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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
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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
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(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
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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
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Each nitrogenous base is covalently linked to the ribose sugar through a bond denoted as the N- -glycosyl bond.
RNA
N- -glycosyl bond
DNA
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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)
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Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings
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LE 5-26a 5 end
Nucleoside Nitrogenous base
Phosphate group Nucleotide
3 end
Pentose sugar
Polynucleotide, or nucleic acid
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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
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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
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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
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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
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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
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Space inside sugarphosphate backbones
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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
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5 end 3 end
4
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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
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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
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Copyright 2005 Pearson Education, Inc. publishing as Benjamin Cummings
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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
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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.
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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
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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.
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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
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Non-polar
polar, uncharged
non-polar
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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
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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
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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
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Polar, uncharged
More charged Or polar
Most likely
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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
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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
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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
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These zig-zag in a flat plane, or turn, like a barrel
Between every 4th amino acid
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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
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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
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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
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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.
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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)
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LE 5-22
Denaturation
Normal protein Renaturation
Denatured protein
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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.
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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
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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
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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
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Fig. 5-UN2b
Fig. 5-UN9
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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
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Physics 2C Quiz 4A Name_MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A plant hangs from a 1cm diameter suction cup affixed to a smooth horizontal surface. What is the maximum weight that
UCSD - PHYS - 2c
Exam 5A Name_MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) If the pressure amplitude of a sound wave doubles, what happens to the decibel level? A) -9.5dB B) -5dB C) 0dB D) 5dB E) 10dB 2)
UCSD - PHYS - 2c
Exam 6a Name_MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) What is the speed of light in a material for which the critical angle with air is 61 degrees? A) 0.8c B) 0.9c C) c D) 1.1c E) 1.2
UCSD - PHYS - 2c
Exam 7a Name_MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) I am wearing glasses to correct my near-sightedness. Which of the following statements most accurately describes the lenses in my
UCSD - PHYS - 2c
Exam 8A Name_MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A double slit system is used to measure the wavelength of light. The system has slit spacing d=15 micron and slit-to-screen dista
UCSD - PHYS - 2c
Exam 9a Name_MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A laser beam shines vertically upwards. What laser power is needed for this beam to support a flat piece of aluminum foil with ma
UCSD - PHYS - 2c
Physics 2C Final Exam A Name_MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Two mols of an ideal gas undergo a reversible adiabatic expansion from 0.0280 cubic meters to 0.0420 cubic meters
UCSD - MATH - 20E
Practice Midterm ExaminationInstructor J. Verstraete Time: 40 minutes No notes allowed All questions carry equal weightQuestion 1.State precisely the - denition of limxa f (x) = L for a function f : Rn R. Then prove using the - denition of limit
UCSD - MATH - 20E
Practice Midterm ExaminationInstructor J. Verstraete Time: 40 minutes No notes allowed All questions carry equal weightQuestion 1.Prove that sin(xy) (x,y)(0,0) x2 + y 2 limdoes not exist.1Question 2.State the denition of dierentiability of
UCSD - MATH - 20E
Practice Midterm ExaminationInstructor J. Verstraete Time: 40 minutes No notes allowed All questions carry equal weightQuestion 1.(a) Show that xy 1 (x2 + y 2 ). 2 (b) Use part (a) and the - denition of limits to show(x,y)(0,0)limxy x2 + y
UCSD - MATH - 20E
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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