(3) You are interested in understanding the genetics of coat color and hair leng
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Question
(3) You are interested in understanding the genetics of coat color and hair length in dachshunds, which have one of 2 primary colors (brown and black) and either long or short hair. You cross a pure-bred black long-hair dachshund with a brown short-hair dachshund, and all of the offspring are black with short hair. When you intercross these offspring, you obtain the following phenotypic ratios: 44 black short hair dachshunds 23 black long hair dachshunds 21 brown short hair dachshunds (a) Perform a statistical test to determine if these data support a hypothesis of Mendelian inheritance based on independent assortment. (4 points) (b) If you reject the null hypothesis of Mendelian inheritance, propose an alternative hypothesis. How specific can you be in your hypothesis? Explain the limits to the conclusions you can make given these data alone. (3 points) (c) Upon doing research, you learn that a locus controlling coat color and a locus controlling hair length are linked with a map distance of 8 cM. Predict the phenotypic ratios that will result from a testeross of the Fl generation (produced above) (3 points) (d) Given these data, predict the phenotypic ratios expected from an intercross of the Fl generation (from above).(4 points) (e) Perform a statistical test to determine if the data provided earlier (44 black & short; 23 black & long; 21 brown& short) support this hypothesis based on linkage. (4 points)Explanation / Answer
Hi,
Let's assume that the coat color is encoded by a gene B. It has 2 varients, B gives black and 'b' gives brown. Similarly, the hair is encoded by gene H, with two varients H coding fro long hair and 'h' coding for small hair. Let us assume that we are crossing pure bred lines. So black long hair = BBHH, and brown short-hair = bbhh.
BBHH x bbhh
gametes = BH bh
SO, F1 = BbHh --> all black long haired puppies.
F2 : BbHh x BbHh
gametes: BH, bH, Bh, bh x BH, bH, Bh, bh
so we get puppies in 9:3:3:1 ratio of black long: black short: brown long: brown short.
But these are not reflected in the actual results (44:23:21:0 = 2:1:1:0). So the mendelism cannot explain this anamoly.
a.
Let us perform statistical analysis to check whether the ratios perform mendelism.
Test: Chi square test
X2 = [expected-observed]^2 / exected
= [9-2]^2/9 + [3-1]^2/3 + [3-1]^2/3 + [1-0]^2/1
= 5.44 + 1.33 + 1.33 + 1
= 9
The degrees of freedom = n - 1
n = no. of parameters
so DF = 4 - 1 = 3
The chi square table at p= 0.05, at DF = 3 is less than the value 9
so we reject the null hypothesis of mendelism.
b. These results can be explained if we assume that the genes lie on same chromsome. Or the recombination is more than what is expeted. Using the available results, we cannot determine with certainity. The linkage of genes can result increase the parental combinations and reduce the recombinants.
c. The genes are linked with a distane of 8cM, this means 8% of progeny will be recombinants and rest will be parental. When F1 is test crossed, BbHh x bbhh, we get
Bh//bH x bh//bh
Bhbh - black short - 46
bHbh - brown long - 46
BHbh - black long - 4
bhbh - brown short - 4
gametes = BH, Bh, bH, bh x bh
BbHh - parental = 46 Blck long hair
Bbhh - recombinant = 4 Black short hair
bbHh -recombinant=4 brown long hair
bbhh - parental =46 brown short
d. Intercourse of F1 =
F1 intercourse: Bh//bH x Bh//bH
Gametes : Bh, bH x Bh,bH without recombination ( parental)
BBhh, black short = 23
BbHh black long = 23
bBHh black long = 23
bbHH brown long = 23
F1 intercourse: Bh//bH Bh//bH
Gametes : BH,bh x BH x bh with recombination
BBHH black long, = 2
BbHh - black long, = 2
BbHh - black long, =2
bbhh - brown short = 2
In total: black long = 48; blackshort= 23; brown long = 23 and brown short = 2
These do not follow as in the given exmpple. This cannot be explained based on linkage.
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