Hydrogen bonds formed with complementary bases FIGURE 8.7 Mutagenic Effects of the Alkylating Agent Nitrosoguanidine (a) Nitrosoguanidine converts guanine bases in Added CH3 alkyl group DNA to methylguanine, which can base-pair with thymine (T) as well as cytosine (C). (b) The altered base-pairing property of methylguanine can result in point mutations following DNA replication. H•...... N. N….. ? What are the three possible outcomes of point mutations? H....... H- H...... N. Alkylating agent H deoxyribose deoxyribose Guanine Methylguanine (sometimes pairs with T) (pairs with C) (a) Wild type Nitroso- Mutant 3' guanidine treatment 5' DNA DNA 3' replication 5' 3' replication 5' 5' 3' AGT C A 5' A G*T тСА 5' 3' 3' 5' 3' 5' Thymine (formerly paired with G*) now serves as template and pairs with adenine (A) Guanine (G) is converted to methylguanine (G*) Methylguanine of template strand pairs with thymine (T) instead of cytosine (C) (b)

Hydrogen bonds formed with complementary bases FIGURE 8.7 Mutagenic Effects of the Alkylating Agent Nitrosoguanidine (a) Nitrosoguanidine converts guanine bases in Added CH3 alkyl group DNA to methylguanine, which can base-pair with thymine (T) as well as cytosine (C). (b) The altered base-pairing property of methylguanine can result in point mutations following DNA replication. H•...... N. N….. ? What are the three possible outcomes of point mutations? H....... H- H...... N. Alkylating agent H deoxyribose deoxyribose Guanine Methylguanine (sometimes pairs with T) (pairs with C) (a) Wild type Nitroso- Mutant 3' guanidine treatment 5' DNA DNA 3' replication 5' 3' replication 5' 5' 3' AGT C A 5' A GT тСА 5' 3' 3' 5' 3' 5' Thymine (formerly paired with G) now serves as template and pairs with adenine (A) Guanine (G) is converted to methylguanine (G*) Methylguanine of template strand pairs with thymine (T) instead of cytosine (C) (b)

Name: What are the three possible outcomes of point mutations? Hydrogen bond
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Description: What are the three possible outcomes of point mutations? Hydrogen bonds formedwith complementary basesFIGURE 8.7 Mutagenic Effects of the Alkylating AgentNitrosoguanidine (a) Nitrosoguanidine converts guanine bases inAddedCH3alkyl groupDNA to methylg

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter28: Dna Metabolism: Replication, Recombination, And Repair

Section: Chapter Questions

Problem 16P: Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins,...

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29.) equalities now called Chargaff's rule. Biochemist Erwin Chargaff was the first to note that, in DNA, [A]=[T] and [G]=[C], A) Using this rule, determine the percentages of all the bases in DNA that is 20% thymine. [A] = [C] = [G] [T] = 20% %3D - B) If a single strand of RNA is 20% uracil, what can you predict about the percentages of the remaining bases and why?
54.5-methylcytosine results from which of the following events? O phosphorylation of cytosine base in DNA O methylation of guanine base in DNA Omethylation of cytosine base in DNA O alkylation of cytosine base in DNA O de-methylation of cytosine in DNA
2.1 Given the following eukaryotic DNA strand, transcribe and translate the DNA into apolypeptide using the 3’ – 5’ strand as the template. You may use drawings, diagrams,colours and annotations to describe how the DNA strand will be synthesized into afunctional protein. 5’ - TATAAAAASSMSBMDATGSBDCCMBDBAATBSMDSTGTGTCCTMSBAG – 3’ (KEY: The letters SBMD are “made up” nucleic acids that depict non-coding regions in theDNA, hypothetically S pairs with B and M pairs with D).
2.1 Given the following eukaryotic DNA strand, transcribe and translate the DNA into apolypeptide using the 3’ – 5’ strand as the template. You may use drawings, diagrams,colours and annotations to describe how the DNA strand will be synthesized into afunctional protein. 5’ - TATAAAAASSMSBMDATGSBDCCMBDBAATBSMDSTGTGTCCTMSBAG – 3’ (KEY: The letters SBMD are “made up” nucleic acids that depict non-coding regions in theDNA, hypothetically S pairs with B and M pairs with D). 2.2. Describe what are missense mutations and its effects on structure and function usinghaemoglobin as an example .
Helicase Unwinding of the E. coli Chromosome Hexameric helicases, such as DnaB, the MCM proteins, and papilloma virus El helicase (illustrated in Figures 16.22 to 16.25), unwind DNA by passing one strand of the DNA duplex through the central pore, using a mechanism based on ATP-dependent binding interactions with the bases of that strand. The genome of E. coli K12 consists of 4,686,137 nucleotides. Assuming that DnaB functions like papilloma virus El helicase, from the information given in Chapter 16 on ATP-coupled DNA unwinding, calculate how many molecules of ATP would be needed to completely unwind the E. coli K 12 chromosome.
A molecule of composition5′-AAAAAAAAAAA-3′3′-TTTTTTTTTTTTT-5′is replicated in a solution containing unlabeled (notradioactive) GTP, CTP, and TTP plus adenine nucleoside triphosphate with all its phosphorus atoms in theform of the radioactive isotope 32P. Will both daughtermolecules be radioactive? Explain. Then repeat thequestion for the molecule5′-ATATATATATATAT-3′3′-TATATATATATATA-5′
2A Draw the complete structure of the trinucleotide 5'-A-G-C-3' (DNA).
5’-CCTAGACCATAACAT-3’ What happens to the amino acid sequence above as a result of each of the following mutations in the DNA: Substitution of A with T at position 4? Addition of T between positions 8 and 9? Substitution of A with T at position 11? Deletion of A at position 12? Substitution of C with T at position 7? State the type of mutation for each of the base changes indicated.
Figure 1 shows the structure of adenine and thymine. (i) (ii) NH adenine C-H thymine Figure 1 Illustrate the potential tautomers of adenine and thymine. Draw a chemical structure of Thyminc-Adenine (T-A) base pair and label the patterns of hydrogen bond acceptors, donors in the major groove of the TA base pair.
STRUCTURE OF DNA Ueing the complementary base pairing rules in DNA, complete the following base pairs: Adenine - Cytosine -i Guanine -i Thymine - : Adenine : Cytosine :: Guanine :: Thymine 1
GT 3 A. Write the structure of the pentapeptide GLDSC. B. What is the complete name of this pentapeptide? 1. Show a tertiary structure of ACGGC after a disulfide bond forms. A sample of an unknown peptide was divided into two aliquots. One aliquot was treated with trypsin; the other was treated with cyanogen bromide. Given the following sequences (N-terminal to C-terminal) of the resulting fragments, deduce the sequence of the original peptide. 2. 3. Trypsin treatment Asn-Thr-Trp-Met-lle-Lys Gly-Tyr-Met-GIn-Phe Val-Leu-Gly-Met-Ser-Arg Cyanogen bromide treatment Gln-Phe Val-Leu-Gly-Met lle-Lys-Gly-Tyr-Met Ser-Arg-Asn-Thr-Trp-Met
Leu-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.