CBSE Questions for Class 12 Medical Biology Principles Of Inheritance And Variation Quiz 14 - MCQExams.com

A) Correct option (D)
B) Explanation of the correct options

• According to pedigree, offspring of normal parents show the trait which means that the trait is recessively present in the normal parents; the trait is recessive.
• In a first filial generation, both son and daughter show the trait.
• As we know that a sex-linked recessive trait can be expressed in either sex, not in both; the X-linked trait is expressed in male progeny while the female progeny serves as a carrier and the Y-linked trait is expressed in sons only.
• The presence of trait in both son and daughter in $$F_1$$ generation confirms that it is an autosomal trait. 
• The second parent generation must have a carrier parent and a recessive homozygous parent to pass it to 1/3rd of progeny.
• Phenylketonuria is an autosomal recessive trait.

A. Correct Option: D

B. Explanation for the correct option:

• Mutation refers to the sudden change in a DNA sequence alerting heritable change in individual phenotype.
•  Mutation in genes occurs due to various physical such as mistakes committed by genes while crossing over or chemical reasons such as exposure to harmful radiation.  
• It also occurs in the chromosome causing change in the structure or arrangement of the chromosome like abnormal numbers of individual chromosomes or chromosome sets. 

• Transposons are the genetic elements that can jump from one location to other in the genome. Also called jumping genes, they were first discovered by Barbara McClintok while working on chromosome breakage in maize. Statement A is correct. 

Sickle cell anaemia is caused by a missense mutation in the Hb$$^b$$ gene that codes for $$\beta$$ chain. Replacement of A by T at the 17th nucleotide of the Hb$$^b$$ gene changes the codon GAG (glutamic acid) to GTG (which encodes valine). The mutated allele Hb$$^s$$ encodes the abnormal haemoglobin molecules which stick to one another and cause stiffness and sickle shape of red blood cells. The sickle-shaped RBC block and damage the vital organs and tissue. The hemoglobin-Beta gene is located on chromosome 11 and the heterozygotes have normal RBCs which means that disease is governed by the recessive gene. Option A is incorrect. The homozygotes for Hb$$^s$$ do not survive which means that homozygous state is fatal; option C is incorrect. The heterozygotes, Hb$$^b$$Hb$$^s$$, show increased resistance to malaria and therefore, have the survival advantage in regions where malaria is a common disease. That’s why sickle cell anaemia has not been eliminated from African population who reside in malaria-prone area. This makes option D correct answer.

• To have both sons and daughters showing the trait, both parents should have at least one copy of the allele. In given pedigree, female parent is shown normal which makes her heterozygous carrier. 
  
  
• The trait is recessive as it is not expressed in a heterozygous female parent. The male parent is homozygous recessive as the trait is recessive. 
  
  
• Daughters get one X chromosome from the father and other from the mother; daughters receiving normal X chromosome from carrier female parent and affected X chromosome from an affected male parent will be normal.
  
  
•  The trait is X-linked recessive and could be colorblindness which is also X-linked recessive trait. 
  
  Thus, the correct answer is option B.

A. Correct option is (C)

 B. Explanation for the correct option.

•  Cross done between unknown progeny and homozygous recessive is called  test cross.
•  This cross is done so as to access whether the organism whose gene are to be found is heterozygous or homozygous dominant.
•  The cross used to identify the genotype behind the phenotype that can be seen.
•   In other words, F1 progeny crossed with homozygous recessive is called cross test.