answer for problem set 8

8/3/2019 Answer for Problem Set 8

http://slidepdf.com/reader/full/answer-for-problem-set-8 1/4

Biology 212: Problem Set #8

Due: March 30, 2011

Names*:

________________________________________________________________ __

________________________________________________________________ __

________________________________________________________________ __

________________________________________________________________ __

________________________________________________________________ __

*By signing here, you are confirming that you and all group members who also

sign have contributed to the completion of this problem set. You are reminded of the Honor Code under which this assignment is administered.



(1). Using the DNA sequence below, the accompanying chart indicating the codon encoding each

amino acid, and your knowledge of transcription and translation, answer the following questions:

3¶-GAATTATTTACGATGTACGTGGGGCCAAAGCTTACAATCAAA-5¶ Template DNA Strand

(a) In the space provided, indicate the sequence and orientation of the mRNA transcribed from

this DNA template. (Assume that this entire region of DNA is transcribed.)

5¶-CUUAAUAAAUGCUACAUGCACCCCGGUUUCGAAUGUUAGUUU-3¶

(b) Find the open reading frame (ORF) within the sequence above. How many amino acids long

is the polypeptide chain encoded by the ORF present within this sequence?

7 amino acids

(c) What is the amino acid sequence encoded by this ORF? (Use the codon chart in Scott

Freeman 15)

N-met-his-pro-gly-phe-glu-cyc- C

(d) Indicate the nucleotide sequence of the anticodon that recognizes the fifth codon of your open

reading frame, clearly showing which are the 5¶ and 3¶ ends of the anticodon sequence.

3¶-AAG-5¶



(d) If poly-G is used as a messenger RNA in an incorporation experiment, glyscin is incorporated

into a polypeptide. If poly-C is used, proline is incorporated. If both poly-G and poly-C are used,no amino acids are incorporated into protein. Why?

Adding poly-G and Poly-C could form complementation (hydrogen bonding), therefore, no

amino acids will make from this complementation.

4. Please use the ³Genetic Code´ chart in Figure 15.8 (page 324, 3rd

edition) in your textbook to answer the following questions. (Note: Position 1 refers to the first base at the

5¶ end of the transcribed strand. The last base in the DNA strand, at the 3¶ end, is at position 16.)

The following is a sequence of one strand of section of DNA isolated from E.coli reads:5¶-GTAGCCTACCCATAGG-3¶

a). Suppose mRNA is transcribed from this DNA using the complementary strand as a

template. What will be the sequence of the mRNA?5¶-GUAGCCUACCCAUAG-3¶

b). What would be made if translated started exactly at the 5¶ end of this mRNA

N-val-ala-tyr-pro-C

What happens to the amino acid sequence or protein produced as a result of each

mutation?

c) Deletion of G at position 1

GUA change to UAG (stop codon)«.would result in a nonsense mutation and no

protein would made.

d) Substitution of A for T at position 12

This is change CCA to CCU, which both code for proline. This silent mutation

answer for problem set 8

Documents