Question

If the sequence of nucleotides in a template stand of DNA is 3-'ATGCTTCCGAAT-5'. Write the sequence in the corresponding region of the transcribed m-RNA.

• A. 5' - TAC GAA GGC CTT - 3'
• B. 5' - UAC GAA GGC UUA - 3'
• C. 3' - UAC GAA GGC UUA - 5'
• D. 3' - TAC GAA GGC CTT - 5'

Answer 1

The Correct Option is B

The process of transcription involves synthesizing an mRNA molecule using a DNA template strand. The RNA polymerase enzyme reads the template strand in the 3' to 5' direction and synthesizes the mRNA strand in the 5' to 3' direction.

The sequence of the mRNA will be complementary to the DNA template strand, with Uracil (U) replacing Thymine (T) in the RNA molecule.

The given DNA template strand is: 

3'-ATGCTTCCGAAT-5'

We need to find the complementary mRNA sequence, remembering the base pairing rules (DNA → RNA):

• A (DNA) → U (RNA)
• T (DNA) → A (RNA)
• G (DNA) → C (RNA)
• C (DNA) → G (RNA)

Let's determine the complementary base for each nucleotide in the template strand, starting from the 3' end of the DNA to build the 5' end of the mRNA:

DNA Template: 3' - A T G C T T C C G A A T - 5'

Complementary mRNA: 5' - U A C G A A G G C U U A - 3'

So, the sequence of the corresponding transcribed mRNA is:

5' - UACGAAGGCUUA - 3'

Now, let's compare this with the given options:

• 5' - TAC GAA GGC CTT - 3' (Incorrect - contains T)
• 5' - UAC GAA GGC UUA - 3' (Correct)
• 3' - UAC GAA GGC UUA - 5' (Incorrect - wrong direction/polarity)
• 3' - TAC GAA GGC CTT - 5' (Incorrect - wrong bases and polarity)

The correct option is 5' - UAC GAA GGC UUA - 3'.

Answer 2

To transcribe the given DNA sequence into mRNA, we follow the base pairing rules. In transcription, the mRNA is synthesized from the DNA template strand, and the base pairing is as follows:

• A (adenine) pairs with U (uracil) in RNA.
• T (thymine) pairs with A (adenine) in RNA.
• C (cytosine) pairs with G (guanine) in RNA.
• G (guanine) pairs with C (cytosine) in RNA.

Given template strand (DNA): 3' - ATGCTTCCGAAT - 5'

Now, transcribing this into mRNA:

A → U, T → A, G → C, C → G, T → A, T → A, C → G, C → G, G → C, A → U, A → U, T → A

So, the corresponding mRNA sequence is: 5' - UAC GAA GGC UUA - 3'