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F08 BIO 202 (Ch 5 pt 2)

Course: BIO 202, Spring 2008
School: SUNY Stony Brook
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Roles 9/9/2008 The of Nucleic Acids There are two types of nucleic acids: Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA) DNA is the genetic material that organisms inherit from their parents. DNA contains the information that cells require to function; genes that code for proteins DNA also provides directions for its own replication LE 5-25 DNA Genomic DNA Synthesis of mRNA in the nucleus mRNA...

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Roles 9/9/2008 The of Nucleic Acids There are two types of nucleic acids: Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA) DNA is the genetic material that organisms inherit from their parents. DNA contains the information that cells require to function; genes that code for proteins DNA also provides directions for its own replication LE 5-25 DNA Genomic DNA Synthesis of mRNA in the nucleus mRNA NUCLEUS CYTOPLASM mRNA Movement of mRNA into cytoplasm via nuclear pore RNA is required for gene expression. messenger RNA (mRNA) is transcribed from a gene, and in the cytoplasm it is translated into proteins with the help of ribosomal RNA and transfer RNA (rRNA and tRNA) Ribosome rRNA, tRNA Synthesis of protein Polypeptide Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Amino acids Nucleotide monomers Nucleic acids are polymers called polynucleotides Each polynucleotide is made of monomers called nucleotides Nucleotides have three components: (1) a nitrogenous (nitrogen containing) base (2) a pentose sugar (3) a phosphate Fig. 5-27 5end Nitrogenous bases Pyrimidines 5 C 3 C Nucleoside Nitrogenous base Cytosine (C) Thymine (T, in DNA) Uracil (U, in RNA) Purines Phosphate group 5 C 3 C 3end (a) Polynucleotide, or nucleic acid (b) Nucleotide Sugar (pentose) Adenine (A) Guanine (G) Sugars Deoxyribose (in DNA) Ribose (in RNA) Nucleoside = nitrogenous base + sugar Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings (c) Nucleoside components: sugars . Nucleotide Monomers There are two families of nitrogenous bases: Pyrimidineshave a single six-memberedring.Purineshave a six-membered ring fused to a five-membered ring Fig. 5-27c-1 Nitrogenous bases Pyrimidines Cytosine (C) Thymine (T, in DNA) Uracil (U, in RNA) Purines Adenine (A) Guanine (G) (c) Nucleoside components: nitrogenous bases Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 1 9/9/2008 Sugars Sugars are poly-hydroxyaldehydes or ketones. Among the most common in biology are the hexoses and pentoses: Glucose is a hexose containing 6 carbon atoms. Fig. 5-27c-2 Sugars Ribose is pentose containing 5 carbon atoms (RNA and DNA). Deoxyribose (in DNA) Ribose (in RNA) (c) Nucleoside components: sugars Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Each nitrogenous base is covalently linked to the ribose sugar through a bond denoted as the N- -glycosyl bond. RNA N- -glycosyl bond DNA Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Remember, RNA contains "U", DNA contains "T". Nucleotide NH2 O Nucleotide Polymers Pyrimidines O NH O N H Uracil (U) O H3C N O NH O Nucleotide polymers are linked together, building a polynucleotide Adjacent nucleotides are joined by covalent bonds that form between the OH group on the 3 carbon of one nucleotide and the phosphate on the 5 carbon on the next These links create a backbone of sugar-phosphate units with nitrogenous bases as appendages N 5 4 3 2 O P O O O Phosphate group 1 N Nitrogenous base N 5-carbon sugar H Cytosine (C) H Thymine (T) Ribose OH HO CH2 O 4 C H H 1C C3 2C H H OH OH 5 Deoxyribose HO CH2 O OH 4 C H H 1C C3 2C H H OH H 5 NH2 N N N Purines N O NH N NH2 N H N H Adenine (A) Guanine (G) Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 2 9/9/2008 LE 5-26a 5 end Nucleoside Nitrogenous base Phosphate group Nucleotide 3 end Pentose sugar Polynucleotide, or nucleic acid Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings The DNA Double Helix 5 T A C G A A G C T A G A C A C A Cartoon of DNA is a double helix. 3 5 T G C C G A A C G A T T C T A G A T T G A 3 5 3 A DNA molecule has two polynucleotides spiraling around an imaginary axis, forming a double helix In the DNA double helix, the two backbones run in opposite 5 3 directions from each other, an arrangement referred to as antiparallel One DNA molecule includes many genes The nitrogenous bases in DNA pair up and form hydrogen bonds: adenine (A) always with thymine (T), and guanine (G) always with cytosine (C) 5 Copyright 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings G C T T C G A T C T G T G T A C A 3 5 T G T 3 Cartoon of double helix 5 3 Copyright 2005 Pearson Education, pairing base Inc. publishing as Benjamin Cummings Space-filling model of double helix Proposed dimensions of the double helix The helical structure of DNA was predicted by Franklin Width of the helix 2.0 nm 3 S P A G S P C C 5 S S P The helical structure of DNA was first Predicted by R. Franklin. Using X-ray crystallography, she determined the dimensions of the DNA double helix Distance between bases 0.34 nm P S S G S C S PS P P P A S S T S P G S P C S C P S G P S Length of one complete turn of helix 3.4 nm T A S P P S G S P A C P S P S S P S G T S T S P P S P P S C G S Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3 S G C P S 5 3 9/9/2008 A-T and G-C base pairing were predicted by Chargaff The base composition of DNA varies from one species to another. DNA base pairing as predicted by Chargraff's rules 5 Cytosine H C H Sugar-phosphate backbone C C N H N H N H Guanine 3 H O C C N C N . In all cellular DNAs, regardless of the species: The number of adenosine residues is equal to the number of thymidine residues (i.e. A=T) The number of guanosine residues is equal to the number of cytidine residues (i.e. G=C). Thus, the sum of the purine residues equals the sum of the pyrimidine residues (i.e. A + G = T +C). C N C C O H N H H N Thymine CH3 C H C N C O H C N H N O H N Adenine H N C C C C N C N Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3 5 Watson and Crick proposed the structure of DNA Width of the helix 2.0 nm 3 5 Base pairing Bases from 2 strands are held together by hydrogen bonds (always A-T and C-G) C-G pairing is stronger than A-T pairing, and CG pairs are held closer than A-T pairs (this is due to the number of hydrogen bonds between them) A-T has 2 bonds, C-G has 3 bonds S P S S G P C C P A G S S Distance between bases 0.34 nm S P S C S P P S P P S T A S S P G S P C C P S G P S Length of one complete turn of helix 3.4 nm S T A S P P S P G S S T A C S P P S Purine-purine pair TOO WIDE Pyrimidine-pyrimidine pair TOO NARROW Purine-pyrimidine pair JUST RIGHT S P S P G S S T P P S P P S C G S S G C P S Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 5 3 Space inside sugarphosphate backbones Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Base pairing is used to: Preserve information (during DNA replication) Repair mistakes (during DNA replication) Transfer information (transcription, translation) One DNA strand serves as a template from which information is read (more about this later in the course) LE 5-27 5 end 3 end Sugar-phosphate backbone Base pair (joined by hydrogen bonding) Old strands Nucleotide about to be added to a new strand 5 end New strands 3 end 5 end Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 5 end 3 end 4 9/9/2008 RNA Primary and Secondary Structure RNA Primary structure: single strand of ribonucleotides linked by phosphodiester bonds. Secondary structure: stem-loop hairpins may form by internal H-bonding. Chemical differences between DNA and RNA BASES: Uracil vs. Thymine Uracil lacks the methyl group at position 5 SUGARS: Ribose vs. Deoxyribose Deoxyriboselacks OH oxygen at position 2 Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings RNA RNA can be both catalyst and template: Secondary structure of RNA Loop Stem An RNA strand can be a template for forming a new strand by base-pairing rules. Hairpin Catalytic RNAs occur in most (if not all) organisms. G G U A C C C A U G Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 3 5 Proteins have many structures, resulting in a wide range of functions Proteins account for more than 50% of the dry mass of most cells Protein functions include structural support, storage, transport, cellular communications, movement, and defense against foreign substances Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 5 9/9/2008 Amino Acid Monomers Cells use 20 different amino acids to make several hundred thousand different proteins. All amino acids have the same basic structure, but each has a unique functional group with distinct physical and functional properties 20 amino acids can produce 20n combinations in a polypeptide of length n For example, there are 2025 possible 25 amino acid combinations, and 20100 (1044) possible 100 amino acid combinations, etc Only the most stable and functional combinations of amino acids, ie proteins, have been conserved through evolution. Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings All amino acids have the same general structure. Remember the periodic table and electron configuration of Non-ionized form 6C, 7N, Amino acids are ionized at physiological pH and 8O. Ionized form H Amino group O C OH Carboxylic acid group Amino group H H3N+ C R Side chain O C O Carboxyl group H2N C R Side chain Amino acids are organic molecules with carboxyl and amino groups Amino acids differ in their properties due to differing side chains, called R groups Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings The carboxyl group is acidic. When it donates a proton, it becomes negatively charged. The amino group is basic. When it accepts a proton, it becomes positively charged. Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Each amino acid has a different side chain. H H3N + Some side chains contain ring structures H O H H3N+ O O O H H3N+ O H H3N+ O H3N+ H O H H3N+ O H2N+ H O H H3N+ C CH2 NH O O C C H C C CH3 O C C O CH H3C CH3 Valine (V) Val C C O CH2 CH H3C CH3 C C C C O O H3C CH H2C CH2 CH2 CH2 CH3 Proline (P) Pro H3N+ C CH2 C C CH2 C O C O Glycine (G) Gly Alanine (A) Ala Leucine (L) Leu Isoleucine (I) Ile OH Hydrophobic, nonpolar R groups Phenylalanine (F) Phe Tyrosine (Y) Tyr Tryptophan (W) Trp Side chains contain carbon and/or hydrogen Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Non-polar polar, uncharged non-polar Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 6 9/9/2008 Some side chains contain hydroxyl groups side Some chains contain sulfur H H3N+ C CH2 CH2 S CH3 Methionine (M) Met C O O H3N+ H C CH2 SH C O H3N+ O H C CH2 OH C O H3N+ O HO H C CH CH3 C O O Sulfur has an atomic number of 16; 6 valence electrons (like oxygen) Cysteine (C) Cys Serine (S) Ser Threonine (T) Thr Polar, uncharged polar, uncharged Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Some side chains are basic Some side chains are acidic H H3N+ C C O NH + NH O H3N+ H C C O H3N+ O H C C O H3N+ O H C CH2 C O O C O H H3N+ C CH2 CH2 C O O CH2 CH2 CH2 CH2 CH2 + NH 3 CH2 CH2 CH2 NH C NH2 O O C O + NH 2 Aspartate (D) Asp Glutamate (E) Glu Histidine (H) His Lysine (K) Lys Arginine (R) Arg - charged + charged Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Some side chains contain amide groups TABLE 3.1 HOW AMINO ACIDS INTERACT WITH WATER In this table, the 20 amino acids are ranked according to how likely they are to interact with water. The ranking is from least likely to most likely. H H3N+ C CH2 C H2N O C H O H3N+ O C CH2 CH2 C H2N O C Highly hydrophobic O O Least likely Isoleucine Valine Leucine Phenylalanine Methionine Uncharged, Non-polar Less hydrophobic Alanine Glycine Cysteine Tryptophan Tyrosine Proline Threonine Serine Interaction with water Asparagine (N) Asn Glutamine (Q) Gln Highly hydrophilic Histidine Glutamate Asparagine Glutamine Aspartate Lysine Arginine Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Polar, uncharged More charged Or polar Most likely Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 7 9/9/2008 A protein is a polymer of amino acids linked to one another by peptide bonds H H2N C C The protein sequence is written from the N-terminal to the C-terminal amino acid. O + H2N Amino group H C C O H 2N OH H O H N H C C O + OH H2O C C H OH H Carboxyl group CH3 Peptide bond CH3 N-terminus H H N H C H O C H N H C CH3 O C H N H C CH2 OH O C H N H C CH2 C O OH OH O C H N H C CH2 H3C O C H N H C CH CH 3 O C H N H C CH2 O C H N H C C-terminus O C OH Peptide bond formation releases a water molecule, so it is called dehydration or condensation. CH2 SH N-terminus C-terminus Gly Ala Ser Asp Phe Val Tyr Cys H2N COOH 1 2 3 4 5 6 7 8 Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings The sequence is also referred to as the primary structure How is the primary structure (amino acid sequence) of a protein determined? The amino acid sequences of polypeptides were originally determined by chemical methods, which involved hydrolyzing the protein into multiple sets of smaller peptide fragments, and then determining the amino acid composition of each individual fragment*. Most of the steps involved in sequencing a polypeptide are now automated Now, amino acid sequence is typically predicted from the DNA sequence of the corresponding gene. Protein Conformation and Function A functional protein is not a linear molecule, instead it consists of one or more polypeptides twisted, folded, and coiled into a unique shape The sequence of amino acids (primary structure) determines a protein's three-dimensional conformation (secondary, tertiary, and quarternary structure) A protein's conformation determines its function * Sanger, Nobel prize Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Secondary Structure pleated sheet Amino acid subunits helix Because both the O and the N are electronegative The coils and folds of secondary structure result from hydrogen bonds between repeating constituents of the polypeptide backbone (C=O....H-N) Typical secondary structures are a coil called an alpha () helix, and a folded structure called a beta () pleated sheet Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 8 9/9/2008 These zig-zag in a flat plane, or turn, like a barrel Between every 4th amino acid Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Ribbon models and space-filling models can depict a protein's conformation In addition to these regular structures (alpha helix and beta pleated sheet), Thereare often regions that have more random or unstructured coils and loops. These region often bridge more structured regions, and provide flexibility. A ribbon model Groove Many proteins assume multiple conformations that influence their activity. For example, most enzymes are globular proteins with many short helical segmentsoriented in different directions, connected by random coils. Groove A space-filling model Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Tertiary structure is determined by interactions among various side chains (R groups) rather than interactions between backbone constituents These interactions between R groups include hydrogen bonds, ionic bonds, hydrophobic interactions, and van der Waals interactions Strong covalent bonds called disulfide bridges may reinforce the protein's conformation Hydrophobic interactions and van der Waals interactions Polypeptide backbone Hydrogen bond Disulfide bridge Ionic bond Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 9 9/9/2008 Quaternary structure results when two or more polypeptide chains form one macromolecule Polypeptide chain Chains Iron Heme Chains Hemoglobin Polypeptide chain Collagen Collagen is a fibrous protein consisting of three polypeptides coiled like a rope. The alpha helix runs the entire length of the protein. Hemoglobin is a globular protein consisting of four polypeptides Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Sickle-Cell Disease: A Simple Change in Primary Structure A slight change in primary structure can affect a protein's conformation and ability to function Sickle-cell disease, an inherited blood disorder, results from a single amino acid substitution in the protein hemoglobin 10 m 10 m LE 5-21b Normal hemoglobin Primary structure Secondary and tertiary structures Val His Leu Thr Pro Glu Glu Sickle-cell hemoglobin 7 Primary structure Secondary and tertiary structures Val His Leu Thr Pro Val Glu 1 2 3 4 5 6 1 2 3 4 5 6 7 subunit Exposed hydrophobic region subunit Quaternary Normal hemoglobin structure (top view) Quaternary structure Sickle-cell hemoglobin Red blood cell shape Normal cells are full of individual hemoglobin molecules, each carrying oxygen. Red blood cell shape Fibers of abnormal hemoglobin deform cell into sickle shape. Function Molecules do not associate with one another; each carries oxygen. Function Molecules interact with one another to crystallize into a fiber; capacity to carry oxygen is greatly reduced. Hydrophobic regions "hide" from water by aggregating with other hydrophobic regions. Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings What Determines Protein Conformation? In addition to primary structure, physical and chemical conditions can affect conformation Alternations in pH, salt concentration, temperature, or other environmental factors can cause a protein to unravel This loss of a protein's native conformation is called denaturation A denatured protein is biologically inactive (why fevers are deadly, also why we preserve food by pickling, salting, changes in temperature) Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings LE 5-22 Denaturation Normal protein Renaturation Denatured protein 10 9/9/2008 The Protein-Folding Problem Scientists use X-ray crystallography and nuclear magnetic resonance (NMR) spectroscopy to determine a protein's conformation. It is hard to precisely predict a protein's conformation solely from its primary structure, but scientists have developed algorithms based on the conformation of wellcharacterized proteins. LE 5-24a X-ray diffraction pattern Photographic film Diffracted X-rays X-ray source X-ray beam Crystal Computers predict the 3D structure of a protein based on X-ray diffraction data plus the known primary amino acid sequence. Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings LE 5-24b The Protein-Folding Problem Nucleic acid Protein Most proteins probably go through several intermediate states on their way to their most stable conformation Chaperonins are protein molecules that assist the proper folding of other proteins Protein mis-folding has been implicated in cystic fibrosis, Alzheimers, Parkinsons. X-ray diffraction pattern 3D computer model Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings LE 5-23b Polypeptide Correctly folded protein Prions are mis-folded proteins that encourage the mis-folding and aggregation of other proteins. Prions are implicated in mad cow disease, scrapie (sheep), Kuru, and Creutzfeldt-Jacob disease. Steps of Chaperonin Action: An unfolded polypeptide enters the cylinder from one end. The cap attaches, causing the cylinder to change shape in such a way that it creates a hydrophilic environment for the folding of the polypeptide. The cap comes off, and the properly folded protein is released. Normal prion protein contains about 40% -helix and little -sheet, whereas misfoldedprion protein is composed of about 30% -helix and 45% -sheet Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings 11 9/9/2008 The Theme of Emergent Properties in the Chemistry of Life: A Review Higher levels of organization result in the emergence of new properties Organization is the key to the chemistry of life Fig. 5-UN2a Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Fig. 5-UN2b Fig. 5-UN9 Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings You should now be able to: 1. List and describe the four major classes of molecules 2. Describe the formation of a glycosidic linkage and distinguish between monosaccharides, disaccharides, and polysaccharides You should now be able to: 5. Distinguish between the following pairs: pyrimidine and purine, nucleotide and nucleoside, ribose and deoxyribose, the 5 end and 3 end of a nucleotide 3. Distinguish between saturated and unsaturated fats and between cis and trans fat molecules 4. Describe the four levels of protein structure Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings Copyright 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings 12
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UC Davis - NPB - 101
UC Davis - NPB - 101
1. Consider a sample of normal human blood that contains approximately 60% plasma. This blood would a. have a hematocrit of about 40. b. would contain about 40% erythrocytes by volume c. would contain about 40% platelets by volume. d. Would con
UC Davis - NPB - 101
1. The lipid bilayer of the cell membrane a. is due primarily to the presence of cholesterol. b. is due to the presence of hydrophilic and hydrophobic groups on the membranes triglycerides. * c. is formed as a result of the arrangement of pho
UC Davis - NPB - 101
NPB 101 (5 Units) Fall 2008 Systemic Physiology Professors: W. Jeffrey Weidner, PhD (Instructor-in- Charge) 195 Briggs Hall (752-3208) wjweidner@ucdavis.edu John C. Wingfield, PhD. 294 Briggs Hall (752 4679) jcwingfield@ucdavis.edu Teaching As
UC Davis - NPB - 101
NPB 101, Autumn 2008.MIDTERM 1. Outline of topics you need to focus on for WINGFIELD lectures for midterm 1. Questions will come from powerpoints NPB.101.Lectures2.2008 to NPB.101.Lectures7.2008. Questions on the midterm will be multiple choice (see
UC Davis - BIS - 101
BIS101-003/Engebrecht Homework01 9/25/08 Suggested problems and homework for week #1 Problems suggested for the whole class these are not to be turned in, but are for practice/study aid. The following problems can be done after the 1/10 lecture: 8th
UC Davis - BIS - 101
Homework assignment #1 Due in your session, the week of 10/2/08-10/8/08 3 problems 1. (16pts) In mice, a recessive mutation in gene T results in tail-less animals and a second unlinked recessive mutation in gene F results in fat animals. Indicate th
UC Davis - BIS - 101
BIS101-003/Engebrecht Homework02 10/02/08 Suggested problems and homework for week #2 Problems suggested for the whole class these are not to be turned in, but are for practice/study aid. The following problems can be done after the meiosis lecture:
UC Davis - BIS - 101
Homework02key2. You have obtained two true-breeding strains of mice, each homozygous for an independently discovered recessive mutation that prevents the formation of hair on the body. The discoverer of one of the mutant strains calls his mutation
UC Davis - BIS - 101
BIS101-003/Engebrecht 10/16/08 Suggested problems and homework for week #3 Problems suggested for the whole class these are not to be turned in, but are for practice/study aid. The following problems can be done after the linkage lectures: Ch. 4: in
UC Davis - BIS - 101
BIS101-003/Engebrecht 1. In a testcross of a female Drosophila heterozygous for three linked recessive genes, the following phenotypes of progeny were obtained:a+ b+ d a b d+ a+ b d+ a b+ d a+ b d a b+ d+ a+ b+ d+ a b d Total442 435 12 11 52 46 1
UC Davis - BIS - 101
BIS101-003/Engebrecht Suggested problems and homework for week #4 Problems suggested for the whole class these are not to be turned in, but are for practice/study aid. The following problems are suggested for viral transduction: Chapter 5 in both 8t