1.The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?
1.aabbccdd
2.AaBbCcDd
3AABBCCDD
4.AaBBccDd
5.AaBBCCdd
a. aabbccdd = 1/4*1/4*1/4*1/4=1/256
b. AaBbCcDd = 1/2*1/2*1/2*1/2= 1/16
c. AABBCCDD = 1/4*1/4*1/4*1/4 = 1/256
d. AaBBccDd = 1/2*1/4*1/4*1/2 = 1/64
e. AaBBCCdd = 1/2*1/4*1/4*1/4 = 1/128
Probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4.
The chance that a son will be color-blind is also 1/2.
9.The genetic map of three genes in Drosophila melanogaster is given below:
a ----------------------b-----------c
10cm 5cm
A cross as given below is made between individuals of the genotypes:
a+--------b+----------- c a----------b------------c+
a+--------b+-----------c X a ---------b-------------c+
The female F1progeny are test-crossed and1000 progeny are obtained. Assuming that there has been no double crossover, what is the expected number of progeny with the genotypes:
A) a+--------b---------- c+ B)a+--------b+----------- c+ C)a+--------b+----------- c
a ---------b-----------c a ---------b-----------c a ---------b-----------c
set (A) B C
a. 100 50 850
b. 50 25 425
c. 100 850 50
d. 0 425 75
Ans. F1 test cross: a+b+c/ a b c+ x a b c/ a b c
(A) The proportion a+ b c+ will be 10/2 % = 5%. In f2 there are 1000 individuals. So the proportion of one having a+ b c+/ a b c will be 5% of 1000=50.
(B) The proportion a+ b+ c+ will be 5/2 % =2.5%.In f2 there are 1000 individuals. So the proportion of one having a+ b+ c+/ a b c will be 25.
(C) Proportion of total parental variety (a+ b+ c and a b c+) = Total progeny- (Proportion of recombinants produced as a result of SCO between a and b gene + Proportion of recombinants produced as a result of SCO between b and c gene)
Total Parental proportion = 1000 - [(10% of1000) + (5% of 1000)]
= 1000 - 150 = 850.a+ b+ c/ a b c is one of the parental variety,so its number will be 850/2 = 425.So the answer will be option B. A 50, B 25 and C 425.
10. In doing a chi-square test on the results of a trihybrid cross, the number of degrees of freedom would be
A. 8 B. 7 C. 4 D. 3 Ans. B (n-1) =8-1=7 11.If two people who are both heterozygous for Huntington's disease (an autosomal dominant trait) marry, what is the probability that they will have three children, all of which are normal?
15.A recessive inherited disease is expressed only in individuals of blood group O and not expressed in blood groups A B and AB. Alleles controlling the disease and blood groups are independently inherited. A normal woman with blood group A and her normal husband with blood group B already had one child with the disease . The woman is pregnant for second time . What is the probability that the second child will also have the disease ?
Ans. 1/2*1/2*1/4 =1/16
Explanation (Recessive allele probability =1/4 normal allele probability 1/2 )
Here given parents are normal that means probability is 1/2*1/2=1/4
The child is recessive means its probability is 1/4 . So, 1/2*1/2*1/4 =1/16
16.Two varieties of maize averaging 48 and 72 inches in 42 are crossed. The F1 progeny is quite uniform averaging 60 inches in height. Of the 500 F 2 plants, with shortest 2 are 48 inches and the tallest 2 are 72 inches. What is the probable number of polygenes involved in this trait.
Ans .4 genes are inovled because 1/4^n=1/250
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1.aabbccdd
2.AaBbCcDd
3AABBCCDD
4.AaBBccDd
5.AaBBCCdd
a. aabbccdd = 1/4*1/4*1/4*1/4=1/256
b. AaBbCcDd = 1/2*1/2*1/2*1/2= 1/16
c. AABBCCDD = 1/4*1/4*1/4*1/4 = 1/256
d. AaBBccDd = 1/2*1/4*1/4*1/2 = 1/64
e. AaBBCCdd = 1/2*1/4*1/4*1/4 = 1/128
Probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4.
2. The possible genotypes of endosperms borne on a heterozygous (Rr)plant will
be -RRR,RRr,rrR,rrr
4.A rare x-linked allele is present in the human population at a frequency of 1 in
10^5 .Among those who carry the allele ,the proportion of males is about -
a.1 in 2 c.1in 4
b.1 in 3 d.1 in 5
Ans. If the X-linked allele is present at a frequency of 1/10000, then if the frequency of the allele is given by q, then: q = 1/10000 = 0.0001 p + q = 1 ,p = 1 - 0.0001 ,p = 0.9999 So, the heterozygotes=2pq = 2 x 0.9999 x 0.0001 = 0.00019998 The recessive homozygotes =q^2 = 0.0001 x 0.0001 = 0.00000001
Since the frequency of males with the allele is equal to the frequency of the X-linked allele in the population, then the frequency of males in the population with the allele is 0.0001
So, the overall frequency of persons with at least one allele is:
0.0001 (hemizygotes, all males) + 0.00000001 (recessive homozygotes, all female) + 0.00019998 (heterozygotes, all female) = 0.00029999
So, the proportion of males with the allele out of all person having at least one allele is 0.0001/0.0002999, which is approximately 1/3, so the answer is B.
6.The likelihood of an individual in a population carrrying two specific alleles of a human Dna marker ,each of which has a frequency of 0.2,will be a.0.4 b.0.32 c. 0.16 d.0.08So, the overall frequency of persons with at least one allele is:
0.0001 (hemizygotes, all males) + 0.00000001 (recessive homozygotes, all female) + 0.00019998 (heterozygotes, all female) = 0.00029999
So, the proportion of males with the allele out of all person having at least one allele is 0.0001/0.0002999, which is approximately 1/3, so the answer is B.
Ans.use square .2 twice and add them together to get the combined likelihood
(.2)^2 + (.2)^2 = .08
8. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. Man XbY Woman XBXb. We know this because she has normal vision (XB). Her father was color-blind and gave her her Xb allele.
A) What is the probability that their daughter will be color-blind?The chance that a son will be color-blind is also 1/2.
9.The genetic map of three genes in Drosophila melanogaster is given below:
a ----------------------b-----------c
10cm 5cm
A cross as given below is made between individuals of the genotypes:
a+--------b+----------- c a----------b------------c+
a+--------b+-----------c X a ---------b-------------c+
The female F1progeny are test-crossed and1000 progeny are obtained. Assuming that there has been no double crossover, what is the expected number of progeny with the genotypes:
A) a+--------b---------- c+ B)a+--------b+----------- c+ C)a+--------b+----------- c
a ---------b-----------c a ---------b-----------c a ---------b-----------c
set (A) B C
a. 100 50 850
b. 50 25 425
c. 100 850 50
d. 0 425 75
Ans. F1 test cross: a+b+c/ a b c+ x a b c/ a b c
(A) The proportion a+ b c+ will be 10/2 % = 5%. In f2 there are 1000 individuals. So the proportion of one having a+ b c+/ a b c will be 5% of 1000=50.
(B) The proportion a+ b+ c+ will be 5/2 % =2.5%.In f2 there are 1000 individuals. So the proportion of one having a+ b+ c+/ a b c will be 25.
(C) Proportion of total parental variety (a+ b+ c and a b c+) = Total progeny- (Proportion of recombinants produced as a result of SCO between a and b gene + Proportion of recombinants produced as a result of SCO between b and c gene)
Total Parental proportion = 1000 - [(10% of1000) + (5% of 1000)]
= 1000 - 150 = 850.a+ b+ c/ a b c is one of the parental variety,so its number will be 850/2 = 425.So the answer will be option B. A 50, B 25 and C 425.
10. In doing a chi-square test on the results of a trihybrid cross, the number of degrees of freedom would be
A. 0.140 B. 0.422 C. 0.250 D. 0.016 Ans. D
12.Given an individual who is heterozygous at 5 loci, how many different gametic genotypes are possible?
A. 25 B. 10 C. 15 D. 32 Ans. 32 (n=5 formula is 2^n=2^5)
13.In a family of 3 children what is the propbability that all 3 are boys?
Ans.1/2*1/2*1/2=1/8
14. In a cross between AABBCcDdEe* AaBbCcDdEe,what is the propability that the offspring will be AABbccddEE ?
AAXAa=1/2, BBXBb=1/2 ,DdxDd=1/4 ,EexEe=1/4
Ans. 1/2x1/2x1/4x1/4x1/4=1/512AAXAa=1/2, BBXBb=1/2 ,DdxDd=1/4 ,EexEe=1/4
15.A recessive inherited disease is expressed only in individuals of blood group O and not expressed in blood groups A B and AB. Alleles controlling the disease and blood groups are independently inherited. A normal woman with blood group A and her normal husband with blood group B already had one child with the disease . The woman is pregnant for second time . What is the probability that the second child will also have the disease ?
Ans. 1/2*1/2*1/4 =1/16
Explanation (Recessive allele probability =1/4 normal allele probability 1/2 )
Here given parents are normal that means probability is 1/2*1/2=1/4
The child is recessive means its probability is 1/4 . So, 1/2*1/2*1/4 =1/16
16.Two varieties of maize averaging 48 and 72 inches in 42 are crossed. The F1 progeny is quite uniform averaging 60 inches in height. Of the 500 F 2 plants, with shortest 2 are 48 inches and the tallest 2 are 72 inches. What is the probable number of polygenes involved in this trait.
Ans .4 genes are inovled because 1/4^n=1/250
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