Download Biochemistry (Unit 1) Exam Review

Biochemistry (Unit 1) Exam Review
1.
Jessica
Fluid Mosaic Model
The cell membrane is selectively permeable, allowing for the uptake of nutrients and the expulsion of
wastes. Describe what would be visible in the cell membrane that has undergone the freeze-fracture
technique.
Answer:
The phospolipid bilayer would be split, exposing the inner and outer phospholipid layers. Found in these
layers would be various proteins (both embedded and 'free floating'); cholesterol; glycoproteins;
glycolipids; aquaporins; and other 'free-floating' molecules.
2.
Hauna
QUESTION: Using your knowledge about proteins, differentiate primary structure from secondary
structure and briefly describe the structures of globular and fibrous proteins.
ANSWER: Primary structure is the number and sequence of amino acids in a polypeptide strand. These
polypeptides usually consist of between 50 to 100 amino acids. The primary structure is determined by the
nucleotide sequence of the gene that codes for the polypeptide.
Secondary structure is the regular, repeating coils and foils in a polypeptide which is caused by hydrogen
bonds between hydrogen and oxygen atoms in the main chain of the polypeptide. If the shape of the
secondary structure is coiled, it is called an alpha-helix. If the shape of the secondary structure is folded, it
is called a
pleated sheet.
Fibrous proteins are long and narrow, while globular proteins are rounded and spherical.
3. Noah
Enzymes are proteins that catalyze biochemical reactions. Describe 3 factors that change the rate of enzyme
activity.
How
are
these
enzymes
activated
and
regulated?
temperature, ph and inhibitors all affect enzyme activity because each enzyme has optimum conditions.
Some enzymes are limited by inorganic cofactors that bind to the substrate or active site.
enzymes regulated by competitive inhibitors that adhere to the substrate or enzyme and prevent it from
binding by blocking it. non-competitive inhibitors bind to the enzyme and change its shape making it
inactive
and
unable
to
bind
to
the
substrate.
This
is
allosteric
regulation.
feedback inhibition occurs when one of the products formed later on in a metabolic pathway allosterically
inhibits an enzyme required earlier on in the metabolic process.
4.
Rae
Nucleic Acid Question:
Compare and contrast the two nucleic acids studied in this course in terms of function, structure, etc.
Answer:
Differences:
Ribonucleic acid (RNA)
1. Used in the "reading" of DNA information.
2. Short sections
3. Single stranded
4. Sugar: Ribose sugar
5. Pyrimidines: cytosine, uracil
Deoxyribonucleic acid (DNA)
1. Contains genetic information for cellular function and heredity.
2. Long coils
3. Double stranded (strands run anti parallel).
4. Sugar: Deoxyribose sugar
5. Pyrimidines: cytosine, thymine.
Similarities
-Both are made up of nucleotides, which is composed of a phosphate group, a sugar, and a base.
-Bases is a pyrimidine or a purine
-Purine is either adenine and guanine
-Phosphodiester bonds hold the nucleotides together
5.
Isidora
Gibbs free energy
1. What will the Gibbs energy be(negative or positive) when ΔH is positive and ΔS is also positive and in
what circumstances will this reaction occur?
a)- ΔG(all the time) b) +ΔG(never) c)?ΔG(in low temperatures) d) not enough information e)?ΔG(?)
2. What is the value of ΔG if ΔH= -30.0kJ, ΔS=22.0kJ and T= 290K will this reaction be spontaneous?
a)6.4x103 and will never happen b)-6.41x103KJ and will always happen c)-6.41x103 and will always
happen d)-6.0x103KJ and will not happen e)-6410KJ and will always happen
3.If ΔG= -2.00x103KJ , ΔH=-45.0KJ, and T= 240K what is ΔS?
a)8.52 J b)-8.52KJc)not enough information d)8.52KJ
Answer key
1.c) 2.b) 3.d)
6. Chelsea
Facilitated Diffusion
Question: A scientist is conducting an experiment in which a glucose molecule is trying to enter the
cytoplasm of a cell. Explain whether or not the molecule is able to enter through the use of facilitated
diffusion and why it will or will not work.
Answer:
- Facilitated diffusion is used to diffuse proteins through the cell membrane. Glucose is too large and is
hydrophobic so it would not be able to embed itself in the phospholipid bilayer, meaning it could not be
carried into the cell membrane.
7. Max
Biological Macromolecules: Carbohydrates
Q1:
a). Two sugar monomers have joined together, forming a disaccharide. Name the bond that links
these two molecules together, and what type of bond this is. In addition, name the reaction that
forms this bond, the byproduct that is produced from said reaction, and the reaction that breaks
this bond.
b). Name the disaccharides formed from the bonding of the following monomers:
glucose + fructose
glucose + galactose
glucose + glucose
A1:
a). The bond that joins the two sugar monomers together is called a glycosidic linkage, which is a
covalent bond. A disaccharide is formed through a condensation reaction, resulting in the release
of H2O. The reaction that breaks a disaccharide into two individual monosaccharides is called a
hydrolysis reaction, which occurs through the addition of H2O.
b). glucose + fructose = sucrose
glucose + galactose = lactose
glucose + glucose = maltose
Q2:
Compare and contrast the structure and function of the polysaccharides starch and cellulose.
A2: (various possible answers)
Starch:
-
Cellulose:
Functions as energy storage in plants.
Its structure is a polymer chain of alpha-glucose molecules
that contains a mixture of amylose (an unbranched chain of
α-glucose held together by α 1-4 glycosidic linkages), and
amylopectin (a branched chain of α-glucose also held
together by α 1-4 glycosidic linkages, and with chains
branching off through α 1-6 glycosidic linkages).
Is easily hydrolyzed into soluble sugars by enzymes, making
it a good source of energy for plants as well as animals.
Can be found in plant cells in the form of concentrated and
insoluble starch granules.
-
-
8.
Functions as structural support in plants.
Made up of unbranched chains of β-glucose joined together
by β 1-4 glycosidic linkages (in which every other monomer
of glucose is inverted, contrary to α 1-4 glycosidic linkages,
in which there is no inversion of the monomers). Parallel
chains are crossed-linked together with hydrogen bonds to
form bundles called microfibrils.
Cellulose microfibrils are a key component in plant structure
and are very strong, and therefore are not easily digested by
most animals.
Salmaan
1. What is the purpose of active transport? What primary substance is used for this purpose and how?
- The purpose of active transport is to move substances in and out of cells, from areas of low concentration
to areas of high concentration. ATP is used for this purpose as the high energy phosphate bonds are broken,
thus releasing the required energy.
2. In the sodium-potassium pump, for each ______ NA+ ions pumped _____ the cell, _____ K+ ions are
pumped _____ the cell.
a) 3, out of, 2, into
b) 2, out of, 3 into
c) 3, into, 2, out of (ANSWER)
d) 2, into, 3, out of
3a. Match the form of active transport to its description:
i) Pinocytosis
ii) Phagocytosis
iii) Receptor-mediated
a) Particles bind to receptors
b) Ingestion of large substances
c) "Cell drinking"
ANSWERS: ic,iib,iiia
3b. What type of active transport do these 3 fit under? _____
Answer: Endocytosis
9.
Halim
What is the difference between a saturated and unsaturated fatty acids?
Saturated fatty acids - Contain the maximum number of hydrogen atoms.
Unsaturated fatty acid - Some double bonded carbon atoms and are not fully saturated with hydrogens.
What is the difference between a phospholipid and triglyceride?
Phospholipid- consist of glycerol attached to two fatty acid chains and a phosphate group (with a polar
nitrogen group attached to the end
Triglyceride- forms when one glycerol bonds with three fatty acid tails
How are fats, oils and waxes different?
Fats- all three fatty acid tails are saturated
Oils- at least one fatty acid tail that is unsaturated
Waxes- Instead of a glycerol backbone, a large, long-chained alcohol backbone is used which is attached to
the three fatty acid tails
10. Peter
Compare and contrast the cell structure of a prokaryotic cell and a eukaryotic cell.
Differences:
Prokaryote
No nucleus
One chromosome (plasmid)
Unicellular
Mitochondria absent
Eukaryote
Nucleus
Multiple chromosomes
Multi-cellular
Mitochondria present
Similarities:
-
Vesicles present
Vacuoles present
Contain DNA
11. Dusty
Road salt leaches into a nearby pond in the spring. Using Osmosis explain the change in the water and
its effect on unicellular organisms.
The road salt would dissolve, raising the solute concentration of the water. This would make the
organisms hypertonic to the water because of the added solute from the road salt. Through osmosis the
water in the cells will follow the concentration difference and move from the low solute concentration
in the cell to the high solute concentration found outside the cell, causing the cell to shrivel and die
12. Atosa – Endomembrane System:
Question: Explain why the endomembrane system is referred to as a “system”, rather than as distinct
parts.
Possible Answer: The organelles of the endomembrane system cooperate with one another to carry
out a variety of tasks in the cell. Whether it's to synthesize proteins, transport molecules into or out of
the membrane, control the movement of lipids, etc. the organelles are related either through direct
contact or through tiny vesicles that connect the contents from one endomembrane structure to
another.