Cytogenetics Questions and Answers – Numerical Problems

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This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Numerical Problems”.

1. What is the probability that Aa BB Cc will form from the cross between Aa Bb Cc X Aa Bb Cc?
a) 1/4
b) 1/8
c) 1/16
d) 1/32
View Answer

Answer: c
Explanation: In this case we have to consider one allele pair at a time. Probability of a heterozygote from F1 cross is ½ and homozygote is ¼. Thus here it will be 1/2X1/4X1/2= 1/16 for Aa BB and Cc respectively.
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2. In F2 progeny you get 1:1 ratio represents two phenotypes, what will be the genotype of the parent?
a) Both parent heterozygous
b) Cross between homozygous dominant and recessive
c) Cross between homozygous recessive and heterozygous
d) Both parent homozygous recessive
View Answer

Answer: c
Explanation: Considering the two phenotypes having genotype Hh and hh (H being dominant), the f1 must have phenotype Hh and hh, and so will the parent also have a phenotype Hn and hh. This is like a repeated test cross.

3. In an experiment with snapdragon flowers you have 10 F1 flowers that are white. What will be the number of red flowers?
a) 20
b) 30
c) 40
d) 10
View Answer

Answer: d
Explanation: According to normal dominant/ recessive relation the ratio between red and white flowers should be 3:1, but in the case of snapdragon where the heterozygous has a different phenotype pink the ratio changes to 1:2:1 for red: pink: white. Thus the expected number of red is 10.
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4. What is the probability of getting a recessive phenotype for all three gene locus for the cross Aa Bb Cc X aa Bb cc?
a) 1/4
b) 1/16
c) 1/2
d) 1/8
View Answer

Answer: b
Explanation: In this case, the probability of each of the locus producing a recessive allele is multiplied. That would be ½ X ¼ X ½ for A, B and C locus respectively.

5. In case of rats the homozygous yellow allele is lethal. If you cross two yellow pats, what will be the ratio of white and yellow rats?
a) 1:2:1
b) 3:1
c) 2:1
d) 1:1
View Answer

Answer: c
Explanation: Homozygous yellow rat YY is lethal, so the parent must be heterozygous Yy. Now, the living progenies will be Yy and yy i.e. yellow and white in the ratio 2:1 and YY will be lethal.
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6. Which of the following crosses can never result in a recessive phenotype?
a) Tt Pp X Tt Pp
b) TT Pp X tt pp
c) tt pp X TT PP
d) tt pp X Tt Pp
View Answer

Answer: c
Explanation: In case of this cross although one is homozygous recessive the other is homozygous dominant, so all the progeny will be heterozygous expressing the dominant trait. In case of TT Pp X tt pp the recessive trait can be seen in case of P locus.

7. You know that H is dominant for height locus and H is recessive. Is the F2 progeny are a mixture of 3:1 dominant: recessive and you pick a dominant plant at random what will be the probability that the F3 will have a mixture of dominant and recessive traits as well?
a) All will show
b) 1/16
c) 1/4
d) 2/3
View Answer

Answer: d
Explanation: Here we are actually selecting the plants which are heterozygous for the trait as in Hh. Only selfing these will give both dominant and recessive trait in next generation. As we are not taking hh into consideration the ratio resulting is 2/3.
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8. Two black female mice are coupled with a brown male mouse. If female A produces 4 black and 3 brown children and female B in its litter produces 10 black children, what is the genotype of A, B and male mouse respectively?
a) BB Bb and bb
b) Bb Bb and Bb
c) Bb BB and bb
d) bb Bb and BB
View Answer

Answer: c
Explanation: Brown mouse is homozygous recessive so this is sort of a test cross for the female mice. As mouse A has both black and brown progeny, it must be heterozygous. And within his litter size, we can say that B is expected to be homozygous BB.

9. If a mother is a non tongue roller and Father is a tongue roller, whose father was a non- roller, what is the probability of having a roller daughter?
a) 1/16
b) 1/2
c) 1/4
d) 1/8
View Answer

Answer: c
Explanation: In this case the father is heterozygous as Rr and mother is homozygous recessive rr. Considering the probability of a girl child being ½ we have the probability of a roller as ½. resulted is the product of these two probabilities ½ x ½ = ¼.
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10. Consider a dyhybrid cross of AaBb with AaBb, how many of the offsprings will breed true?
a) 1/16
b) 2/16
c) 6/16
d) 4/16
View Answer

Answer: d
Explanation: In this case, all of AA BB, AA bb, aa BB and aa bb will breed true. This makes 4 out of 16 being true breeding. In a Punett square these are along the diagonals.

11. In a dihybrid cross which of the following will be of the least occurrence?
a) All will occur equally
b) Pure breeding
c) Homozygous recessive
d) Heterozygous dominant
View Answer

Answer: c
Explanation: According to the dihybrid phenotypic ratio of 9:3:3:1 the 1/16th of the probability is for homozygous recessive phenotype, making it the least occurring one.

12. What is the probability of getting a Red flower by crossing a pink and white snapdragon flower?
a) 1/4
b) 2/4
c) 2/3
d) None
View Answer

Answer: d
Explanation: In snapdragons the heterozygote is pink. Crossing a heterozygote with a homozygous recessive will never result in a homozygous dominant progeny to express red colouration.

13. In bats the progeny with Qq die. If you cross two wild type bats which are homozygous for dominant and recessive form of the trait, what will be the expected ratio of the F1?
a) All heterozygote
b) 3:1
c) 1:2:1
d) No ratio
View Answer

Answer: d
Explanation: Cross between QQ and qq will always result in Qq which is lethal. This prevents generation of any heterozygote or rather any progeny by mating of these two bats.

14. What is the probability of getting Aa Bb Cc from the cross aa BB cc X AA bb CC?
a) 1/4
b) 1/8
c) 1
d) 4
View Answer

Answer: c
Explanation: In this case, all three locus is homozygous dominant for one and homozygous recessive for the other. Thus all the gametes are heterozygous. So considering the calculation it will be 1 X 1 X 1 for A, B and C locus respectively.

Sanfoundry Global Education & Learning Series – Cytogenetics.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter