Question

You are testing learning and memory in two strains (A and B) of hamsters by determining which hamsters learn and remember how to navigate a maze better. You discover that a strain of hamsters raised in a diverse environment has enhanced memory compared to a strain of hamsters raised in a bare cage. You look at a specific gene that is involved in brain development and is activated during learning and memory. If you use bisulfite treatment and subsequent sequencing to study the methylation pattern is of the promoter for this gene, you find that at the same region of the promoter sequence:

Strain A has the following bisulfite treated sequence: GTACGTTAAACGATCG
Strain A has the following untreated sequence: GTACGTTAAACGATCG
Strain B has the following bisulfite treated sequence: GTATGTTAAATGATTG
Strain B has the following untreated sequence: GTACGTTAAACGATCG
Based on the sequences above, greater levels of transcription for this learning and memory gene more likely occur in which of the strains?

Answer 1

The correct answer is - Strain B.

Step-by-step explanation:

The methylation process is adding the methyl group to the DNA molecule that can alter the activity of the DNA segment without any change in the sequence. Methylation presence in the promotor in a gene can repress the transcription process.

By the sequence given it is clear that strain B is not treated and less likely to methylated whereas strain B is more likely to methylated so the greater transcribe sequence would be - Strain B.