1.Calculate the chi-square value forthe following data set: phenotypic class AB
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Question
1.Calculate the chi-square value forthe following data set:
phenotypic class
AB
ab
Ab
aB
Observed
43
55
48
54
Expected
50
50
50
50
a.1.88
b.0.72
c.1.09
d.0.36
e.1.97
2.A population consisting of 1876 AA individuals, 1793 Aa individuals, and 5640 aa individuals is established on a remote island. The allele frequencies in this population are:
a. p=0.4, q=0.6
b.p=0.2, q=0.8
c.p^2=0.09, 2pq=0.21,q^2=0.49
d.p=0.3, q=0.7
e.p^2=0.04, 2pq=0.16, q^2=0.64
3.White wool in sheep is controlled by a dominant allele R, and black wool by th recessive allele r. In an isolated population of 6530 sheep, 514 are black, 4981 are heterozygous, 1035 are homozygous white. When will this population be expected to reach Hardy-Weinberg equilibrium?
a.in the next generation
b.in two generations
c.it already is in equlilbrium
d.impossible to tell
e. it will not reach equilibrium
4.A population has the following genotypic disturbutions:
A^M A^M=1787, A^M A^O=2011, A^OA^O=2331. IS THIS POPULATION AT H-W EQUILIBRIUM
a.YES, with allel frequencis pf p=0.45562 and q=0.54437
b.yes, with allele frequencies of p=0.54437 and q=0.37862
c.no, with allele frequencies of p=0.45562 and q=0.37862
d.no, with expected genotypic ratios of p^2=1272, 2pq=3040, and q^2=1826
e.yes, with xpectd genotypic frequencies of p^2=1778, 2pq=2015, and q^2=2336
5.
From the following agarose gel electrophoresis result, which of the DNA fragments numbered 1 to 5, is the smallest fragment?
- _________ 1
__________2
__________3
_________4
_________5
a.1
b.2
c.3
d.4
e.5
6.Which of the following is NOT for DNA sequencing?
a.a primer
b.DNA polymerase
c.nucleotides
d.restriction endonucleases
e.fluorescent or radioactive labels
Explanation / Answer
1). a. 1.88
Formula for chi-square test = Sum of (observed value - expected value)2 / Expected value.
Chi-square value = (43-50)2 / 50 + (55-50)2/ 50 + (48-50)2 / 50 + (54-50)2/ 50
= 0.98+ 0.5+ 0.08 + 0.32 = 1.88.
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