BICD100 GENETICS
Prof. Chisholm
MIDTERM 1
Approximate grade scheme:
>222 = A
168.5222 = B
114168 = C
59113 = D
058 = F
(the other midterms and final will follow similar grade curves)
Descriptive statistics (as of 2/1/08)
Minimum
25.0
25% Percentile
132.875
Median
167.75
75% Percentile
210.0
Maximum
278.0
Mean of 5 best scores:
264.4
Mean
168.17
Std. Deviation
54.206
Midterm 1 distribution
2
0
4
6
8
1
0
10
20
30
40
Bin center
number of exams
BICD 100 GENETICS Winter 2008 MIDTERM 1 KEY
1
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View Full DocumentQuestion 1
A highly prized breed of chickens has a mixture of light and dark feathers, a plumage
known as ‘Erminette’.
In 1964 Frederick Hutt decided to investigate the inheritance of
the Erminette phenotype.
When male and female Erminette chickens were crossed,
their F1 progeny consisted of 130 with Erminette plumage, 62 pure white plumage, and
64 pure black plumage.
(a) Assuming that this trait is determined by a single gene, formulate a hypothesis for
the genetic basis of the Erminette phenotype.
Calculate the chisquared statistic for
the observed data to test your hypothesis. (20 points)
The simplest model is that the Erminette phenotype results from incomplete
dominance.
Denote the Dark allele as A and the light as a.
If Erminette is Aa,
then the cross Aa x Aa will give AA:Aa:aa in 1:2:1 ratio.
The Expected numbers
based on this model are 64:128:64.
!
2
= (4/128)+ (4/64) = 0.093.
(Credit for
saying codominance, which is a viable interpretation)
10 points for the ratio and expected numbers.
10 points for
!
2
.
(b) What is the approximate P value of the chisquared?
Should you accept or reject
your hypothesis, using a signiﬁcance level or critical value of 5%? Give the degrees
of freedom in your calculation.
(10 points)
There are three classes so two
degrees of freedom, so 0.975 <
!
2
< 0.9.
The critical value of
!
2
is 5.99.
The
hypothesis cannot be rejected (‘accept hypothesis’ ﬁne).
(c) A geneticist objected that Hutt’s results might also be explained by a twogene
model, in which the white phenotype is epistatic to black or Erminette.
What
phenotypic ratio does this hypothesis predict?
Calculate a chisquared value for the
observations given the twogene model and comment on the relative merits of the
one gene vs two gene models. (20 points).
An epistasis model would predict a
9:3:4 ratio of Erminette: black: white.
The Erminette phenotype would be due
to dominant alleles at two loci (A_B_).
Black would be A_bb.
aa__ would be
white.
The expected values would be 144:48:64 and the chi squared would be
(196/144) +(196/48) = 5.44.
Given a critical value of 5.99 we cannot reject the
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 Winter '08
 Nehring
 Genetics, Sex linkage, Arthur Rr, Erminette

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