Uploaded by oweniogiri

Cell Biology & Chemistry Homework: Principles Based Problems

advertisement
Ch1 Homework A (Principles Based Problems)
15 questions
Course Info
Instructor Name
Student Name
Question 1 of 15
A eukaryotic cell contains many compartmentalized organelles that organize its interior. Each organelle is a discrete unit
enclosed by a membrane and has a specialized function. Other structures, such as ribosomes or the cytoskeleton, are not
enclosed by a membrane.
The generalized animal cell shown is colored so that the features can easily be distinguished from each other. Label each
structure with its function. Choose the best answer for each structure.
Answer Bank
storage of
genetic information
protein transport out of cell
protein synthesis
cellular respiration;
ATP synthesis
membrane synthesis;
protein modification
© Macmillan Learning
Question 2 of 15
How does the surface‑to‑volume ratio affect the rate of cellular metabolism?
Metabolic rate is limited by surface area.
The rate of metabolism is maximized when volume is maximized.
Larger surface‑to‑volume ratios equate to slower metabolic rates.
The rate of metabolism does not depend on the surface‑to‑volume ratio.
The spherical bacterium Neisseria gonorrhoeae is responsible for the disease gonorrhea. Calculate the surface‑to‑volume ratio,
given that the diameter of the bacterium is
is
4 3
𝜋𝑟 .
3
0.5 𝜇m. Note that the surface area of a sphere is 4𝜋𝑟2 and the volume of a sphere
𝑆
:
𝑉
µm−1
How many times greater is the surface‑to‑volume ratio of Neisseria gonorrhoeae compared to that of a globular amoeba, a large
eukaryotic cell with a diameter of
150 𝜇m?
times greater:
© Macmillan Learning
Question 3 of 15
Complete the table, which illustrates the energy sources and carbon sources for various metabolic strategies. Then, complete
the sentences.
Answer Bank
chemoautotroph
photoheterotroph
chemoheterotroph
bear
photoautotroph
dandelion
must obtain at least some carbon from organic materials.
include organisms that can use sunlight to generate energy.
can fix atmospheric carbon dioxide.
© Macmillan Learning
Question 4 of 15
Which statements identify differences between proteomics and genomics?
Genomics starts at the DNA level and makes inferences about the product, whereas proteomics begins with
the product.
Microarrays are used for the analysis of DNA in genomics, but cannot be utilized in proteomics.
Proteomics involves the study of genes, whereas genomics involves the study of proteins.
Mass spectrometry is a technique that is used in proteomics, but not in genomics.
Genomics is used to identify the entire human genome, whereas proteomics is only used to identify specific
proteins in organisms.
© Macmillan Learning
Question 5 of 15
Assume you have a crude lysate sample that you obtained from mechanically homogenizing E. coli cells. You centrifuged the
supernatant from the sample at a medium speed
supernatant at high speed
(80,000 𝑔) for 1 h.
(20,000 𝑔) for 20 min, collected the supernatant, and then centrifuged the
What procedure should you follow to isolate the ribosomes from this sample?
Layer the
from the high‑speed centrifugation over
medium. Centrifuge at a very high
speed (150,000 g) for 3 h and collect the
© Macmillan Learning
Question 6 of 15
Ibuprofen is an over‑the‑counter drug that blocks a class of prostaglandins that cause inflammation and pain.
Ibuprofen is available as a racemic mixture of (R)‑ibuprofen and (S)‑ibuprofen. In living organisms, an isomerase catalyzes the
chiral inversion of the (R)‑enantiomer to the (S)‑enantiomer. The reverse reaction does not occur at an appreciable rate.
In the figure, sites a, b, and c represent binding sites in the isomerase enzyme that converts the (R)‑enantiomer to the
(S)‑enantiomer. All three sites recognize the corresponding functional groups of the (R)‑enantiomer of ibuprofen. However,
sites a and c do not recognize the corresponding functional groups of the (S)‑enantiomer of ibuprofen.
Place the substituent groups to represent the (R)‑enantiomer and the (S)‑enantiomer.
Answer Bank
CH3
The (S)‑enantiomer of ibuprofen is
H
COO −
100 times more efficacious for pain relief than is the (R)‑enantiomer. Drug companies
sometimes make enantiomerically pure versions of drugs that were previously sold as racemic mixes, such as esomeprazole
(Nexium) and escitalopram (Lexapro).
Since (S)‑ibuprofen is more effective, why do drug companies not sell enantiomerically pure (S)‑ibuprofen?
The presence of the (R)‑enantiomer prevents chiral inversion, and inactivation, of the (S)‑enantiomer.
The conversion molecule binds the (S)‑enantiomer and keeps it out of circulation.
It is unnecessary because an enzyme converts the less effective enantiomer to the effective enantiomer.
It is unnecessary because both enantiomers relieve pain at equal levels in a racemic mixture.
© Macmillan Learning
Question 7 of 15
Three important biomolecules are depicted in their ionized forms at physiological pH.
You may need to reference a table (https://sites.google.com/site/chempendix/amino-acids) of amino acid structures.
Guanosine triphosphate (GTP) is an energy‑rich nucleotide that serves as a precursor to RNA.
Methionine enkephalin is an endogenous compound in the human brain, sometimes called the brain's own opiate.
Phosphatidylcholine is a component of many membranes.
Classify the chemical consituents that are part of each molecule.
Guanosine triphosphate (GTP)
Methionine enkephalin
Phosphatidylcholine
Answer Bank
glycerol
one phosphate
three phosphoryl groups
palmitate
guanine
methionine
choline
α‑D‑ribose
two glycines
phenylalanine
© Macmillan Learning
Question 8 of 15
Researchers isolated an unknown substance, X, from rabbit muscle. They determined its structure from the following
observations and experiments.
(a) Qualitative analysis showed that X was composed entirely of C, H, and O. A weighed sample of X was completely oxidized,
H2 O and CO2 produced were measured. This quantitative analysis revealed that X contained 40.00% C, 6.71% H, and
53.29% O by weight.
and the
(b) The molecular mass of X, as determined by mass spectrometry, was
90.00 atomic mass units (u).
(c) Infrared spectroscopy showed that X contained one double bond.
(d) X dissolved readily in water, and the solution demonstrated optical activity when tested in a polarimeter.
(e) The aqueous solution of X is acidic.
What is the empirical formula of X?
What is the structure of X?
© Macmillan Learning
Question 9 of 15
Use the diagram which represents a spontaneous reaction to
answer the questions.
Is the reaction endothermic or exothermic?
exothermic
endothermic
What is the activation energy of the reaction?
activation energy:
kJ/mol
© Macmillan Learning
Question 10 of 15
The diagram represents a spontaneous reaction. Use the
diagram to answer the questions.
Is the reaction endothermic or exothermic?
endothermic
exothermic
What is the activation energy of the reaction?
𝐸a =
kJ/mol
© Macmillan Learning
Question 11 of 15
Each beaker contains two types of particles, represented as different colors. Arrange the three beakers and their contents in order
of increasing entropy.
Greatest entropy
Least entropy
Answer Bank
© Macmillan Learning
Question 12 of 15
Which of the reactions are spontaneous (favorable)?
C(s) + H2 O(l) ⟶ CO(g) + H2 (g)
Δ𝐺 = 90.8 kJ/mol
AgCl(s) ⟶ Ag+ (aq) + Cl− (aq)
Δ𝐺 = 55.6 kJ/mol
2 H2 O(g) ⟶ 2 H2 (g) + O2 (g)
Δ𝐺 = 457 kJ/mol
2 Mg(s) + O2 (g) ⟶ 2 MgO(s)
Δ𝐺 = −1137 kJ/mol
CH4 (g) + 2 O2 (g) ⟶ CO2 (g) + 2 H2 O(l)
Δ𝐺 = −820 kJ/mol
NH3 (g) + HCl(g) ⟶ NH4 Cl(s)
Δ𝐺 = −91.1 kJ/mol
© Macmillan Learning
Question 13 of 15
For the reaction
2 NH3 (g) −
↽⇀
− 3 H2 (g) + N2 (g)
the equilibrium concentrations were found to be
[NH3 ] = 0.250 M, [H2 ] = 0.360 M, and [N2 ] = 0.800 M. What is the
equilibrium constant for this reaction?
𝐾 eq =
© Macmillan Learning
Question 14 of 15
Suppose that the gene for a protein
500 amino acids in length undergoes a mutation. If the mutation causes the synthesis of a
mutant protein in which just one of the
500 amino acids is incorrect, the protein may lose its biological function.
How can a small change in a protein’s sequence inactivate the protein?
The active site may base pair incorrectly with the complementary DNA strand.
The catalytic site of the protein may be malformed, preventing complementary fit.
RNA may not recognize the altered DNA sequence, and bases may pair incorrectly.
It may alter the shape of the protein’s molecular chaperone, preventing proper protein folding.
The resulting protein may fold incorrectly and may not attain its native conformation.
© Macmillan Learning
Question 15 of 15
Tardigrades, also called water bears or moss piglets, are small animals that
can grow to about
0.5 mm in length. Terrestrial tardigrades typically live in
the moist environments of mosses and lichens. Some of these species are
capable of surviving extreme conditions.
Some tardigrades can enter a reversible state called cryptobiosis, typically
triggered by water loss, in which metabolism completely stops. In this state,
various tardigrade species have withstood dehydration, extreme temperatures
from
−200 °C to +150 °C, pressures from 6000 atm to a vacuum, anoxic
conditions, and the radiation of space. When conditions become hospitable,
and when water is available, the animal becomes active again.
Photo credit: Schokraie E, Warnken U, Hotz-Wagenblatt A, Grohme MA, Hengherr S, Förster F, et al.
(2012) Comparative proteome analysis of Milnesium tardigradum in early embryonic state versus
adults in active and anhydrobiotic state. PLoS ONE 7(9): e45682. https://doi.org/10.1371/journal.pone.0045682 (http://)
Do tardigrades in an ametabolic state meet the definition of life?
No, metabolism and energy use are prerequisites for life.
No, entropy will increase without energy input from normal body function.
Yes, biochemical mechanisms exist for restoring the normal state.
Yes, an animal in a cryptobiotic state can still reproduce by spores.
Yes, DNA replication and maintenance still takes place.
© Macmillan Learning
Download