Concept explainers
Interpretation: The hydrogen bond donors and acceptors are to be determined in each of the four bases present in DNA.
Concept Introduction:
Answer to Problem 1P
The donors in the hydrogen bond are NH and NH2 groups, while the acceptors are the oxygen atom of the carbonyl group, and nitrogen atom that is not bonded to a hydrogen atom.
Explanation of Solution
The hydrogen bonding between guanine (G) - cytosine (C), and adenine (A) − thymine (T) is shown below:
From the hydrogen bonding in the above figure, it can be seen that −NH2 group of adenine forms bond with -CO group of thymine, and -NH group of thymine forms bond with the nitrogen atom of the adenine. Further, -NH and −NH2 groups of guanine forms bond with cytosine, and −NH2 forms bond with -CO group of guanines.
Hence, the donors in hydrogen bonding are -NH and −NH2groups, while the acceptors are the oxygen atom of -CO groups, and nitrogen atom involved in the ring.
NH and NH2 groups are the donors in the hydrogen bond, and oxygen atom of the carbonyl group, and nitrogen atom that is not bonded to a hydrogen atom are the acceptors.
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Chapter 1 Solutions
Biochemistry
- NOTE: PLEASE SEND. CLEARER IMAGE PLEASE THANK YOU SOMUCH 2. Structural representation of tetrapeptide Phe-Met-Tyr-Asnarrow_forwardб. Draw all possible products of a single hydrolysis being completed on the following polypeptide: Ala-Ser-Met-Serarrow_forwardBe sure to answer all parts. Give the amino acid sequence of an octapeptide that contains the amino acids Lys, Pro, Tyr, Ile (2 equiv) , Gln, Trp, Phe, and forms the following fragments when partially hydrolyzed with HCl: Lys–Ile-Phe-Pro, Phe-Pro-Gln-Tyr, and Trp-Ile-Lys.arrow_forward
- a. Suppose that you have the peptide Ala-Gly-Tyr-His-Leu and you treat it with FDNB and then 6M HCl. Draw the structures of all the products that you will have in solution (assume all reactions to go to completion).arrow_forwardLeu-Trp-Phe-Met-Ala-Ile-Val- Draw the structure of the peptide at pH7.4. and Indicate the hydrogen bonds formed in the alpha helix.arrow_forward. Provide an explanation for the fact that a-D-mannose is more stable than B-D-mannose, whereas the opposite is true for glucose.arrow_forward
- . Draw Haworth projections for the following: (a) CHO in a-furanose form. Name the sugar. H-C-OH | HO-C-H H-C-OH CH,OH (b) The L isomer of (a) (c) a-D-GlcNAc (d) a-D-Fructofuranosearrow_forwardPeptide mass determination. You have isolated a proteinfrom the bacterium E. coli and seek to confirm its identityby trypsin digestion and mass spectrometry. Determinationof the masses of several peptide fragments has enabled youto deduce the identity of the protein. However, there is adiscrepancy with one of the peptide fragments, whichyou believe should have the sequence MLNSFK and an(M 1 H)1 value of 739.38. In your experiments, yourepeat edly obtain an (M 1 H)1 value of 767.38. What isthe cause of this discrepancy and what does it tell youabout the region of the protein from which this peptide isderived?arrow_forwardtrue or false1. In the structural characteristics of a nucleotide, the phosphate subunit is bonded to the sugar subunit.2. In the structural characteristics of a nucleotide, the nitrogenous base subunit is bonded to the phosphate subunit.arrow_forward
- Cyclic forms of D-glucose: Haworth formula The furanose and pyranose forms of D-glucose can be presented using Haworth formula. The following templates are used for this formula. Note the numbering of the carbon atoms and the anomeric carbon (this carbon defines the alpha and beta isomer). Furanose template (furan) Pyranose template (pyran) anomeric carbon anomeric carbon Compare the Fisher and Haworth formula of D-glucofuranose and D-glucopyranose. Notice that the groups attached to C-1, C-2, and C-3 on the left side of the Fisher formula are written above the plane of the furanose template. Fisher formula Haworth formula H FOH H-2 HOH,č 5 -OH FOH- H HO-3H 4 он H 1 H-4 H-5 H OH 3 6 CH2OH H OH Alpha-D-glucofuranose Alpha-D-glucofuranose HOH2C- он OH но H FOH- OH Но H H. OH H ČH2OH Beta-D-glucofuranose Beta-D-glucofuranosearrow_forwardPlease give a clear handwritten answer only with explanation..Draw 5’-A-A-3’ ribonucleotide with 5’ phosphate and 3’ hydroxylarrow_forwardOn the trail of carbons. Tissue culture cells were incubated with glutamine labeled with 15NN in the amide group. Subsequently, IMP was isolated and found to contain some 15N.N. Which atoms in IMP were labeled?arrow_forward
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning