principles of genetics

 Part A: Questions from the Drosophila melanogaster cross practical
1). In the first cross wild-type (wt) female flies and male flies with vestigial wings (vg) were
crossed. Our class results for the F2 generation of the Drosophila melanogaster cross are the
follows:
F2 Males Females
Normal wings 592 691
Vestigial wings 162 120
In the beginning I checked the possibility of vestigial or long wings alleles location on the X
chromosome. In this case we can write down crosses results in the next view.
P: Xwt Xwt x XvgY
F1: Xwt Xvg x XwtY
F2:
If vestigial wings were X-linked recessive trait, we would have only normal winged female in F2
generation. If vestigial wings were X-linked dominant trait, we would have equal number of flies
of four phenotypic classes in F2 generation. But in accordance with our class results, females
have normal and vestigial wings. There are smaller number of vestigial winged flies than normal
winged flies. Thus, vg and long wings alleles are not X-linked and are located on autosomes.
2) To investigate if the trait is X-linked, reciprocal cross might be carried out. In the case of Xlinked nature of vestigial wings gene, this trait would no behave equally in reciprocal crosses:
Xwt Y
Xwt Xwt Xwt
normal winged female
Xwt Y
normal winged male
Xvg Xwt Xvg
normal winged female or
vestigial winged female
Xvg Y
vestigial winged male
♀♂
♀
♀
♂
♂
DO MY ASSIGNMENT SUBMIT
WWW.ASSIGNMENTEXPERT.COM
Sample: Biology - Principles of Genetics
1
 Reciprocal cross
P: Xwt Xwt x XvgY P: Xvg Xvg x XwtY
F1: Xwt Xvg , XwtY F1: Xwt Xvg , XvgY
 normal winged flies normal vestigial
Male flies with vestigial wings will appear in reciprocal cross. But if the vestigial wings were
coded by autosomal gene, the results of reciprocal crosses would equal and all flies would have
normal wings
Reciprocal cross
P: wt wt x vgvg P: vg vg x wt wt
F1: wt vg F1: wt vg
 normal winged flies normal winged flies
3) We defined the autosomal location of our gene in previous step
P: wt wt x vg vg
F1: wt vg x wt vg
F2:
We can observe two phenotypic classes in the F2 progeny: flies with normal and rudimentary
(vestigial) wings. Thus, two alleles of one gene are involved in the cross. Due to smaller quantity
of flies with vestigial wings we can conclude that vestigial wings is a recessive trait and normal
wings is a dominant trait. In this case segregation ratio 3:1 should be observed in F2 progeny,
with 3 normal winged flies : 1 vestigial winged flies
wt vg
wt wt wt
normal winged flies
wt vg
normal winged flies
vg wt vg
normal winged flies
vg vg
vestigial winged flies
♂
♂
♂
♂
♂ ♂
♀
♀ ♀
♀
♀ ♀
♀
♀
♀♂
♂
♂
DO MY ASSIGNMENT SUBMIT
WWW.ASSIGNMENTEXPERT.COM
2
4) Our null hypothesis is: in the F2 progeny we will observe 3:1 phenotypic distribution (3
normal winged flies : 1 vestigial winged flies)
Alternative hypothesis: In the F2 progeny we will not observe 3:1 phenotypic distribution (3
normal winged flies : 1 vestigial winged flies)
Chi square test will help us to accept or reject the null hypothesis.
To calculate the chi squared value we will use the next formula (O-E)2
/E, where O is the
observed number of flies in each category, and E is the expected number of flies.
Observed Expected (O-E)2
/ E
Normal winged flies 592 + 691 = 1283 1565/4*3 = 1173,75 10.17
Vestigial winged flies 162 + 120 = 282 1565/4 = 391,25 30.51
Total 1565 1565 40.68
We have two phenotypic classes, giving us 1 degree of freedom. The total chi squared value of
the cross is too large.
I found significance level for chi squared value of normal winged flies (10.17). Entering
the Chi square distribution table with 1 degree of freedom and reading along the row I found our
value of x2 (10.17) lies between 6.635 and 10.827. The corresponding probability is
0.01






Enjoy big discounts
Get 20% discount on your first order



ORDER NOW