Ulowing information to answer the questioní(s) below A part of an mRNA molecule

Question

Ulowing information to answer the questioní(s) below A part of an mRNA molecule with the following sequence is being read by a r charged transfer RNA molecules shown in Figure 144 (with their an Two of them can correctly match the miENA so that a dipeptide can form. 8 read by a ribosome: 5-CCG-ACG-3 (mRNA). The ticodons shown in the 3 to 5 direction) are avamad are avalable. RNA Anticodon Amino Acid GGC CGU UGC CCG ACG CGG Proline Alanine Threonine Glycine Cysteine Alanine Figure 14.4 53) 53) The dipeptide that will form w ill be A) alanine-alanine. B) glycine-cysteine. C) proline-threonine. D) cysteine-alanine. E) threonine-glycine. 54) 54) The anticodon loop of the first tRNA that will complement this mRNA is A) 3'-UGC-5 B) 5'-UGC-3 C) 5-ACG-3 D) 3'-GGC-5 E) 5-GGC-3

Explanation / Answer

Ans. #1. tRNA binds complementarily to mRNA codon with its anti-codon loop. Its sequence is also complementary to mRNA with antiparallel polarity.

            mRNA: 5’- CCG-ACG- 3’

            tRNA: 3’-GGC-5’          , and 3’-UGC-5’           ; [note: U (but not T) is complementary to A in RNA].

So, the two tRNA are - 3’-GGC-5’ (1st anticodon)      , and 3’-UGC-5’ (3rd codon from top in list).

#53. The genetic code dictionary gives amino acid based on mRNA codon.

1st mRNA codon = CCG = Proline (Pro)

2nd mRNA codon = ACG = Threonine (Thr)

Thus, the sequence of dipeptide is- Pro- Thr

Correct option- C

#54. The first anticodon loop that complements with given mRNA = 3’-GGC-5’

Correct option – D.

#50. mRNA = 5’- AUGUCUUCGUUAUCCUUG -3’

Amino acid sequence = MSSLSL (one letter code) = Met- Ser- Ser- Leu- Ser- Leu

Correct option- A.

See genetic code dictionary for depicting amino acid encoded by a codon.

Or, use the tool - http://www.attotron.com/cybertory/analysis/trans.htm

#51. Protein sequence = Phe-Pro-Lys-Gly-Phe-Pro = FPKGFP (One letter code)

mRNA = 5’- UUU-CCC-AAA-GGG-UUU-CCC -3’

See genetic code dictionary for depicting amino acid encoded by a codon.

Or, use the tool - http://www.molbiol.ru/eng/scripts/01_19.html

Note: The sequence of coding strand is exactly the same as mRNA in sequence as well as polarity, except U in mRNA is replaced by T in DNA.

So, simply replace all U with T without changing polarity to generate the coding DNA strand.

Thus,

coding DNA strand is- 5’- TTT-CCC-AAA-GGG- TTT-CCC -3’

Thus, correct option is- A.

Now.

Coding strand is – 3’- CCC-TTT-GGG-AAA-CCC-TTT- 5’                   (Polarity inversed)

Template DNA is – 5’- GGG-AAA-CCC-TTT-GGG-AAA-3’

Note: The sequence of template DNA (to which RNA pol binds during transcription) is complementary to coding strand with polarity reversed.

#52. Correct option. C. 3’-> 5’ along the template strand.

RNA polymerase binds to the template strand. It transcribes the template strand while moving in 3’ -> 5’ direction. Thus, newly synthesized mRNA is obtained as 5’-3’.

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