BIO 2220 Assignment: Mendelian Genetics and Cell Replication Transcription Translation Questions

BIO 2220 Assignment: Mendelian Genetics and Cell Replication Transcription Translation Questions

Chapter 11 Mendelian Patterns of Inheritance Genetic Disorder: Inability to Process Phenylalanine 1 in 10,00 people cannot break down phenylalanine Outline 11.1 Gregor Mendel 11.2 Mendel’s Laws 11.3 Mendelian Patterns of Inheritance & Human Disease 11.4 Beyond Mendelian Inheritance 3 Section 11.1 Gregor Mendel Concept of Blending Inheritance:  Parents of contrasting appearances produce offspring of intermediate appearance  Blending was a popular concept during Mendel’s time: Larmarck—inheritance of acquired characteristics Mendel’s findings challenged the blending concept  He proposed the Particulate Theory of Inheritance • Inheritance involves reshuffling of factors (genes) from generation to generation 4 Gregor Mendel • Austrian monk  Studied science and mathematics at the University of Vienna  Conducted breeding experiments with the garden pea Pisum sativum  Carefully gathered and documented quantitative data from his experiments • Formulated fundamental laws of heredity in the early 1860s  Had no knowledge of cells or chromosomes  Did not own or use a microscope 5 Gregor Mendel a. b. Gregor Mendel • The garden pea:  Organism used in Mendel’s experiments  A good choice for several reasons: • Easy to cultivate • Short generation • Normally self-pollinating, but can be crosspollinated by hand • True-breeding varieties were available • Simple, objective traits 7 Flower Structure stamen anther filament stigma style ovules in ovary Cutting away anthers Brushing on pollen from another plant All peas are yellow when one parent produces yellow seeds and the other parent produces green seeds. carpel Garden Pea Anatomy Flower Structure stamen anther filament stigma style ovules in ovary carpel a. 9 Fig. 11-2a Garden Pea Anatomy Cutting away anthers Brushing on pollen from another plant All peas are yellow when one parent produces yellow seeds and the other parent produces green seeds. 10 Fig. 11.2 Fig. 11-2 11 Outline 11.1 Gregor Mendel 11.2 Mendel’s “Laws” 11.3 Mendelian Patterns of Inheritance & Human Disease 11.4 Beyond Mendelian Inheritance 12 Section 11.2 Mendel’s Laws Mendel performed cross-breeding experiments  Used “true-breeding” (homozygous) plants •

Tall plants always had tall offspring and short plants always had short offspring  Chose varieties that differed in only one trait (monohybrid cross)  Performed reciprocal crosses • Parental generation = P • First filial generation offspring = F1 • Second filial generation offspring = F2  Formulated the Law of Segregation 13 Law of Segregation/Separation of Alleles  Each individual has two factors (alleles) for each trait  The factors (alleles) segregate (separate) during gamete (sperm & egg) formation  Each gamete contains only one factor (allele) from each pair of factors  Fertilization gives the offspring two factors for each trait 14 Mendel’s

Monohybrid Cross Pure breeding Fig. 11.3 15 Relationship Between Observed Phenotype & F2 Offspring Fig. 11.5 16 Fig. 11.4 Classical Genetics and Mendel’s Cross  Each trait in a pea plant is controlled by two alleles (alternate forms of a gene)  Dominant allele (upper case) masks the expression of the recessive allele (lower-case)  Alleles occur on a homologous pair of chromosomes at a particular gene locus • Homozygous = identical alleles • Heterozygous = different alleles 17 Classical View of Homologous Chromosomes Fig. 11.5 18 Genotype Versus Phenotype Genotype  Refers to the two alleles an individual has for a specific trait  If identical, genotype is homozygous  If different, genotype is heterozygous Phenotype  Refers to the physical appearance of the individual 19 Independent Assortment of Homologous Chromosomes  Each pair of factors (alleles) segregates (sorts) independently of the factors (alleles) for other traits  All possible combinations of factors (alleles) can occur in the gametes 20 Mendel’s Laws • A dihybrid cross uses true-breeding (homozygous) plants differing in two traits • Mendel tracked each trait through two generations (next slide).  Started with true-breeding plants differing in two traits  The F1 plants showed both dominant characteristics  F1 plants self-pollinated  Observed phenotypes among F2 plants 21 Mendel’s Dihybrid Cross Fig. 11.6 22 Independent Assortment & Segregation during Meiosis Mendel’s laws hold because of meiosis. Fig. 11-7 23 Mendel’s Laws of Probability: Punnett Square •

Important information for writing discussion questions and participation

Welcome to class

Hello class and welcome to the class and I will be your instructor for this course. This is a -week course and requires a lot of time commitment, organization, and a high level of dedication. Please use the class syllabus to guide you through all the assignments required for the course. I have also attached the classroom policies to this announcement to know your expectations for this course. Please review this document carefully and ask me any questions if you do. You could email me at any time or send me a message via the “message” icon in halo if you need to contact me. I check my email regularly, so you should get a response within 24 hours. If you have not heard from me within 24 hours and need to contact me urgently, please send a follow up text to

I strongly encourage that you do not wait until the very last minute to complete your assignments. Your assignments in weeks 4 and 5 require early planning as you would need to present a teaching plan and interview a community health provider. I advise you look at the requirements for these assignments at the beginning of the course and plan accordingly. I have posted the YouTube link that explains all the class assignments in detail. It is required that you watch this 32-minute video as the assignments from week 3 through 5 require that you follow the instructions to the letter to succeed. Failure to complete these assignments according to instructions might lead to a zero. After watching the video, please schedule a one-on-one with me to discuss your topic for your project by the second week of class. Use this link to schedule a 15-minute session. Please, call me at the time of your appointment on my number. Please note that I will NOT call you.

Please, be advised I do NOT accept any assignments by email. If you are having technical issues with uploading an assignment, contact the technical department and inform me of the issue. If you have any issues that would prevent you from getting your assignments to me by the deadline, please inform me to request a possible extension. Note that working fulltime or overtime is no excuse for late assignments. There is a 5%-point deduction for every day your assignment is late. This only applies to approved extensions. Late assignments will not be accepted.

If you think you would be needing accommodations due to any reasons, please contact the appropriate department to request accommodations.

Plagiarism is highly prohibited. Please ensure you are citing your sources correctly using APA 7th edition. All assignments including discussion posts should be formatted in APA with the appropriate spacing, font, margin, and indents. Any papers not well formatted would be returned back to you, hence, I advise you review APA formatting style. I have attached a sample paper in APA format and will also post sample discussion responses in subsequent announcements.

Your initial discussion post should be a minimum of 200 words and response posts should be a minimum of 150 words. Be advised that I grade based on quality and not necessarily the number of words you post. A minimum of TWO references should be used for your initial post. For your response post, you do not need references as personal experiences would count as response posts. If you however cite anything from the literature for your response post, it is required that you cite your reference. You should include a minimum of THREE references for papers in this course. Please note that references should be no more than 5 years old except recommended as a resource for the class. Furthermore, for each discussion board question, you need ONE initial substantive response and TWO substantive responses to either your classmates or your instructor for a total of THREE responses. There are TWO discussion questions each week, hence, you need a total minimum of SIX discussion posts for each week. I usually post a discussion question each week. You could also respond to these as it would count towards your required SIX discussion posts for the week.

I understand this is a lot of information to cover in 5 weeks, however, the Bible says in Philippians 4:13 that we can do all things through Christ that strengthens us. Even in times like this, we are encouraged by God’s word that we have that ability in us to succeed with His strength. I pray that each and every one of you receives strength for this course and life generally as we navigate through this pandemic that is shaking our world today. Relax and enjoy the course!

Hi Class,

Please read through the following information on writing a Discussion question response and participation posts.

Contact me if you have any questions.

Important information on Writing a Discussion Question

• Your response needs to be a minimum of 150 words (not including your list of references)
• There needs to be at least TWO references with ONE being a peer reviewed professional journal article.
• Include in-text citations in your response
• Do not include quotes—instead summarize and paraphrase the information
• Follow APA-7th edition
• Points will be deducted if the above is not followed

Participation –replies to your classmates or instructor

• A minimum of 6 responses per week, on at least 3 days of the week.
• Each response needs at least ONE reference with citations—best if it is a peer reviewed journal article
• Each response needs to be at least 75 words in length (does not include your list of references)
• Responses need to be substantive by bringing information to the discussion or further enhance the discussion. Responses of “I agree” or “great post” does not count for the word count.
• Follow APA 7th edition
• Points will be deducted if the above is not followed

• Remember to use and follow APA-7th edition for all weekly assignments, discussion questions, and participation points.
• Here are some helpful links
• The is a great resource

The Punnett square is a table that lists all possible genotypes resulting from a cross – All possible sperm genotypes are lined up on one side – All possible egg genotypes are lined up on the other side – Every possible zygote genotypes are placed within the squares 24 Punnett Square Punnett Square Allows the easy calculation of probability, genotypes, and phenotypes among the offspring 25 Fig. 11.8 Mendel’s Laws: Testcrosses

 Individuals with recessive phenotype always have the homozygous recessive genotype  However, individuals with dominant phenotype have an unknown genotype. There are two possibilities: • Homozygous dominant, or • Heterozygous dominant  Perform a testcross to determine the genotype of an individual having the dominant phenotype 26 Heterozygous One-trait Testcrosses Homozygous Fig. 11-9 27 Mendel’s Laws: Two-trait Testcross  An individual with both dominant phenotypes is crossed with an individual with both recessive phenotypes.  If the individual with the dominant phenotypes is heterozygous for both traits, the expected phenotypic ration is 1:1:1:1. 28 Two-trait Testcross L = long wings l = vestigial (short) wings G = gray bodies g = black bodies Gametes: LG Lg lG lg Page 201 29 Outline 11.1 Gregor Mendel 11.2 Mendel’s Laws 11.3 Mendelian Patterns of Inheritance & Human Disease 11.4 Beyond Mendelian Inheritance 30 Mendel’s Laws & Human Genetic Disorders • Genetic disorders are medical conditions caused by alleles inherited from parents • Autosome – Chromosomes that are not sex chromosomes (X or Y) • Genetic disorders caused by genes on autosomes are called autosomal disorders  Some genetic disorders are autosomal recessive • An individual with AA does NOT have the disorder • An individual with Aa does NOT have the disorder, but is a carrier • An individual with aa has the disorder  Some genetic disorders are autosomal dominant • An individual with AA has the disorder • An individual with Aa has the disorder 31 • An individual with aa does NOT have the disorder Carrier Female Male Child affected; neither parent is affected. Parents are carriers. Page 195 Child is unaffected but both parents are affected 32 Autosomal Recessive Pedigree Generations I aa II IV Fig. 11-10 Aa A? III A? Aa A? * Aa Aa aa aa A? A? A? Key aa = affected Aa = carrier (unaffected) AA = unaffected A? = unaffected (one allele unknown) Autosomal recessive disorders • Most affected children have unaffected parents. • Heterozygotes (Aa) have an unaffected phenotype. • Two affected parents will always have affected children. • Close relatives who reproduce are more likely to have affected children. • Both males and females are affected with equal frequency. 33 Autosomal Dominant Pedigree Generations Aa Aa I * II III Aa aa Aa Aa A? aa Autosomal dominant disorders Fig. 11-11 aa aa aa aa aa aa Key AA = affected Aa = affected A? = affected (one allele unknown) aa = unaffected • Affected children will usually have an affected parent. • Heterozygotes (Aa) are affected. • Two affected parents can produce an unaffected child. • Two unaffected parents will not have affected children. • Both males and females are affected with equal frequency. 34 Mendel’s Laws: Autosomal Recessive Disorders • Methemoglinemia • Cystic fibrosis 35 Autosomal Recessive Disorders: Methemoglobinemia • Relatively harmless disorder • Accumulation of methemoglobin in blood causes skin to appear bluish-purple • Methylene blue temporarily removes the tint from skin 36 Autosomal Recessive Disorders: Cystic Fibrosis • Mucus in bronchial tubes and pancreatic ducts is thick and viscous. • Extremely harmful disorder and often fatal disorder. Genetic Disorder: Inability to Process Phenylalanine Autosomal Dominant Disorders • Osteogenesis imperfecta • Huntington disease • Hereditary spherocytosis 39 Autosomal Dominant Disorders Osteogenesis Imperfecta •Weakened, brittle bones, short height, loose joints, tooth problems, etc. •Most cases (90%) are caused by mutation in two genes of type I collagen (COL1A1 or COL1A2) Classic blue sclerae of a person with osteogenesis imperfecta 40 https://en.Wikipedia.org/wiki/Osteogenesis_imperfecta Huntington Disease The gene that synthesizes the protein huntingtin (MW = ) is located on chromosome 4. Symptoms • uncontrolled movement of the arms, legs, head, face and upper body • decline in thinking and reasoning skills, memory, concentration, judgment as well as loss of ability to plan and organize. • Altered mood, depression, anxiety, unprovoked anger and irritability 41 Hereditary Spherocytosis • Red blood cells become spherical. • The RBCs are fragile and burst easily because of the weakened cytoskeleton. • Caused by mutations in membrane and cytoskeletal genes. Fig. 17.20. World of Cell, 8th ed. 42 Blood Sample Testing and Pedigree Analysis Fig. 11A 43 Page 208 44 Page 208 45 Page 213 46 Outline 11.1 Gregor Mendel 11.2 Mendel’s Laws 11.3 Mendelian Patterns of Inheritance & Human Disease 11.4 Beyond Mendelian Inheritance 47 Section 11.4 Beyond Mendelian Inheritance • Multiple allelic traits • Incomplete dominance/incomplete penetrance • Pleiotropic effects • Polygenic inheritance • X-linked inheritance 48 Section 11.4 Multiple Alleles (Multiple Allelic Traits) • The gene exists in several allelic forms (but each individual only has two alleles) • ABO blood types  The alleles: • IA = A antigen on red blood cells; plasma contains anti-B antibody • IB = B antigen on red blood cells; plasma contains anti-A antibody • i = Neither A nor B antigens on red blood cells; plasma contains both anti-A and anti-B antibodies • The ABO blood type is also an example of codominance  More than one allele is fully expressed  Both IA and IB are expressed in the presence of the other 49 ABO Blood Type Phenotype A B AB O Genotype I A I A, I Ai IBIB, IBi IA IB ii Example IA i IA IAIA IAi i IAi ii 50 Incomplete Dominance  Heterozygote expresses a phenotype intermediate between that of either homozygote • Homozygous red has red phenotype • Homozygous white has white phenotype • Heterozygote has pink (intermediate) phenotype  Phenotype reveals genotype without a test cross 51 Incomplete Dominance Fig. 11.14 Fig. 11-14 52 Human Examples of Incomplete Dominance Familial Hypercholesterolemia (FH) • Homozygotes for the mutant allele develop fatty deposits in the skin and tendons and may have heart attacks during childhood • Heterozygotes may suffer heart attacks during early adulthood • Homozygotes for the normal allele do not have the disorder • Brown and Goldstein—receptor mediated endocytosis (pg. 94) 53 Human Examples of Incomplete Penetrance Incomplete penetrance (expression) • The dominant allele may not always lead to the dominant phenotype in a heterozygote • Many dominant alleles exhibit varying degrees of penetrance • Example: polydactyly – Extra digits on hands, feet, or both – Not all individuals who inherit the dominant polydactyly allele will exhibit the trait 54 Three Methods of Endocytosis Chapter 5, pg. 94. See Mescher Fig. 2-7, pg.27. 55 Figure 2-7 Pleiotropy: Marfan’s Syndrome Pleiotropy occurs when a single mutant gene affects two or more distinct and seemingly unrelated traits. Example • Marfan’s syndrome is pleiotropic and results in the following phenotypes:  disproportionately long arms, legs, hands, and feet  a weakened aorta  poor eyesight See next slide 57 Marfan Syndrome Connective tissue defects Skeleton Chest wall deformities Long, thin fingers, arms, legs Scoliosis (curvature of the spine) Flat feet Long, narrow face Loose joints Heart and blood vessels Mitral valve prolapse Enlargement of aorta Eyes Lens dislocation Severe nearsightedness Lungs Skin Collapsed lungs Stretch marks in skin Recurrent hernias Dural ectasia: stretching of the membrane that holds spinal fluid Aneurysm Aortic wall tear 58 Sickled Red Blood Cell Page 199 1,600X (colorized SEM) Polygenic Inheritance  Occurs when a trait is governed by two or more genes having different alleles  Each dominant allele has a quantitative effect on the phenotype  These effects are additive  Results in continuous variation of phenotypes within a population  The traits may also be affected by the environment  Examples • Human skin color • Height • Eye color 60 Polygenic Inheritance Several pairs of genes control the trait. Fig. 11.16 61 Skin Color: A Polygenic Trait X-Linked Inheritance In mammals • The X and Y chromosomes determine gender • Females are XX • Males are XY 63 X-Linked Inheritance The term X-linked is used for genes that have nothing to do with gender • X-linked genes chromosome. are carried on the X • The Y chromosome does not carry these genes • Discovered in the early 1900s by a group at Columbia University, headed by Thomas Hunt Morgan. – Performed experiments with fruit flies » They can be easily and inexpensively raised in simple laboratory glassware » Fruit flies have the same sex chromosome pattern as humans 64 Fig. 11.18 Fig. 11-18 65 X – Linked Inheritance X-linked Inheritance 67 Fig. 11.17 Examples of X-linked Recessive Disorders Color blindness) • The allele for the blue-sensitive protein is autosomal • The alleles for the red- and green-sensitive pigments are on the X chromosome. (See next slide) 68 Color Blindness: X-Linked Recessive Pedigree XBY XBXb grandfather XbY XBXB daughter XBY XbXb XbY XBY XBXB XBXb XbY XBXB XBXb XbXb XbY XbY grandson Key = Unaffected female = Carrier female = Color-blind female = Unaffected male = Color-blind male X-Linked Recessive Disorders • More males than females are affected. • An affected son can have parents who have the normal phenotype. • For a female to have the characteristic, her father must also have it. Her mother must have it or be a carrier. • The characteristic often skips a generation from the grandfather to the grandson. • If a woman has the characteristic, all of her sons will have it. 69 Examples of X-linked Recessive Disorders Menkes syndrome • Caused by a defective allele on the X chromosome • Disrupts movement of the copper in and out of cells. • Phenotypes include kinky hair, poor muscle tone, seizures, and low body temperature Muscular dystrophy • Wasting away of the muscle • Caused by the absence of the muscle protein dystrophin. Without dystrophin, Ca2+ leaks into the cell and damages/kills the muscle cells. 70 Muscular Dystrophy fibrous tissue abnormal muscle normal tissue 71 Examples of X-linked Recessive Disorders Adrenoleukodystrophy • A fatal X-linked recessive disorder • Failure of a carrier protein to move either an enzyme or very long chain fatty acid into peroxisomes. • Causes progressive loss of vision, hearing, speech Hemophilia • Absence or minimal presence of clotting factor VIII or clotting factor IX • Blood of affected person either does not clot or clots very slowly. 72 End 73 Mendel’s Laws Punnett Square  Allows us to easily calculate probability, of genotypes and phenotypes among the offspring  Punnett square in next slide shows a 50% (or ½) chance • The chance of E = ½ • The chance of e = ½  An offspring will inherit: • • • • The chance of EE =½  ½=¼ The chance of Ee =½  ½=¼ The chance of eE =½  ½=¼ The chance of ee =½  ½=¼ 74 l = vestigial (short) wings G = gray body Page 201 75 Page 201 76 77 Molecular Biology of the Gene Chapter 12 Outline • 12.1 The Genetic Material • 12.2 Replication of DNA • 12.3 The Genetic Code of Life • 12.4 First Step: Transcription • 12.5 Second Step: Translation 2 The Genetic Basis of Eye, Hair, and Skin Coloration • All life on Earth contains the four bases of DNA: A,G, C, and T. • MC1R gene • contributes to skin, hair, and eye color in humans • Some of the cells in skin become specialized pigment-making cells called melanocytes. • Humans have variations in their MC1R that come from their ancestry. 3 12.1 The Genetic Material • Requirements for the genetic material: • • • • store genetic information stable replicated accurately undergo mutations to provide genetic variability • Researchers showed that DNA can fulfill all these functions. 4 12.1 The Genetic Material • Frederick Griffith investigated virulence of Streptococcus pneumoniae. • Concluded that virulence could be passed from a dead strain to a nonvirulent living strain • Transformation • Further research by Avery et al. • Discovered that DNA is the transforming substance • DNA from dead cells incorporated into the genome of living cells 5 The Genetic Material • Griffith’s Transformation Experiment • Mice were injected with two strains of pneumococcus, an encapsulated (S) strain and a non-encapsulated (R) strain. • The S strain is virulent (the mice died); it has a mucous capsule and forms “shiny” colonies. • The R strain is not virulent (the mice lived); it has no capsule and forms “dull” colonies. • Frederick Griffith investigated virulence of Streptococcus pneumoniae. • Concluded that virulence could be passed from a dead strain to a nonvirulent living strain • Transformation capsule Injected live S strain has capsule and causes mice to die. Injected live R strain has no capsule and mice do not die. Injected heatkilled S strain does not cause mice to die. Injected heat-killed S strain plus live R strain causes mice to die. Live S strain is withdrawn from dead mice. 6 12.1 The Genetic Material • DNA and proteins were the candidates for the hereditary material. • Proteins contain 20 amino acids that can be sequenced in different ways. • DNA and RNA each contain only four types of nucleotides. 7 12.1 The Genetic Material • Further research by Avery et al. • Discovered that DNA is the transforming substance • DNA from dead cells incorporated into the genome of living cells • Avery, McLeod, and McCarty’s Experiment • Scientists argued that DNA lacked variability to be able to store genetic information. • Avery et al. used enzymes: DNase, RNase and protease • Only enzyme to have an effect – DNase, which prevented “transformation” 8 The Genetic Material • Hershey and Chase’s Experiment • Used a virus T phage, composed of radioactively labeled DNA and coat proteins (in two separate experiments) to infect E.coli bacteria. 9 10 The Genetic Material • Transformation of organisms today: • The result is the so-called genetically modified organisms (GMOs). • Invaluable tool in modern biotechnology today • Commercial products that are currently much used • Green fluorescent protein (GFP) used as a marker • A jellyfish gene codes for GFP. • The jellyfish gene is isolated and then transferred to a bacterium, or the embryo of a plant, pig, or mouse. • When this gene is transferred to another organism, the organism glows in the dark. 11 Nucleotide Composition of DNA Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. NH2 adenine (A) C N C HC C CH N O HO P O O 5 4 CH2 N C thymine (T) N O HN nitrogen-containing base CH C O CH3 C N O O C H H C 3 OH HO H C1 C H 2 H P O C C HN 4 CH2 O C H H C1 C H 2 H sugar = deoxyribose NH2 N O HO P C O O 5 4 CH2 O N CH C CH N N O O C H H C 3 OH a. Purine nucleotides C cytosine (C) N CH H2N C phosphate 5 H C 3 OH O guanine (G) O HO H C1 C H 2 H P O b. Pyrimidine nucleotides O 5 4 CH2 O C H H C 3 OH H C1 C H 2 H DNA Composition in Various Species (%) Species A T G C Homo sapiens (human) 31.0 31.5 19.1 18.4 Drosophila melanogaster (fruit fly) Zea mays (corn) Neurospora crassa (fungus) 27.3 25.6 23.0 27.6 25.3 23.3 22.5 24.5 27.1 22.5 24.6 26.6 Escherichia coli (bacterium) Bacillus subtilis (bacterium) 24.6 28.4 24.3 29.0 25.5 21.0 25.6 21.6 c. Chargaff’s data 12 The Genetic Material • Chargaff’s Rules: • The amounts of A, T, G, and C in DNA: • Are constant among members of the same species • Vary from species to species • In each species, there are equal amounts of: • A and T • G and C • All of this suggests that DNA uses complementary base pairing to store genetic information. • Each human chromosome contains, on average, about 140 million base pairs. • The number of possible nucleotide sequences is 4140,000,000. 13 X-Ray Diffraction of DNA She found that if a concentrated, viscous solution of DNA is made, it can be separated into fibers. Under the right conditions, the fibers can produce an X-ray diffraction pattern Rosalind Franklin diffraction pattern diffracted X-rays a. X-ray beam Crystalline DNA b. c. A colleague of Franklins, Wilkins, showed her photo to James Watson, who understood its significance about DNA’s structure. 14 Watson and Crick Model of DNA 3.4 nm 0.34 nm 2 nm Double helix model b. Sugar-phosphate backbones Hydrogen-bonded bases make up the rungs. d. The two DNA strands are antiparallel. C a. G 5′ end sugar-phosphate backbone T 3′ end P Information stored in DNA must be read in the 5’ to 3’ direction so DNA is replicated in a 5’ to 3’ direction. A C G S P S A T P P T S 3′ end A Complementary base pairing – purine is always bonded to a pyrimidine (A with T, G with C). P S G P 5′ end They received a Nobel Prize in 1962. C P complementary base pairing c. C G P sugar hydrogen bonds 12.5a(DNA model): © Photodisk Red/Getty RF; 12.5d: © A. Barrington Brown/Science Source 15 Outline • 12.1 The Genetic Material • 12.2 Replication of DNA • 12.3 The Genetic Code of Life • 12.4 First Step: Transcription • 12.5 Second Step: Translation 16 12.2 Replication of DNA • DNA replication is the process of copying a DNA molecule. • Semiconservative replication: • Contains 1 parent strand and 1 daughter strand 17 Replication of DNA • Replication requires the following steps: • Complementary base pairing between a new nucleotide and a nucleotide on the template strand • DNA primase places short primers on the strands to be replicated. • Polymerase recognizes RNA and begins DNA synthesis. • The two strands are replicated differently: leading and lagging strands. • Joining of nucleotides in the lagging strand by DNA ligase form the new strand. • Each daughter DNA molecule contains one old and one new strand. 18 Replication of DNA • Replication requires 3 steps: • Unwinding • DNA helicase – unwinds and separates • Single-stranded binding proteins (SSB) – prevent separated DNA from reforming into helix • Complementary base pairing • • • • • Needs a primer – short strand of RNA DNA primase places short primers on the strands DNA polymerase recognizes primer and begins synthesis DNA polymerase also proofreads new strand and can make corrections Contains leading and lagging strands (replication of lagging strands are Okazaki fragments) • Joining • DNA polymerase converts RNA primer into DNA • DNA ligase joins all Okazaki fragments together Enzymes in DNA Replication 20 Replication of DNA • Prokaryotic Replication • • • • • Bacteria have a single circular loop of DNA. Replication moves around the circular DNA molecule in both directions. It produces two identical circles. The process begins at the origin of replication. Replication takes about 40 minutes, but the cell divides every 20 minutes. • A new round of replication can begin before the previous round is completed. 21 Replication of DNA • Eukaryotic Replication • DNA replication begins at numerous points along each linear chromosome. • DNA unwinds and unzips into two strands. • Each old strand of DNA serves as a template for a new strand. • Complementary base pairing forms a new strand paired with each old strand. • Requires enzyme DNA polymerase 22 Replication of DNA • Eukaryotic Replication • Replication bubbles spread bidirectionally, until they meet. • The complementary nucleotides are joined to form new strands. • Replication is semiconservative: ▪ Polymerase is unable to replicate the ends of chromosomes (telomeres). • Telomeres are copied by the enzyme telomerase. • Unregulated telomerase activity negatively affects cell function. 23 Prokaryotic versus Eukaryotic Replication Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. origin replication is complete replication is occurring in two directions a. Replication in prokaryotes replication fork replication bubble parental strand new DNA duplexes b. Replication in eukaryotes daughter strand 24 Replication of DNA • Accuracy of Replication • DNA polymerase is very accurate, yet makes a mistake about once per 100,000 base pairs. • Capable of identifying and correcting errors 25 Outline • 12.1 The Genetic Material • 12.2 Replication of DNA • 12.3 The Genetic Code of Life • 12.4 First Step: Transcription • 12.5 Second Step: Translation 26 12.3 The Genetic Code of Life