019
Using two genetic crosses, you will investigate the inheritance of three genes. As with the lab practicals, you will examine a two factor cross (Cross A) and a single factor cross (Cross B). For each cross, all of the flies in the parental generation are either homozygous or hemizygous; i.e. there are no heterozygous parents.
In the two factor cross, one gene influences the colour of the fly abdomen and the mutant phenotype has a dark abdomen compared to the wild type. The second gene in the two factor cross influences the size of bristles on the fly body. The mutant here has stubble bristles compared to the wild type.
In the single factor cross, you will look at a gene involved in eye pigment colour. The mutant phenotype produces flies with apricot coloured eyes. The referral work involves no experimental work – you are only required to analyse the data that you have been presented with.
NOTE: c2 tests cannot be performed with an expected value of 0. If you have devised a null hypothesis which produces a class with an expected value of 0, you will have to revise your null hypothesis.
Two factor cross
- Dark abdomen, stubble bristle female x wild type male
Single factor cross
- Wild-type female x apricot eye male
SFB1003 Report of results for genetic crosses with Drosophila
Marks will be allocated as indicated, with emphasis on clear logical reasoning and explanations!
Cross A: Dark abdomen, stubble bristle female x wild type male
Observed results
Phenotypes of F1 generation
wild type
| normal bristle, dark abdomen,
| stubble bristle, normal colour | stubble bristle, dark abdomen | |
Male | 15 | 0 | 0 | 0 |
Female | 25 | 0 | 0 | 0 |
Total | 40 | 0 | 0 | 0 |
Phenotypes of F2 generation – Individual group results
wild type (normal bristle, normal colour) | normal bristle, dark abdomen,
| stubble bristle, normal colour | stubble bristle, dark abdomen | |
Male | 23 | 0 | 0 | 3 |
Female | 12 | 0 | 0 | 2 |
Total | 35 | 0 | 0 | 5 |
Phenotypes of F2 generation – Class results
wild type (normal bristle, normal colour) | normal bristle, dark abdomen,
| stubble bristle, normal colour | stubble bristle, dark abdomen | |
Male | 293 | 0 | 0 | 104 |
Female | 323 | 0 | 0 | 80 |
Total | 616 | 0 | 0 | 184 |
On the basis of these results explain what you think the genetic basis of these results is.
- a) Choose symbols for the genes, and their alleles, defining what you have chosen.
- b) Draw up a full crossing scheme showing parents, gametes, F1, gametes from F1
- c) do a Punnett square for the F2.
- d) From this Punnett square derive the ratios expected of the possible phenotypes in the F2.
[20 marks]
c2 test of goodness of fit of observed F2 phenotype ratios to expected
You predicted a certain ratio of phenotypes in the section above. Use this prediction to work out what numbers of each type of fly you expected in your cross. Compare this to the observed number, using the c2 test. You may use the individual group results or those of the class.
First construct a ‘Null hypothesis’: ‘That …….
Insert your c2 table here:
Degrees of freedom (df) (number of different classes – 1):
Critical value of c2 for a probability of 5% and the df above:
Accept/reject (delete as appropriate) null hypothesis because:
What do you conclude from the c2 test?
[20 marks]
Further work
Suggest other experiments you would like to do, to confirm your conclusions
[10 marks]
Cross B: Wild-type female x apricot eye male
Observed results
Phenotypes of F1 generation
wild type eye | apricot eye | |
Male | 37 | 0 |
Female | 23 | 0 |
Total | 60 | 0 |
Phenotypes of F2 generation
wild type eye | apricot eye | |
Male | 14 | 16 |
Female | 30 | 0 |
Total | 44 | 16 |
On the basis of these results explain whether you think the apricot eye allele is dominant or recessive to wild type and whether it is sex-linked or autosomal. Give your reasoning!
[10 marks]
On the basis of these results explain what you think the genetic basis of these results is.
- a) Choose symbols for the genes, and their alleles, defining what you have chosen.
- b) Draw up a full crossing scheme showing parents, gametes, F1, gametes from F1
- c) do a Punnett square for the F2.
- d) From this Punnett square derive the ratios expected of the different phenotypes in the F2.
[20 marks]
c2 test of goodness of fit of observed F2 phenotype ratios to expected
You predicted a certain ratio of apricot and wild type phenotypes in the section above. Use this prediction to work out what numbers of each type of fly you expected in your cross. Compare this to the observed number, using the c2 test
First construct a ‘Null hypothesis’: ‘That …….
Insert your c2 table here:
Degrees of freedom (df) (number of different classes – 1):
Critical value of c2 for a probability of 5% and the df above:
Accept/reject (delete as appropriate) null hypothesis because:
What do you conclude from the c2 test?
[20 marks]