Chapter 9: DNA and its role in heredity

Concept 9.1: DNA structure Read page 172 and answer the question. Contains C, H, N, O and P (NO Sulfur) NEVER leaves the nucleus Antiparallel Double helix

Concept 9.1: DNA structure Historical context Avery, MacLeod, and McCarty: Discovered DNA was the material of inheritance Used principles of transformation 1944

Concept 9.1: DNA structure Historical context Erwin Chargaff: Discovered the base pairing rules 1950

Concept 9.1: DNA structure Historical context Alfred Hershey and Martha Chase Experiment: Is DNA nucleic acid or protein? 1952

Concept 9.1: DNA structure Historical context Rosalind Franklin: Best x-ray crystallography photograph (photo 51) 1953

Concept 9.1: DNA structure Historical context James Watson and Francis Crick: Discovery of overall structure of DNA Used photo-51 to determine DNA is helical and antiparallel February 1953

Concept 9.1: DNA Structure Four key features define DNA structure: DNA is a double stranded helix of uniform diameter Two strands of DNA are anti-parallel The outer edges of the nitrogenous bases are exposed in the major and minor grooves (surfaces of the A-T and G-C base pairs are chemically distinct in order to allow other molecules like proteins to recognize them) Considered to be right-handed

Concept 9.1: DNA Structure Structure of DNA is essential to function: Storage of genetic information Precise replication during the cell division cycle Susceptibility to mutations Expression of the coded information as phenotypes

9.1 Questions Explain why in a DNA molecule of any species the amount of adenine is equal to thymine (A=T) and cytosine is equal to guanine (C=G), but (A + T) does not equal (C + G). Given the following DNA sequence, what would the sequence of the complementary strand look like? 5’ – GCTAACTGTGATCGTATAAGCTGA – 3’ Why would the majority of the scientists in the 1940’s believe that DNA was made of protein and not nucleic acid? What does “5’” end and “3’” end represent in the DNA molecule?

9.1 Answers The purine adenine (A) always pairs with the pyrimidine thymine (T), and the purine guanine (G) always pairs with the pyrimidine cytosine (C). Therefore in a section of DNA A=T and G=C. The amount of G and A or C and T is independent from one another and so the total does not equal. The complementary strand is: 3’ – CGATTGACACTAGCATATTCGACT – 5’ Most genomes contain DNA that has both nucleic acid and protein. DNA wraps around and is organized by protein and protein is what is used to manipulate and express the DNA. Scientists also knew that most protein is functional in various cell processes and so it was almost assumed that protein would also make up the genome. The 5’ end and 3’ ends are free ends of the chain

Concept 9.2: DNA Replication

Concept 9.2: DNA Replication Semiconservative model Parental strands will become the templates Two main steps: Parental DNA is unwound and the two sides are separated New nucleotides will form complementary strands

Concept 9.2: DNA Replication DNA is unwound at replication fork by DNA helicase and stabilized by single-stranded binding proteins Topoisomerase prevents over winding of the DNA Primase makes primer and lays it down at the 5’ end of the leading strand and the 5’ end of each Okazaki fragment DNA polymerase moves in at the location of primer and adds nucleotides to elongate the complementary strand and will remove the primer (leading strand is made continuously but lagging strand is made in the Okazaki fragments) DNA ligase joins the sugar-phosphate backbone of the Okazaki fragments to form the 3’  5’ lagging strand Telomerase adds telomeres at the end of each strand DNA polymerase repairs any errors: proofread method or mismatch repair method

Concept 9.2: DNA Replication

Concept 9.2: DNA Replication Scientists have developed lab techniques to take advantage of the DNA replication process: PCR (polymerase chain reaction)

9.2 questions Explain the role of Okazaki fragments in the synthesis of the lagging strand. Differentiate between proofreading and mismatch repair. Explain how PCR amplifies a particular sequence of DNA. Is the replication process linear or exponential? How is this application beneficial?

9.2 answers Okazaki fragments are short, discontinuous stretches of sequence that are formed when the lagging strand is replicated. Because only a portion of the DNA is opened up at the end of replication fork at any given time, and because DNA polymerase adds nucleotides to the 3’ end of the previous nucleotide, continuous growth is not possible. Proofreading occurs during synthesis of the DNA molecule, whereas mismatch repair occurs after synthesis. PCR amplifies a sequence of DNA by means of a lab technique that uses a sample of double-stranded DNA, two artificially synthesized primers that are complementary to the ends of the sequence to be amplified, the four dNTPs, and a DNA polymerase. During each cycle of a PCR reaction, the double-stranded DNA is separated and replicated, producing two double-stranded copies. These copies are then amplified during the next cycle. The amount of DNA amplifies exponentially.

Concept 9.3: Mutations

Concept 9.3: Mutations Two general categories of mutations: Somatic mutation (body cells) Germline mutation (sex cells)

Concept 9.3: Mutations Results from mutations: Silent mutation: does not affect gene function (most mutations in large genomes) Loss-of-function mutation: loss of gene expression or produces non functional protein Gain-of-function mutation: leads to a protein with an altered function (common in cancer) Conditional mutation: can only become altered under environmental conditions, like change in temperature

Concept 9.3: mutations Small scale mutations: Point mutation Addition or deletion of a single nucleotide base or the substitution of one base for another Caused by DNA replication errors or by environmental mutagens (radiation or chemicals) Example: Sickle cell anemia

Concept 9.3: mutations Large scale mutations: Chromosomal alterations Deletions: removal of a part of a chromosome Duplication: part of a chromosome broke off and reconnected on a different chromosome Inversion: a part of a chromosome is removed and inserted in a reversed order Translocation: reciprocal exchange between two different chromosomes Caused by errors in DNA replication, meiosis, random disruption, mutagens

9.3 questions Two individuals have a mutation in gene X but at different sites. The mutation affects the first individual adversely, and the second individual experiences no effect. Explain this observation. Are all mutations inherently bad? What about those in somatic cells? Starting with chromosome ABCDEFC and the non homologous chromosome LMNOPQR, illustrate the following: Deletion of segment C Duplication of segment C Duplication of segment D and a deletion of segment E An inversion of segment, going from A to C A reciprocal translocation between CDEFG and LMNO

9.3 answers The mutation in gene X in the first individual must have occurred in an essential region of the gene that is required for its function. The mutation in gene X in the second individual may be a silent mutation or it may be in a region that is nonessential for the function of that protein. Mutations may be beneficial, neutral, or harmful to organisms. Somatic cell mutations will only affect the individual but germ cell mutations will be passed to the next generation. As follows ABDEFC ABCCDEFG ABCDDEG CBADEFG ABLMNO and CDEFGPQR