Molecular Basis of Inheritance — NEET UG practice

1. Recall how histones are organized.

Given that DNA packaging begins with a core protein structure, eight histone molecules come together to form a histone octamer.

Hence, statement A is correct.

2. Check the charge on histones.

Since histones must attract the negatively charged DNA, they themselves are positively charged basic proteins, not negatively charged.

Hence, statement B is incorrect.

3. Check the amino acid composition of histones.

Given that a positive charge on a protein comes from basic amino acids, histones are indeed rich in lysine and arginine.

Hence, statement C is correct.

4. Check the charge on DNA in the nucleosome.

Since DNA carries negative charge from its phosphate backbone, it is the negatively charged DNA (not positively charged) that wraps around the histone octamer to form the nucleosome.

Hence, statement D is incorrect.

5. Check what helps higher-order packaging of chromatin.

Given that further compaction of chromatin beyond the nucleosome level needs extra support, this is provided by non-histone chromosomal (NHC) proteins.

Hence, statement E is correct.

Conclusion: The correct statements are A, C and E, so the answer is A.

1. Recall the disease linked to a mutated beta globin gene.

Given that a single amino acid change in the beta chain of haemoglobin causes sickle cell anaemia, we need to find that specific change.

2. Identify the normal codon at the sixth position.

Since the normal beta globin gene has the codon for glutamic acid at the sixth position:

$$\text{Normal codon} = GAG \ (\text{codes for glutamic acid})$$

3. Identify the mutant codon.

Given that a single base substitution changes adenine to uracil in the mRNA, the codon changes as follows:

$$GAG \rightarrow GUG$$

This new codon GUG codes for valine instead of glutamic acid.

4. Connect this change to the disease effect.

Since replacing glutamic acid with valine makes haemoglobin molecules polymerize under low oxygen conditions, the red blood cells become sickle-shaped.

Conclusion: The mutant sixth codon is GUG, so the answer is A.

1. Recall the genes present in the lac operon and their products.

Given that the lac operon consists of one regulatory gene and three structural genes, each codes for a specific protein:

$$𝑖 \text{ gene} \rightarrow \text{repressor protein}$$ $$𝑧 \text{ gene} \rightarrow \beta\text{-galactosidase}$$ $$𝑦 \text{ gene} \rightarrow \text{permease}$$ $$𝑎 \text{ gene} \rightarrow \text{transacetylase}$$

2. Match the $𝑧$ gene to its product.

Since the $𝑧$ gene specifically codes for the enzyme that breaks down lactose into glucose and galactose, this enzyme is beta-galactosidase.

Hence, the answer is C, beta-galactosidase.