Cytogenetics Questions and Answers – Sex Linked Inheritance

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This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Sex Linked Inheritance”.

1. All genes in sex chromosomes are sex linked.
a) True
b) False
View Answer

Answer: b
Explanation: There is a homologous region in the sex chromosome X and Y. Thus genes present there shows normal inheritance pattern.
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2. An X linked recessive gene would appear to __________________
a) Be expressed in both males and females equally
b) Skip generations
c) Be lethal
d) Gradually degrade
View Answer

Answer: b
Explanation: Sex linked genes especially the sex linked recessive gene skips generations before expression. It is expressed more in males than female and is not necessarily lethal.

3. A Y linked gene ___________________
a) Is expressed only when homozygous
b) Is present near the telomere of long arm in human
c) Is carried by mother
d) Expressed only in men
View Answer

Answer: d
Explanation: A Y linked gene is expressed only in men and never in females as females lack a Y chromosome. It is not expressed in homozygous, rather in hemizygous for as its present only in a single copy. The region near that telomere is common for both sex chromosomes, so it is not sex linked.
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4. Genes which are present in the homologous region of X and Y chromosomes are called______________
a) Autosomal
b) Sex linked
c) Partially sex linked
d) Unlinked
View Answer

Answer: c
Explanation: Genes present in the homologous region of X and Y chromosomes are known as partially sex linked. Although they are present on sex chromosome their inheritance pattern is like normal genes.

5. Choose the odd one out.
a) Color blindness
b) Haemophilia
c) Duchenne muscular dystrophy
d) Faulty tooth enamel
View Answer

Answer: d
Explanation: While the first three diseases are X linked recessive diseases, the last i.e. faulty tooth enamel and decolonization is a dominant trait.
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6. X linked dominant trait tends to be milder in heterozygous females. State if it is true or false.
a) True
b) False
View Answer

Answer: a
Explanation: As females have two copies of X gene, if only one of them is mutant there is a 50% chance of the mutant X gene being turned off in a cell due to dosage compensation. Thus, the effect of the mutant gene product is diluted.

7. In a family of four including a normal mother, a normal father, a color blind son and a normal son; who do you think has the defective X gene other than the affected son?
a) The mother
b) The father
c) The normal son
d) The effected son only
View Answer

Answer: a
Explanation: Males have only one X chromosome, so if the father or normal son carried the defective X gene they would show the phenotypic effect. Since the female has two copies of X chromosome, only in homozygous form the recessive gene like colorblindness is expressed. So mother here is a carrier.
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8. A colorblind mother and a normal father give birth to a Klinefelter child with normal vision. Where do you think the non-disjunction has taken place?
a) 1st meiotic division of mother
b) 2nd meiotic division of the father
c) 2nd meiotic division of mother
d) 1st meiotic division of father
View Answer

Answer: d
Explanation: The non disjunction must be in the male parent as the child was not color blind. This is possible only if one X chromosome was normal which here came from father. Also as the disjunction produced XY and not two copies of same chromosome it must have been in 1st division.

9. If colorblindness was a dominant trait, which of the following would not be true?
a) It would be expressed in heterozygous females
b) It would be expressed in males
c) It would never be turned off
d) It would not skip generations
View Answer

Answer: c
Explanation: Due to dosage compensation X chromosomes are randomly turned off, so the mutant X could as well be turned off in heterozygote. Also it will affect every generation as it would be dominant. Option a and b are also true.
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10. What is the probability that the son of a color blind father would be color blind?
a) 0
b) 1/4
c) 1/2
d) 1
View Answer

Answer: a
Explanation: Sons get the Y chromosome from father and the X from mother. So unless the mother is also a carrier, there is no probability of the son being colorblind.

11. Y linked inheritance is _______________ inheritance.
a) Criss cross
b) Straight
c) Loop
d) Jumping
View Answer

Answer: b
Explanation: The inheritance pattern of Y linked gene is straight as it moves from father to son. Only males can inherit these traits.

12. Which of the following is wrong?
a) Y chromosome lacks dosage compensation
b) X linked genes are inherited as criss cross
c) Y linked gene like haemophilia passes from father to son
d) X linked recessive genes are carried by females
View Answer

Answer: c
Explanation: It is true that Y linked genes pass from father to son, but Haemophilia is not a Y linked gene. It is a recessive X linked gene.

13. Which of the following is a Y linked gene?
a) Faulty dental enamel
b) Haemophilia
c) Hypertrichosis
d) Phenyl ketone urea
View Answer

Answer: c
Explanation: Among the following options only hypertrichosis is a Y linked gene. It is the presence of hair in the ear. It follows the Y linked rule of inheritance.

14. Which of the following is partially sex linked?
a) Complete colorblindness
b) Hypertrichosis
c) Faulty tooth enamel
d) Haemophilia
View Answer

Answer: a
Explanation: While red green colorblindness is an X linked recessive trait. Complete colorblindness however is a partially sex linked gene in man.

15. A turner child born to normal parents is haemophilic. Where did the non-disjunction occur?
a) Mother
b) Father
c) Zygote
d) Grand mother
View Answer

Answer: b
Explanation: The only possibility here is that the mother was a carrier. The egg received this defective X chromosome from mother and no sex chromosome from the father. Thus, there was non disjunction in the father. The resultant child was AA+ X’O. (X’=mutant).

Sanfoundry Global Education & Learning Series – Cytogenetics.

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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