General orrganic and biological chemistry structures off liffe 5th by karen timberlake21 worked examples

Sample Problem 21.1 Nucleotides
For each of the following nucleotides, identify the components and whether the nucleotide is found in DNA only, RNA only, or both DNA and RNA:
a.

deoxyguanosine monophosphate (dGMP)

b.

adenosine monophosphate (AMP)

Solution
a.

This nucleotide of deoxyribose, guanine, and a phosphate group is only found in DNA.

b.

This nucleotide of ribose, adenine, and a phosphate group is only found in RNA.

Study Check 21.1
What is the name and abbreviation of the DNA nucleotide of cytosine?

Answer
deoxycytidine monophosphate (dCMP)

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 21.2 Bonding of Nucleotides
Draw the condensed structural formula for an RNA dinucleotide formed by joining the 3′ OH group of adenosine monophosphate and the 5′ phosphate group of cytidine monophosphate.

Solution
The dinucleotide is drawn by connecting the 3′ hydroxyl group on the adenosine 5′ monophosphate with the 5′ phosphate group on the cytidine monophosphate.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.2 Bonding of Nucleotides
Continued

Study Check 21.2
What type of linkage connects the two nucleotides in Sample Problem 21.2?

Answer
A 3′,5′ phosphodiester linkage connects the 3′ OH group in the ribose of AMP and the 5′ carbon in the ribose of CMP.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.3 Complementary Base Pairs
Write the complementary base sequence for the following segment of a strand of DNA:
5′A C G A T C T 3′

Solution

The complementary base pairs are A–T and C–G. The complementary strand is written in the opposite direction, from the 3′ end to the 5′ end.

Study Check 21.3
What sequence of bases is complementary to a DNA segment with a base sequence of 5′ G G T T A A C C 3′?

Answer
3′ C C A A T T G G 5′

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.4 Direction of DNA Replication
In an original DNA strand, a segment has the base sequence 5′ A G T C C G 3′.
a.

What is the sequence of nucleotides in the daughter DNA strand that is complementary to this segment?

b.

Why would the complementary sequence in the daughter DNA strand be synthesized as Okazaki fragments that require a DNA ligase?

Solution
a.

Only one possible nucleotide can pair with each base in the original segment. Thymine will pair only with adenine, whereas cytosine pairs only with guanine to give the complementary base sequence: 3′
T C A G G C 5′.

b.

Because DNA polymerase only adds nucleotides in the 5′ to 3′ direction, DNA on the lagging strand is synthesized as short Okazaki fragments, which are joined by DNA ligase.

Study Check 21.4
How many daughter strands are formed during the replication of DNA?

Answer
Two daughter strands are formed, one from each strand of the DNA double helix.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.5 RNA Synthesis
The sequence of bases in a part of the DNA template strand is 3′ C G A T C A 5′. What corresponding mRNA is produced?

Solution
To form the mRNA, the bases in the DNA template are paired with their complementary bases: G with C, C with G, T with A, and A with U.

Study Check 21.5
What is the DNA template strand segment that codes for the mRNA segment with the nucleotide sequence
5′ G G G U U U A A A 3′?

Answer

3′ C C C A A A T T T 5′

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.6 Transcription
Describe why transcription will or will not take place in each of the following conditions:
a.

There are no transcription factors bound at the promoter region.

b.

An activator binds to a transcription factor.

Solution
a.

Transcription will not take place as long as there are no transcription factors to bind and activate the RNA polymerase.

b.

Transcription will take place when the activated transcription factor binds to the promoter region on the DNA, which makes the RNA polymerase begin transcription.

Study Check 21.6
Where in a cell does pre-mRNA processing take place?

Answer
The processing of pre-mRNA to mature mRNA takes place in the nucleus before the mRNA can move into the cytosol and the ribosomes.

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.7 Protein Synthesis
Use three-letter and one-letter abbreviations to write the amino acid sequence for the peptide from the mRNA sequence of 5′ UCA AAA GCC CUU 3′.

Solution
Each of the codons specifies a particular amino acid. Using Table 21.6, we write a peptide with the following amino acid sequence:

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.7 Protein Synthesis
Continued

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.

Sample Problem 21.7 Protein Synthesis
Continued

Study Check 21.7
Use three-letter and one-letter abbreviations to write the amino acid sequence for the peptide from the mRNA sequence of 5′ GGG AGC AGU GAG GUU 3′.

Answer
Gly–Ser–Ser–Glu–Val, GSSEV

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.8 Mutations
An mRNA has the sequence of codons 5′ CCC AGA GCC 3′. If a point mutation in the DNA changes the mRNA codon of AGA to GGA, how is the amino acid sequence affected in the resulting protein?

Solution
The mRNA sequence 5′ CCC AGA GCC 3′ codes for the following amino acids: proline, arginine, and alanine. When the point mutation occurs, the new sequence of the mRNA codons is 5′ CCC GGA
GCC 3′, which codes for proline, glycine, and alanine. The basic amino acid arginine is replaced by the nonpolar amino acid glycine.

Study Check 21.8
How might the protein made from this mRNA be affected by this mutation?

Answer
Because the point mutation replaces a polar basic amino acid with a nonpolar neutral amino acid, the tertiary structure may be altered sufficiently to cause the resulting protein to be less effective or
nonfunctional.

General, Organic, and Biological Chemistry: Structures of Life, 5/e

Karen C. Timberlake

© 2016 Pearson Education, Inc.


Sample Problem 21.9 Viruses
Why are viruses unable to replicate on their own?

Solution
Viruses contain only packets of DNA or RNA, but not the necessary replication machinery that includes enzymes and nucleosides.

Study Check 21.9
What are the essential parts of a virus?

Answer
nucleic acid (DNA or RNA) and a protein coat

General, Organic, and Biological Chemistry: Structures of Life, 5/e
Karen C. Timberlake

© 2016 Pearson Education, Inc.