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Chapter 1 1)
In an aerobic bacterium, where do you think most of the proteins responsible for cellular respiration are located? In the plasma membrane. According to the theory of mitochondrial origin outlined in this chapter, the plasma membrane of the engulfed bacterium would become the inner mitochondrial membrane, where most of the proteins involved in cellular respiration are located.
2)
The protozoan Didinium feeds on other organisms by engulfing them. Why are bacteria, in general, unable to feed on other cells in this way? Didinium engulfs prey by changing its shape, and for this it uses its cytoskeleton. Bacteria have no cytoskeleton and cannot easily change their shape because they are generally surrounded by a tough cell
wall.
Chapter 4 1)
Some of the enzymes that oxidize sugars to yield usable cellular energy (for example, ATP) are regulated by phosphorylation. For these enzymes, would you expect the inactive form to be the phosphorylated form or the dephosphorylated form? Explain your answer.
In general, the inactive form is the phosphorylated form. The main purpose of glycolysis and the citric acid cycle is to generate ATP; thus, the enzymes are inactive when the concentration of ATP is high and active when it is low. It makes sense that cells would
not want to have to phosphorylate their enzymes to turn them on when ATP levels are already low, because phosphorylation requires ATP.
2)
Using the example of the p53 protein, explain how different combinations of covalent modifications can lead to a wide variety of protein functions.
In a protein with a complex regulatory protein code, such as p53, the covalent attachment
or removal of modifying groups can change the protein's behavior, its activity or stability, its binding partners, or its location within a cell. In the case of p53, there are at least 20 different locations (amino acids) that can be modified through such processes as phosphorylation, ubiquitylation, and acetylation.
3)
Indicate whether the following statements are true or false. If a statement is false, explain why it is false.
A. The amino acids in the interior of a protein do not interact with the ligand and do
not play a role in selective binding.
B. Antibodies are Y shaped and are composed of six different polypeptide chains.
C. ATPases generate ATP for the cell.
D. Hexokinase recognizes and phosphorylates only one of the glucose stereoisomers.
A.
False. The interior amino acids form a structural scaffold that maintains the
specific orientation for those that directly interact with the ligand. Changes to
these interior amino acids can change the protein shape and render it nonfunctional.
B. False. Although antibodies are Y shaped, they are composed of four, not six,
polypeptide chains. There are two heavy chains and two light chains.
C. False. ATPases hydrolyze ATP; they do not produce it. These enzymes enable the
cell to harness the chemical energy stored in the high-energy phosphate bonds.
D. True.
4)
Indicate whether the following statements are true or false. If a statement is false, explain why it is false.
A. Feedback inhibition is defined as a mechanism of down-regulating enzyme activity by the accumulation of a product earlier in the pathway.
B. If an enzyme's allosteric binding site is occupied, the enzyme may adopt an alternative conformation that is not optimal for catalysis.
C. Protein phosphorylation is another way to alter the conformation of an enzyme
and serves exclusively as a mechanism to increase enzyme activity.
D. GTP binding proteins typically have GTPase activity, and the hydrolysis of GTP
transforms them to the "off" conformation
A. False. Feedback inhibition occurs when an enzyme acting early in a metabolic pathway is inhibited by the accumulation of a product late in the pathway. The inhibitory product binds to a site on the enzyme that lowers its catalytic activity.
B. True.
C. False. Although phosphorylation of a protein can change its conformation, this modification may be either as a positive or a negative regulator of enzyme
activity, depending on the protein in question.
D. True.
5)
Indicate whether the following statements are true or false. If a statement is false, explain why it is false.
A. Generally, the total number of nonpolar amino acids has a greater effect on
protein structure than the exact order of amino acids in a polypeptide chain.
B. The "polypeptide backbone" refers to all atoms in a polypeptide chain, except for
those that form the peptide bonds.
C. The chemical properties of amino acid side chains include charged, uncharged
polar, and nonpolar.
D. The relative distribution of polar and nonpolar amino acids in a folded protein is
determined largely by hydrophobic interactions, which favor the clustering of
nonpolar side chains in the interior
A. False. The order in which amino acids are linked is unique for each protein and is the most important factor in determining overall protein structure.
B. False. Peptide bonds are planar amide bonds that are central to the polypeptide backbone formation. The atoms in the amino acid side chains are not considered
to be part of the backbone.
C. True.
D. True 6)
You wish to produce a human enzyme, protein A, by introducing its gene into bacteria.
The genetically engineered bacteria make large amounts of protein A, but it is in the form
of an insoluble aggregate with no enzymatic activity. Which of the following procedures might help you to obtain soluble, enzymatically active protein? Explain your reasoning.
A. Make the bacteria synthesize protein A in smaller amounts.
B. Dissolve the protein aggregate in urea, then dilute the solution and gradually remove the urea.
C. Treat the insoluble aggregate with a protease.
D. Make the bacteria overproduce chaperone proteins in addition to protein A.
E. Heat the protein aggregate to denature all proteins, then cool the mixture.
Choices A, B, and D are all worth trying. Some proteins require molecular chaperones if they are to fold properly within the environment of the cell. In the absence of chaperones, a partly folded polypeptide chain has exposed amino acids that can form noncovalent bonds with other regions of the protein itself and with other proteins, thus causing nonspecific aggregation of proteins.
7)
GPCRs activate G proteins by reducing the strength of GDP binding, allowing GDP to dissociate and GTP, which is present in much higher concentrations, to bind. How do you suppose the activity of a G protein would be affected by a mutation that caused its affinity for GDP to be reduced without significantly changing its affinity for GTP?
Normally, GDP is tightly bound by the alpha subunit, which keeps the G protein in its inactive state until release of GDP is stimulated by interaction with an appropriate GPCR. Therefore, this
mutant G protein would be constantly active. Each time the alpha subunit hydrolyzed GTP to GDP, the GDP would spontaneously dissociate, allowing GTP to bind and reactivate the alpha subunit.
8)
Antibodies are Y-shaped molecules that carry two identical binding sites. Scientists have discovered or developed antibodies that specifically bind to the extracellular domains of particular receptor tyrosine kinases, such as Herceptin. When cells expressing a particular RTK are exposed to an antibody that binds to RTK, one of three outcomes occurs: 1) the antibody might cause activation, 2) the antibody may block activation and 3) the antibody may not affect the function of the RTK.
Describe a molecular mechanism that could explain each outcome.
Activation - the antibody has identical antigen binding domains so it's possible that each antigen
binding domain could bind a receptor domain and cause dimerization of the RTKs.
Block activation - the antibody could somehow block signal binding or block dimerization.
No effect - the antibody could bind somewhere else on the receptor domain that does not affect signal binding or dimerization.
Chapter 5
1)
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A. DNA molecules, like proteins, consist of a single, long polymeric chain that is assembled from small monomeric subunits.
B. The polarity of a DNA strand results from the polarity of the nucleotide subunits.
C. There are five different nucleotides that become incorporated into a DNA strand.
D. Hydrogen bonds between each nucleotide hold individual DNA strands together.
A. False. DNA is double-stranded. It is actually is made of two polymers that are
complementary in sequence.
B. True.
C. False. There are four different nucleotides that are used to make a DNA polymer:
adenine, thymine, guanine, and cytosine. A fifth nucleotide, uracil, is found
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exclusively in RNA molecules, replacing thymine nucleotides in the DNA
sequence.
D. False. Nucleotides are linked covalently through phosphodiester bonds.
Hydrogen-bonding between nucleotides from opposite strands holds the DNA
molecule together.
2)
Because hydrogen bonds hold the two strands of a DNA molecule together, the
strands can be separated without breaking any covalent bonds. Every unique DNA
molecule "melts" at a different temperature. In this context, Tm, melting temperature, is the point at which two strands separate, or become denatured. Order the DNA sequences listed below according to relative melting temperatures
(from lowest Tm to highest Tm). Assume that they all begin as stable doublestranded
DNA molecules. Explain your answer.
A. GGCGCACC
B. TATTGTCT
C. GACTCCTG
D. CTAACTGG
The order in which the DNA molecules would denature as the temperature
increases is:
1—B; 2—D; 3—C; 4—A
All the DNA molecules are the same length, so only the A + T and G + C content
determine their relative Tm. Molecules with higher G + C content will be more stable than molecules with a high A + T content. This is because there are three hydrogen bonds
between each G-C base pair but only two between each A-T base pair. More energy
(heat) is required to disrupt a larger number of hydrogen bonds.
3)
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A. Each strand of DNA contains all the information needed to create a new doublestranded
DNA molecule with the same sequence information.
B. All functional DNA sequences inside a cell code for protein products.
C. Gene expression is the process of duplicating genes during DNA replication.
D. Gene sequences correspond exactly to the respective protein sequences produced
from them.
A. True.
B. False. Some sequences encode only RNA molecules, some bind to specific
regulatory proteins, and others are sites where specific chrosomosomal protein
structures are built (for example, centromeric and telomeric DNA).
C. False. Gene expression is the process of going from gene sequence to RNA
sequence, to protein sequence.
D. False. This statement is false for two reasons. First, genes often contain intron
sequences. Second, genes always contain nucleotides flanking the protein-coding
sequences that are required for the regulation of transcription and translation.
4)
Consider the structure of the DNA double helix.
A. You and a friend want to split a double-stranded DNA molecule so you each have half.
Is it better to cut the length of DNA in half so each person has a shorter length, or to separate the strands and each take one strand? Explain.
B. In the original 1953 publication describing the discovery of the structure of DNA, Watson and Crick wrote, "It has not escaped our notice that the specific pairings
we have postulated immediately suggest a possible copying mechanism for the genetic material." What did they mean?
A. It is better to separate the strands and each take a single strand, because all of the information found in the original molecule is preserved in a full-length single
strand but not in a half-length double-stranded molecule.
B. Watson and Crick meant that the complementary base pairing of the strands
allows a single strand to contain all of the information necessary to direct the
synthesis of a new complementary strand.
5)
In principle, what would be the minimum number of consecutive nucleotides
necessary to correspond to a single amino acid to produce a workable genetic code? Assume that each amino acid is encoded by the same number of nucleotides. Explain your reasoning
Because there are 20 amino acids used in proteins, each amino acid would have to
be encoded by a minimum of three nucleotides. For example, a code of two
consecutive nucleotides could specify a maximum of 16 (42) different amino acids, excluding stop and start signals. A code of three consecutive nucleotides
has 64 (43) different members and thus can easily accommodate the 20 amino
acids plus a signal to stop protein synthesis.
6)
Consider two different species of yeast that have similar genome sizes. Is it likely that they contain the same number of genes? A similar number of chromosomes?
A similar genome size indicates relatively little about the number of genes and virtually nothing about the number of chromosomes.
7)
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A. Comparing the relative number of chromosome pairs is a good way to determine
whether two species are closely related.
B. Chromosomes exist at different levels of condensation, depending on the stage of
the cell cycle.
C. Eukaryotic chromosomes contain many different sites where DNA replication can be initiated.
D. The telomere is a specialized DNA sequence where microtubules from the mitotic
spindle attach to the chromosome so that duplicate copies move to opposite ends of the dividing cell.
A. False. There are several examples of closely related species that have a drastically
different number of chromosome pairs. Two related species of deer—Chinese and Indian muntjac—have 23 and 3, respectively.
B. True.
C. True.
D. False. The telomere is a specialized DNA sequence, but not for the attachment of
spindle microtubules. Telomeres form special caps that stabilize the ends of linear
chromosomes.
8)
For each of the following sentences, choose one of the options enclosed in square brackets to make a correct statement about nucleosomes.
A. Nucleosomes are present in [procaryotic/eucaryotic] chromosomes, but not in
[procaryotic/eucaryotic] chromosomes.
B. A nucleosome contains two molecules each of histones [H1 and H2A/H2A and
H2B] as well as of histones H3 and H4.
C. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately [twice/three times/four times].
D. Nucleosomes are aided in their formation by the high proportion of [acidic/basic/polar]
amino acids in histone proteins.
E. Nucleosome formation compacts the DNA into approximately [one-third/onehundredth/ one-thousandth] of its original length.
A. Nucleosomes are present in eucaryotic chromosomes, but not in procaryotic chromosomes.
B. A nucleosome contains two molecules each of histones H2A and H2B, as well as histones H3
and H4.
C. A nucleosome core particle contains a core of histone with DNA wrapped around it approximately twice.
D. Nucleosomes are aided in their formation by the high proportion of basic amino acids in histone proteins.
E. Nucleosome formation compacts DNA into approximately one-third of its
original length.
9)
Evidence suggests that the replication of DNA packaged into heterochromatin occurs later than the replication of other chromosomal DNA. What is the simplest possible explanation for this phenomenon? The DNA double helix in heterochromatin may be so tightly packed and condensed that it
is inaccessible to the proteins that bind replication origins, including the DNA replication
machinery. It may take extra time to remodel the chromatin to make it more accessible to the proteins required to initiate and perform DNA replication.
Chapter 7
NOTE: For a cell's genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called (transcription).
Various kinds of RNA are produced, each with different functions. (mRNA) molecules code for proteins, (tRNA) molecules act as adaptors for protein synthesis, (rRNA) molecules are integral components of the ribosome, and (snRNA) molecules are important in the splicing of RNA transcripts.
1)
List three ways in which the process of eucaryotic transcription differs from the process of bacterial transcription
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Any three of the following are acceptable.
1. Bacterial cells contain a single RNA polymerase, whereas eucaryotic cells
have three.
2. Bacterial RNA polymerase can initiate transcription without the help of
additional proteins, whereas eucaryotic RNA polymerases need general
transcription factors.
3. In eucaryotic cells, transcription regulators can influence transcriptional
initiation thousands of nucleotides away from the promoter, whereas
bacterial regulatory sequences are very close to the promoter.
4. Eucaryotic transcription is affected by chromatin structure and
nucleosomes, whereas bacteria lack nucleosomes.
2)
In eucaryotic cells, general transcription factors are required for the activity of all promoters transcribed by RNA polymerase II. The assembly of the general transcription factors begins with the binding of the factor __TFIID_ to DNA, causing a marked local distortion in the DNA. This factor binds at the DNA sequence called the ______TATA______ box, which is typically located 25 nucleotides upstream from the transcription start site. Once RNA polymerase II has been brought to the promoter DNA,
it must be released to begin making transcripts. This release process is facilitated by the addition of phosphate groups to the tail of RNA polymerase by the factor ___TFIIH____. It must be remembered that the general transcription factors and RNA polymerase are not sufficient to initiate transcription in the cell and are affected by proteins bound thousands of nucleotides away from the promoter. Proteins that link the distantly bound transcription regulators to RNA polymerase and the general transcription factors
include the large complex of proteins called the____mediator____. The packing of DNA into chromatin also affects transcriptional initiation, and histone _____deacetylase______ is an enzyme that can render the DNA less accessible to the general transcription factors.
3)
Name three covalent modifications that can be made to an RNA molecule in eucaryotic cells before the RNA molecule becomes a mature mRNA.
A poly A tail is added.
2. A 5 cap is added.
′
3. Introns can be spliced out.
4)
The length of a particular gene in human DNA, measured from the start site for transcription to the end of the protein-coding region, is 10,000 nucleotides, whereas the length of the mRNA produced from this gene is 4000 nucleotides.
What is the most likely reason for this difference?
The gene contains one or more introns.
5)
Genes in eukaryotic cells often have intronic sequences coded for within the DNA. These sequences are ultimately not translated into proteins. Why?
Intronic sequences are removed from RNA molecules by the spliceosome, which works in the nucleus.
6)
Why is the old dogma "one gene—one protein" not always true for eucaryotic
genes?
The transcripts from some genes can be spliced in more than one way to give mRNAs containing different sequences, thus encoding different proteins. A single euKaryotic gene may therefore encode more than one protein.
7)
Is this statement true or false? Explain your answer. "Since introns do not contain protein coding
information, they do not have to be removed precisely (meaning, a nucleotide here and there should not matter) from the primary transcript during RNA splicing."
False. Although it is true that the sequences within the introns are mostly dispensable, the introns must still be removed precisely because an error of one or two nucleotides would shift the reading frame of the resulting mRNA molecule and change the protein it encodes.
8)
You have discovered a gene (Figure Q7-27A) that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in Figure Q7-27B.
The lines connecting the exons that are included in the mRNA
indicate the splicing. From your experiments, you know that protein translation begins in exon 1. For all forms of the mRNA, the encoded protein sequence is the
same in the regions of the mRNA that correspond to exons 1 and 10. Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8. Which of
the following statements about exons 2 and 3 is the most accurate? Explain your
answer. Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder
(that is, 0, 1, or 2).
9)
The following DNA sequence includes the beginning of a sequence coding for a
protein. What would be the result of a mutation that changed the C marked by an
asterisk to an A?
5 -AGGCTATGAATGGACACTGCGAGCCC
....
′
The change creates a stop codon (TGA, or UGA in the mRNA) very near the beginning of the protein-
coding sequence and in the correct reading frame (the beginning of the coding sequence is indicated by the ATG). Thus, translation would terminate after only four amino acids had been joined together, and the complete protein would not be made.
10)
Below is a segment of RNA from the middle of an mRNA.
5′-UAGUCUAGGCACUGA-3′
If you were told that this segment of RNA was part of the coding region of an mRNA for a
large protein, give the amino acid sequence for the protein that is encoded by this segment of mRNA. Write your answer using the one-letter amino
acid code.
SLGT is the correct answer. (Reading frame two is the only reading frame that
does not contain a stop codon.)
11)
One strand of a section of DNA isolated from the bacterium E. coli reads:
5′-GTAGCCTACCCATAGG-3′
A. Suppose that an mRNA is transcribed from this DNA using the
complementary strand as a template. What will be the sequence of the mRNA in this region (make sure you label the 5′ and 3′ ends of the mRNA)?
B. How many different peptides could potentially be made from this
sequence of RNA, assuming that translation initiates upstream of this
sequence?
C. What are these peptides? (Give your answer using the one-letter amino acid code.)
A. 5 -GUAGCCUACCCAUAGG-3
′
′
B. Two. (There are three potential reading frames for each RNA. In this case,
they are
GUA GCC UAC CCA UAG ...
UAG CCU ACC CAU AGG ...
AGC CUA CCC AUA GG? ... .
The center one cannot be used in this case, because UAG is a stop codon.)
C. VAYP
SLPIG
Note: PTHR will not be a peptide because it is preceded by a stop codon.
NOTE: Once an mRNA is produced, its message can be decoded on ribosomes. The ribosome is
composed of two subunits: the large subunit, which catalyzes the formation of the peptide bonds
that link the amino acids together into a polypeptide chain, and the small subunit, which matches the tRNAs to the codons of the mRNA. During the chain elongation process of translating an mRNA into protein, the growing polypeptide chain attached to a tRNA is bound to the P-site of the ribosome. An incoming aminoacyl-tRNA carrying the next amino acid in
the chain will bind to the A-site by forming base pairs with the exposed codon in the mRNA. The
peptidyl transferase enzyme catalyzes the formation of a new peptide bond between the growing polypeptide chain and the newly arriving amino acid. The end of a protein-coding message is signaled by the presence of a stop codon, which binds the protein called release factor. Eventually, most proteins will be degraded by a large complex of proteolytic enzymes called the proteasome.
12) Give a reason why DNA makes a better material than RNA for the storage of genetic information,
and explain your answer.
Three possible answers are:
1. The deoxyribose sugar of DNA makes the molecule much less susceptible than RNA to breakage, because of the lack of the hydroxyl group on carbon 2 of the ribose sugar.
2. DNA is double-stranded and therefore the complementary strand provides a template from which damage can be repaired accurately.
3. The use of "T" in DNA instead of "U" (as in RNA) protects against the effect of deamination, a common form of damage. Deamination of T produces an aberrant base (methyl C), whereas deamination of U generates C, a normal base. The presence of an abnormal base eases the
cell's job of recognizing the damaged strand.
Chapter 8
NOTE:
The genes of a bacterial operon are transcribed into a single mRNA. Many bacterial promoters contain a region known as an operator, to which a specific transcription regulator binds. Genes in which transcription is prevented are said to be repressed. The interaction of
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small molecules, such as tryptophan, with allosteric DNA-binding proteins, such as the tryptophan repressor, regulates bacterial genes. Genes that are being constitutively expressed are being transcribed all the time.
1)
An allosteric transcription regulator called HisP regulates the enzymes for histidine biosynthesis in the bacterium E. coli. Histidine modulates HisP activity.
On binding histidine, HisP alters its conformation, markedly changing its affinity for the regulatory sequences in the promoters of the genes for the histidine
biosynthetic enzymes.
A. If HisP functions as a gene repressor, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine is abundant? Explain your answer.
B. If HisP functions as a gene activator, would you expect that HisP would bind more tightly or less tightly to the regulatory sequences when histidine levels are low? Explain your answer.
If HisP functions as a gene repressor, it would bind more tightly to the regulatory sequences when histidine is abundant, because the histidine biosynthetic genes should be turned off when the cell has enough histidine.
B. If HisP functions as a gene activator, it should bind more tightly to the regulatory sequences when histidine levels are low. When histidine levels are low, the cell needs to synthesize more histidine.
2)
The CAP activator protein and the Lac repressor both control the Lac operon (see
Figure Q8-23). You create cells that are mutant in the gene coding for the Lac repressor so that these cells lack the Lac repressor under all conditions. For these
mutant cells, state whether the Lac operon will be switched on or off in the following situations and explain why.
A. in the presence of glucose and lactose
B. in the presence of glucose and the absence of lactose
C. in the absence of glucose and the absence of lactose
D. in the absence of glucose and the presence of lactose
A. Operon off. CAP will not bind in the presence of glucose.
B. Operon off. Although normally the Lac repressor would bind in the
absence of lactose, the lack of the Lac repressor in this case does not
matter because the presence of glucose means that the CAP protein will
not bind and activate transcription.
C. Operon on. Normally in the absence of both glucose and lactose, the
operon would be off. However, because the cells lack the Lac repressor,
the cells cannot sense the absence of lactose. Because the CAP protein
will bind and activate transcription, the operon will be on.
D. Operon on. The CAP protein will bind and activate transcription because
of the presence of glucose. It does not matter whether the Lac repressor
gene is mutant, because there is lactose available.
3)
You have discovered an operon in a bacterium that is turned on only when
sucrose is present and glucose is absent. You have also isolated three mutants that
have changes in the upstream regulatory sequences of the operon and whose
behavior is summarized in the Table Q8-24. You hypothesize that there are two
gene regulatory sites in the upstream regulatory sequence, A and B, which are
affected by the mutations. For this question, a plus (+) indicates a normal site and
a minus (-) indicates a mutant site that no longer binds its transcription regulator.
A. If mutant 1 has sites A- B+, which of these sites is regulated by sucrose and
which by glucose?
B. Give the state (+ or -) of the A and B sites in mutants 2 and 3.
C. Which site is bound by a repressor and which by an activator?
A. Site A is regulated by sucrose, and site B by glucose.
B. Mutant 2 (A+ B-); mutant 3 (A- B-) or (A- B+).
C. Site A is bound by an activator and site B by a repressor.
NOTE:
During transcription in eucaryotic cells, transcriptional regulators that bind to
DNA thousands of nucleotides away from a gene's promoter can affect a gene's transcription. The mediator is a complex of proteins that links distantly bound transcription regulators with the proteins bound closer to the transcriptional start site. Transcriptional activators can also interact with histone acetylases, which alter chromatin by modifying lysines in the tail of histone proteins to allow greater accessibility to the underlying DNA. Gene repressor proteins can reduce the efficiency of transcription initiation by attracting histone deacetylases.
Sometimes, many contiguous genes can become transcriptionally inactive as a
result of chromatin remodeling, like the heterochromatin found in interphase chromosomes.
4)
From the sequencing of the human genome, we believe that there are approximately 24,000 protein-coding genes in the genome, for which there are an
estimated 1500-3000 transcription factors. If every gene has a tissue-specific and signal-
dependent transcription pattern, how can such a small number of transcriptional regulatory proteins generate a much larger set of transcriptional
patterns?
Transcription regulators are generally used in combinations, thereby increasing
the possible regulatory repertoire of gene expression with a limited number of proteins.
NOTE: RNAi is triggered by the presence of foreign doublestranded
RNA molecules, which are digested by the dicer enzyme into shorter
fragments approximately 23 nucleotide pairs in length.
5)
The gene for a hormone necessary for insect development contains binding sites
for three transcription regulators called A, B, and C. Because the binding sites for
A and B overlap, A and B cannot bind simultaneously. You make mutations in the
binding sites for each of the proteins and measure hormone production in cells
that contain equal amounts of the A, B, and C proteins. Figure Q8-47 summarizes
your results. In each of the following sentences, choose one of the phrases within
square brackets to make the statement consistent with the results.
A. Protein A binds to its DNA-binding site [more tightly/less tightly] than
protein B binds to its DNA-binding site.
B. Protein A is a [stronger/weaker] activator of transcription than protein B.
C. Protein C is able to prevent activation by [protein A only/protein B
only/both protein A and protein B]
A. Protein A binds to its DNA-binding site more tightly than protein B binds
to its DNA-binding site.
B. Protein A is a weaker activator of transcription than protein B.
C. Protein C is able to prevent activation by both protein A and protein B.
6)
The Drosophila Eve gene has a complex promoter containing multiple binding
sites for four transcription regulators: Bicoid, Hunchback, Giant, and Krüppel.
Bicoid and Hunchback are activators of Eve transcription, whereas Giant and
Krüppel repress Eve transcription. Figure Q8-48A shows the patterns of expression of these regulators MIGHT BE WRONG
A. transcriptional regulator binding site leads to gene product lost B. i) Kruppel is the reporter of eve stripe 2, deletion of kruppel binding sites should cause a broadening of the eve stripe 2 expression domain, and this shift should be to be posterior since kruppel is normally expressed just posterior to the position of eve stripe 2, in a wild-type embryo
ii) Bicoid protein act as transcriptional activator
Chapter 11
1)
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Not all words or phrases will be used; each word or phrase should be used only once.
The specialized functions of different membranes are largely determined by the
proteins
they contain. Membrane lipids are
amphipathic
molecules, composed of a hydrophilic portion and a hydrophobic portion. All cell membranes have the same
lipid bilayer
structure, with the
fatty acid tails
of the phospholipids facing into the interior of the membrane and the __
hydrophilic head groups
___ on the outside.
2)
Glycolipids are found on the surface of healthy cells, and contribute to the cell's defense
against chemical damage and infectious agents.
A. In which organelle are sugar groups added to membrane lipids?
B. By what mechanism are glycolipids transported to the plasma membrane and
presented to the extracellular environment?
A. The Golgi apparatus.
B. Membranes that contain newly synthesized glycolipids bud from the Golgi
apparatus to form vesicles. These vesicles then fuse with the plasma membrane.
The glycolipids that were facing the lumen of the Golgi will now face the
extracellular environment
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3)
Although membrane proteins contribute roughly 50% of the total mass of the membrane,
there are about 50 times more lipid molecules than there are protein molecules in cellular
membranes. Explain this apparent discrepancy.
Membrane proteins are much larger molecules than the membrane lipids. Thus, fewer are required to represent the same total mass contributed by the lipid components of the membrane. By this estimation, the molecular weight of the average membrane protein is
50 times that of the average membrane lipid.
4)
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A. Lipid-linked proteins are classified as peripheral membrane proteins because the
polypeptide chain does not pass through the bilayer.
B. A protein can be embedded on the cytosolic side of the membrane bilayer by
employing a hydrophobic α helix.
C. A protein that relies on protein-protein interactions to stabilize its membrane
association is classified as a peripheral membrane protein because it can be
dissociated without the use of detergents.
D. Membrane proteins that pump ions in and out of the cell are classified as
enzymes.
A. False. Lipid-linked proteins are classified as integral membrane proteins because
although they are not transmembrane proteins, they are covalently bound to
membrane lipids and cannot be dissociated without disrupting the membrane's
integrity.
B. False. An embedded protein employs an amphipathic helix. The hydrophobic side
interacts with the fatty acid tails of the membrane lipids and the hydrophilic
portion interacts with the aqueous components of the cytosol.
C. True.
D. False. Membrane proteins that pump ions in either direction across the membrane
are in the functional class of transporters.
5)
Sodium dodecyl sulfate (SDS) and Triton X-100 are both detergents that can be used to
lyse cells.
A. If the the goal is to study the activity of membrane proteins after cell lysis, explain
why SDS would not be a good choice.
B. How does Triton X-100 work in cell lysis, and why is it a better choice of
detergent to help you extract proteins?
A. SDS is a strong ionic detergent. When cells are exposed to SDS membrane,
proteins are not only extracted from the membrane, they are completely unfolded.
After denaturation, they cannot be studied as functional molecules.
B. Triton X-100 has a smaller nonpolar portion and a polar but uncharged end,
allowing it to mimic more closely the type of solvation effect of the membrane
lipids. Triton X-100 forms a shell around the hydrophobic portion of the protein
without disrupting the existing structure. This makes it possible to then place the
protein into a new, synthetic membrane bilayer for study.
6)
There are several ways that membrane proteins can associate with the cell membrane. Membrane proteins that extend through the lipid bilayer are called _
transmembrane
___
proteins and have __
hydrophobic
___ regions that are exposed to the interior of the bilayer. On the other hand, membrane-associated proteins do not span the bilayer and instead associate with the membrane through an α helix that is __
amphipathic
____. Other proteins are __
covalently
__ attached to lipid molecules that are inserted in the membrane. __
Peripheral
___ membrane proteins are linked to the membrane through noncovalent interactions with other membrane-bound proteins.
7)
Cholesterol can either make membranes more rigid or more fluid. Explain how this is possible.
There are two properties of phospholipids that affect how tightly they pack together: the length of the hydrocarbon chain and the number of double bonds. The degree of packing, in turn, influences the relative mobility of these molecules in the membrane. The effect of cholesterol is dependent upon both temperature and concentration. The presence of cholesterol at normal physiological concentrations helps maintain membrane fluidity at lower temperatures by solvating the long hydrocarbon tails of phospholipids. At higher temperatures, cholesterol stabilizes the membrane (observed as an increase in the melting temperature).
8)
Double bonds in hydrocarbon tails have what effect on the rigidity of fats and phospholipid bilayers? Double bonds decrease the ability of hydrocarbon tails to pack together, which makes the bilayer less stiff
9)
Anemia, a condition that results in individuals with a low red blood cell count, can be
caused by a number of factors. Why do individuals with defects in the spectrin protein
often have this condition?
Spectrin is the primary protein in the cortex of red blood cells. A defect in the spectrin
protein directly affects the strength and shape of the cortex. Red blood cells that contain
mutated spectrin molecules have an irregular shape and are prone to lysis as a result of
cortical fragility, leading to a smaller population of red blood cells.
10)
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A. When a mouse cell is fused with a human cell, the movement of the respective
membrane proteins is restricted to their original locations at the time of fusion.
B. Epithelial cell membranes are asymmetric, and proteins from the apical side of the
cell membrane cannot diffuse into the basal side of the membrane.
C. The longest carbohydrates found on the surfaces of cells are linked to lipid
molecules.
D. The only role of the carbohydrate layer on the cell surface is to absorb water,
which creates a slimy surface and prevents cells from sticking to each other.
Spectrin is the primary protein in the cortex of red blood cells. A defect in the spectrin
protein directly affects the strength and shape of the cortex. Red blood cells that contain
mutated spectrin molecules have an irregular shape and are prone to lysis as a result of
cortical fragility, leading to a smaller population of red blood cells.
11)
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A. When a mouse cell is fused with a human cell, the movement of the respective
membrane proteins is restricted to their original locations at the time of fusion.
B. Epithelial cell membranes are asymmetric, and proteins from the apical side of the
cell membrane cannot diffuse into the basal side of the membrane.
C. The longest carbohydrates found on the surfaces of cells are linked to lipid
molecules.
D. The only role of the carbohydrate layer on the cell surface is to absorb water,
which creates a slimy surface and prevents cells from sticking to each other.
A. False. After about 1 hour, the mouse and human proteins present on the surface of
the fused cell are found evenly dispersed throughout the plasma membrane.
B. True.
C. False. The very long, branched polysaccharides that are attached to integral
membrane proteins are much longer than the oligosaccharides covalently attached
to membrane lipids.
D. False. Although the absorption of water is an important role of the carbohydrates
on the surface of the plasma membrane, a second critical role is that of cell-cell
recognition important in immune responses, wound healing and other processes
that rely on cell-type-specific interactions.
12)
Cell membranes are fluid, and thus proteins can diffuse laterally within the lipid bilayer.
However, sometimes the cell needs to localize proteins to a particular membrane domain.
Name three mechanisms that a cell can use to restrict a protein to a particular place in the
cell membrane.
1. The protein can be attached to the cell cortex inside the cell.
2. The protein can be attached to the extracellular matrix outside the cell.
3. The protein can be attached to other proteins on the surface of a different cell.
4. The protein can be restricted by a diffusion barrier, such as that set up by specialized
junctional proteins at a tight junction.
Chapter 12
1)
Indicate whether the following statements are true or false. If a statement is false, explain
why it is false.
A. CO2 and O2 are water-soluble molecules that diffuse freely across cell
membranes.
B. The differences in permeability between artificial lipid bilayers and cell
membranes arise from variations in phospholipid content.
C. Transporters are similar to channels, except that they are larger, allowing folded
proteins as well as smaller organic molecules to pass through them.
D. Cells expend energy in the form of ATP hydrolysis so as to maintain ion
concentrations that differ from those found outside the cell.
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A. True.
B. False. The primary difference between cell membranes and artificial membranes
is that cell membranes have proteins responsible for creating a selective
permeability, which varies with the location and function of the membrane.
C. False. Transporters work by changing conformation after specific binding of the
solute to be transported. Channels exclude molecules on the basis of size and
charge, but do not depend on specific recognition of the molecules moving
through.
D. True.
2)
A molecule moves down its concentration gradient by __
passive
__
transport but requires __
active
____ transport to move up its
concentration gradient. Transporter proteins and ion channels function in
membrane transport by providing a __
hydrophilic
___ pathway through the
membrane for specific polar solutes or inorganic ions. ____
Transporter proteins
____ are highly selective in the solutes they transport, binding the solute at a specific site
and changing its conformation so as to transport the solute across the membrane.
On the other hand, ___
ion channels
______ discriminate between solutes mainly on the
basis of size and electrical charge.
3)
Facilitated diffusion can be described as the favorable movement of one solute down its concentration gradient being coupled with the unfavorable movement of a second solute up its concentration gradient.
B. Transporters undergo transitions between different conformations, depending on whether the substrate-binding pocket is empty or occupied.
C. The electrochemical gradient for K+ across the plasma membrane is small. Therefore, any movement of K+ from the inside to the outside of the cell is driven solely by its concentration gradient.
D. The net negative charge on the cytosolic side of the membrane enhances the rate of glucose import into the cell by a uniporter.
A. False. This describes coupled transport, which is one type of active transport. Facilitated diffusion can also be called passive transport, in which a solute always moves down its concentration gradient.
B. True.
C True.
D. False. Glucose is an uncharged molecule, and its import is not directly affected by the voltage difference across the membrane if glucose is being transported alone.
If the example given were the Na/glucose symporter, we would have to consider the charge difference across the membrane.
4)
It is thought that the glucose transporter switches between two conformational states in a
completely random fashion. How is it possible for such a system to move glucose across the
membrane efficiently in a single direction?
Although the opening of the glucose transporter on one side of the membrane or the other
is random, the binding of glucose into the binding site of the transporter is not a random
event. The affinity between the glucose molecule and the transporter governs the binding
event: transporter + glucose Û transporter-glucose At high glucose concentrations the complex formation is favored; at low glucose
concentrations dissociation of glucose from the transporter is favored. So, as long as there
is a large concentration gradient, efficient transport can occur by the simple rules of binding equilibria
5)
If ATP production is blocked in an animal cell, the cell will swell up. Explain this
observation.
ATP is required to power the Na+-K+ pump, which is necessary for maintaining osmotic
balance. The pump requires ATP hydrolysis to drive its pumping cycle. So, in the
absence of ATP production, the Na+ concentration inside the cell will increase. This is
followed by passive diffusion of water across the membrane, causing the cell to swell.
6)
You have prepared lipid vesicles (spherical lipid bilayers) that contain Na+-K+ pumps as the sole membrane protein. All of the Na+-K+ pumps are oriented in such a way that the portion of the molecule that normally faces the cytosol is on the inside of the vesicle and the portion of the molecule that normally faces the extracellular space is on the outside of the vesicle. Assume that each pump transports one Na+ ion in one direction and one K+ ion in the other direction during each pumping cycle (see Figure Q12-16 for how the Na+-K+ pump normally functions in the plasma membrane).
A. Without any ATP to provide energy for the Na+-K+ pumps, no ions will be
pumped.
B. The pumps will use the energy from ATP hydrolysis to transport Na+ out of the
vesicles and K+ into the vesicles. (The pumps will stop working either when the
amount of ATP inside the vesicle is depleted or when the K+ outside the vesicles
is depleted.)
C. The pump will bind a molecule of Na+, causing the ATPase activity to hydrolyze
ATP and transfer the phosphate group onto the pump. A conformational change
will occur, leading to the release of Na+ from the vesicle. However, because there
is no K+ outside the vesicle, the pump will get stuck at that step and subsequent
steps of the cycle will not occur.
7)
The movement of glucose into the cell, against its concentration gradient, can be powered
by the co-transport of Na+ into the cell. Explain this movement with respect to the net
entropy of the system (i.e. thermodynamics).
The movement of Na+ ions from an area that has a high Na+ concentration to a new area
of low Na+ concentration is energetically favorable because the net entropy in the system
is increasing. As long as the difference in Na+ ion concentration across the membrane is
large, the entropic factor will be sufficient to drive the import of glucose into the cell,
which represents a decrease in entropy with respect to the population of glucose
molecules inside the cell.
8)
Indicate whether the statements below are true or false. If a statement is false,
explain why it is false.
A. Gap junctions are large pores that connect the cytosol to the extracellular space.
B. Aquaporin channels are found in the plasma membrane, allowing the rapid
passage of water molecules and small ions in and out of cells.
C. The ion selectivity of a channel completely depends solely on the charge of the
amino acids liningthe pore inside the channel.
D. Most ion channels are gated, which allow them to open and close in response to a
specific stimulus rather than allowing the constant, unregulated flow of ions.
A. False. Gap junctions are used to connect the cytosol of adjacent cells, allowing the
sharing of ions and small metabolites. Because gap junctions are large channels, if
they were open while facing the extracellular environment, the ability of the
plasma membrane to serve as a permeability barrier would be greatly reduced.
B. False. Charged molecules (even protons, which are very small), are not able to
pass through aquaporins.
C. False. Selectivity depends on three parameters: the diameter, shape and charge of
the ion trying to pass through the pore of the channel.
D. True. 9)
Indicate whether the statements below are true or false. If a statement is false, explain
why it is false.
A. Neurotransmitters are small molecules released into the synaptic cleft after the
fusion of synaptic vesicles with the presynaptic membrane.
B. Action potentials are usually mediated by voltage-gated Ca2+ channels.
C. Voltage-gated Na+ channels become automatically inactivated shortly after
opening, which ensures that the action potential cannot move backward along the
axon.
D. Voltage-gated K+ channels also open immediately in response to local
depolarization, reducing the magnitude of the action potential.
A. True.
B. False. Action potentials are usually mediated by voltage-gated Na+ channels.
C. True.
D. False. Voltage-gated K+ channels respond more slowly than the voltage-gated Na+
channels. Because they do not open until the action potential reaches its peak,
they do not affect its magnitude. Instead, they help to restore the local membrane
potential quickly while the voltage-gated Na+ channels are in the inactivated
conformation.
10)
Studies on the squid giant axon were instrumental in our current understanding of how
action potentials are generated. You decide to do some experiments on the squid giant
axon yourself.
A. You remove the cytoplasm in an axon and replace it with an artificial cytoplasm
that contains twice the normal concentration of K+ by adding KOAc, where OAc-
is an anion that is impermeable to the membrane. In this way you double the
internal concentration of K+ while maintaining the bulk electrical balance of the
cytoplasmic solution. Will this make the resting potential of the membrane more
or less negative?
B. You add NaCl to the extracellular fluid and effectively double the amount of
extracellular Na+ cation. How does this affect the action potential?
C. You replace half of the NaCl in the extracellular fluid with choline chloride.
(Choline is a monovalent cation much larger than Na+. Note that the presence of
choline will not impede the flow of Na+ through its channels.) How will this
affect the action potential?
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A. Increasing the concentration of K+ in the cytoplasm of the squid axon will make
the membrane potential more negative. Doubling the amount of K+ increases the
driving force for K+ to move out of the cell, leaving the inside of the cell more
negative and thus decreasing the membrane potential. (Remember, from the
Nernst equation, the driving force for an ion across a membrane is proportional to
the ratio of the concentration of the ion on the outside to the concentration of the
ion on the inside.)
B. Doubling the amount of Na+ in the extracellular fluid will increase the height of
the peak of the action potential. Again, this is because now the driving force for
Na+ to enter the cell is greater than it was before. Thus, when Na+ channels open,
the flux of Na+ ions is now greater. (Remember that flux is the number of ions
entering per second.)
C. The action potential in this case will reach a height that is less than that normally
achieved. (Choline is added in this case to maintain bulk electrical neutrality.
Because Na+ channels are not permeable to choline, they do not contribute to the
electrochemical gradient.) You have now halved the concentration of Na+ and
thus decreased the driving force for Na+ to enter the cell
Chapter 15 1)
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. The _
cytosol
___ makes up about half of the total cell volume of a typical eukaryotic cell.
Ingested materials within the cell will pass through a series of compartments called ___
endosomes
___ on their way to the __
lysosomes
_, which contains digestive enzymes and will ultimately degrade the particles and macromolecules taken into the cell and will also degrade worn-out organelles. The _
GA
__ has a cis and trans face and receives proteins and lipids from the _
ER
_, a system of interconnected sacs and tubes of membranes that typically extends throughout the cell.
2)
For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below. Plasma membrane proteins are inserted into the membrane in the _
ER
__. The address information for protein sorting in a eukaryotic cell is contained in the __________ of the proteins. Proteins enter the nucleus in their __
folded
__ form. Proteins that remain in the
cytosol do not contain a __
signal
__. Proteins are transported into the Golgi apparatus via _
vesicles
__. The proteins transported into the endoplasmic reticulum by __
ER
__ are in their __
unfolded
__form.
3)
FILL in the blank: Proteins are transported out of a cell via the secretory or exocytic pathway. Fluid and macromolecules are transported into the cell via the endocytic pathway. All proteins being transported out of the cell pass through the endoplasmic reticulum and the Golgi apparatus. Transport vesicles link organelles of the endomembrane system. The formation of disulfide bonds in the endoplasmic reticulum stabilizes protein structure.
4)
Name a type of protein modification that can occur in the ER but not in the cytosol.
Proteins in the ER can undergo disulfide bond formation. (This does not occur in the cytosol because of its reducing environment.)
(Signal-sequence cleavage and N-linked glycosylation are also acceptable answers.)
5)
For each of the following sentences, choose one of the two options enclosed in
square brackets to make a correct statement.
New plasma membrane reaches the plasma membrane by the
[regulated/constitutive] exocytosis pathway. New plasma membrane proteins
reach the plasma membrane by the [regulated/constitutive] exocytosis pathway.
Insulin is secreted from pancreatic cells by the [regulated/constitutive]
exocytosis pathway. The interior of the trans Golgi network is [acidic/alkaline].
Proteins that are constitutively secreted [aggregate/do not aggregate] in the trans
Golgi network.
constitutive
constitutive
regulated
acidic
do not aggregate
6)
Name the membrane-enclosed compartments in a eukaryotic cell where each of the functions listed below takes place.
A. photosynthesis
B. transcription
C. oxidative phosphorylation
D. modification of secreted proteins
E. steroid hormone synthesis
F. degradation of worn-out organelles
G. new membrane synthesis
H. breakdown of lipids and toxic molecules
a. Photosynthesis = chloroplast b. Transcription = nucleus
c. Oxidative phosphorylation = mitochondrion
d. Modification of secreted proteins = Golgi apparatus and rough endoplasmic reticulum (ER)
e. Steroid hormone synthesis = smooth ER
f. Degradation of worn-out organelles = lysosome
g. New membrane synthesis = ER
h. Breakdown of lipids and toxic molecules = peroxisome
7)
What would happen in each of the following cases? Assume in each case that the protein involved is a soluble protein, not a membrane protein.
A. You add a signal sequence (for the ER) to the N-terminal end of a normally cytosolic protein.
B. You change the hydrophobic amino acids in an ER signal sequence into charged amino acids.
C. You change the hydrophobic amino acids in an ER signal sequence into other hydrophobic amino acids.
D. You move the N-terminal ER signal sequence to the C-terminal end of the protein.
A. The protein will now be transported into the ER lumen.
B. The altered signal sequence will not be recognized and the protein will remain in the cytosol.
C. The protein will still be delivered into the ER. It is the distribution of hydrophobic amino acids
that is important, not the actual sequence.
D. The protein will not enter the ER. Because the C-terminus of the protein is the last part to be made, the ribosomes synthesizing this protein will not be recognized by the signal-recognition particle (SRP) and hence not carried to the ER. 8)
Using genetic engineering techniques, you have created a set of proteins that contain two (and only two) conflicting signal sequences that specify different compartments. Predict which signal would win out for the following combinations. Explain your answers.
A. Signals for import into the nucleus and import into the ER.
B. Signals for export from the nucleus and import into the mitochondria.
C. Signals for import into mitochondria and retention in the ER.
A. The protein would enter the ER. The signal for a protein to enter the ER is recognized as the protein is being synthesized and the protein will end up either in the ER or on the ER membrane. Cytosolic nuclear transport proteins recognize proteins destined for the nucleus once those proteins are fully synthesized and fully folded.
B. The protein would enter the mitochondria. For a nuclear export signal to work, the protein would have to end up in the nucleus first and thus would need a nuclear import signal for the nuclear export signal to be used.
C. The protein would enter the mitochondria. To be retained in the ER, the protein needs to enter the ER. Because there is no signal for ER import, the ER retention signal would not function.
9)
Indicate on your drawing the extracellular space, the cytosolic face, and the plasma membrane, as well as the N- and C-terminus of the protein.
A. deleting the first signal sequence
B. changing the hydrophobic amino acids in the first, cleaved sequence to charged amino acids
C. changing the hydrophobic residues in every other transmembrane sequence to charged residues, starting with the first, cleaved signal sequence
A. Deleting the first signal sequence completely would convert the next membrane-spanning domain into an internal start-transfer signal and would invert the orientation of the protein
B. Changing the hydrophobic amino acids to charged amino acids destroys the ability of the sequence both to act as a signal sequence and to become a membrane-spanning sequence. Therefore, the adjacent membrane-spanning domain will now become an internal start-transfer sequence and the protein will be inverted, as seen above in part A. The mutated signal sequence would not get cleaved off, because it would remain on the cytoplasmic side of the membrane and signal peptidase is found only inside the ER
C. Mutating every other membrane-spanning region so that they are now charged (and thus cannot span the membrane) would decrease the number of transmembrane regions and increase the size of the loops between membrane-spanning regions
10)
In a cell capable of regulated secretion, what are the three main classes of proteins that must be separated before they leave the trans Golgi network?
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Those destined for lysosomes
Those destined for secretory vesicles
Those destined for immediate delivery to the cell surface
11)
Name three possible fates for an endocytosed molecule that has reached the endosome.
Recycled to the original membrane
Destroyed in the lysosome
Transcytosed across the cell to a different membrane
12)
If a lysosome breaks, what protects the rest of the cell from lysosomal enzymes?
The lysosomal enzymes are all acid hydrolases, which have optimal activity at the low pH (about 5.0) found in the interior of lysosomes. If a lysosome were to break, the acid hydrolases would find themselves at pH 7.2, the pH of the cytosol, and would therefore do little damage to cellular constituents.
13)
Two protein kinases, PK1 and PK2, work sequentially in an intracellular signaling pathway. You create cells that contain inactivating mutations in the genes that encode either PK1 or PK2 and find that these cells no longer respond to a particular extracellular
signal. You also create cells containing a version of PK1 that is permanently active and find that the cells behave as though they are receiving the signal even when the signal is
not present. When you introduce the permanently active version of PK1 into cells that have an inactivating mutation in PK2, you find that these cells also behave as though they are receiving the signal even when no signal is present.
a. From these results, does PK1 activate PK2, or does PK2 activate PK1? Explain your answer.
b. You now create a permanently active version of PK2 and find that cells containing this version behave as though they are receiving the signal even when the signal is not present. What do you predict will happen if you introduce the permanently active version of PK2 into cells that have an inactivating mutation in PK1?
a. Normally, PK2 activates PK1. We are told that PK1 and PK2 normally work sequentially in an intracellular signaling pathway. If PK1 is permanently activated, a response is seen independently of whether or not PK2 is present. If PK1 activated PK2, no response should be seen if PK1 were activated in the absence of PK2.
b. You would predict that no response to the signal would be observed. This is because PK2 normally needs to activate PK1 for the cells to respond to the signal. When PK2 is permanently activated in the absence of PK1, PK1 is not there to relay the signal.
Chapter 16
1)
When adrenaline binds to adrenergic receptors on the surface of a muscle cell, it activates a G protein, initiating an intracellular signaling pathway in which the activated α
subunit activates adenylyl cyclase, thereby increasing cAMP levels in the cell. The cAMP
molecules then activate a cAMP-dependent kinase (PKA) that, in turn, activates enzymes that result in the breakdown of muscle glycogen, thus lowering glycogen levels.
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You obtain muscle cells that are defective in various components of the signaling pathway. Referring to Figure Q16-29, indicate how glycogen levels would be affected in the presence of adrenaline in the following cells. Would they be higher or lower than in normal cells treated with adrenaline?
A. cells that lack adenylyl cyclase
B. cells that lack the GPCR
C. cells that lack cAMP phosphodiesterase
D. cells that have an α subunit that cannot hydrolyze GTP but can interact
properly with the β and γ subunits
A. HIGHER
B HIGHER
C. LOWER
D. LOWER
2)
The rod photoreceptors in the eye are extremely sensitive to light. The cells sense
light through a signal transduction cascade involving light activation of a GPCR that activates a G protein that activates cyclic GMP phosphodiesterase. How would you expect the addition of the following drugs to affect the light-sensing ability of the rod cells? Explain your answers.
A. a drug that inhibits cyclic GMP phosphodiesterase
B. a drug that is a nonhydrolyzable analog of GTP
A. DECREASE: A drug that inhibits cyclic GMP phosphodiesterase would decrease any light response in the rod cell. Normally, cyclic GMP is continuously being produced in the eye. The perception of light by a rod cell normally leads to the activation of cyclic GMP phosphodiesterase, which then hydrolyzes cyclic GMP molecules. This causes Na+ channels to
close, which changes the membrane potential and alters the signal sent to the brain. If cyclic GMP phosphodiesterase were blocked, levels of cyclic GMP would remain high and there would
be no cellular response to light.
B. PROLONGED RESPONSE:
A drug that is a nonhydrolyzable analog of GTP would lead to a prolonged response to light. This is because a nonhydrolyzable analog of GTP would prevent the G protein from turning itself off by hydrolyzing its bound GTP to GDP. Continued activation of the G protein would keep cyclic GMP phosphodiesterase levels higher than normal, leading to a prolonged period of lowered levels of cyclic GMP. This in turn would cause Na+ channels to
be closed for longer than normal, leading to a prolonged change in the membrane potential and an extended light response.
3)
Antibodies are Y-shaped molecules that have two identical binding sites. Suppose that you have obtained an antibody that is specific for the extracellular domain of an RTK. When the antibody binds to the RTK it brings together two RTK molecules. If cells containing the RTK were exposed to the antibody, would you expect the kinase to be activated, inactivated or unaffected?
The RTK will probably become activated on binding the antibody molecule. this is because the signal induced dimerization usually activates RTKs. When RTK molecules are brought together their cytoplasmic
kinase domains become activated and each receptor phosphorylates the other.
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4)
Two protein kinases, PK1 and PK2, work sequentially in an intracellular signaling pathway. You create cells that contain inactivating mutations in the genes that
encode either PK1 or PK2 and find that these cells no longer respond to a
particular extracellular signal. You also create cells containing a version of PK1
that is permanently active and find that the cells behave as though they are
receiving the signal even when the signal is not present. When you introduce the permanently active version of PK1 into cells that have an inactivating mutation in PK2, you find that these cells also behave as though they are receiving the signal
even when no signal is present.
A. From these results, does PK1 activate PK2 or does PK2 activate PK1?
Explain your answer.
B. You now create a permanently active version of PK2 and find that cells
containing this version behave as though they are receiving the signal even
when the signal is not present. What do you predict will happen if you
introduce the permanently active version of PK2 into cells that have an
inactivating mutation in PK1?
A. PK2 activates PK1.
B. predict that no response to signal would be observed
5)
Your friend is studying mouse fur color and has isolated the GPCR responsible for determining its color, as well as the extracellular signal that activates the receptor. She finds that, on addition of the signal to pigment cells (cells that produce the pigment determining fur color), cAMP levels rise in the cell. She
starts a biotech company, and the company isolates more components of the
signaling pathway responsible for fur color. Using transgenic mouse technology, the company genetically engineers mice that are defective in various proteins involved in determining fur color. The company obtains the following results.
Normal mice have beige (very light brown) fur color.
Mice lacking the extracellular signal have white fur.
Mice lacking the GPCR have white fur.
Mice lacking cAMP phosphodiesterase have dark brown fur.
Your friend has also made mice that are defective in the α subunit of the G
protein in this signaling pathway. The defective α subunit works normally except
that, once it binds GTP, it cannot hydrolyze GTP to GDP. What color do you
predict that the fur of these mice will be? Why?
Dark brown fur, because the mice with defective alpha subunit should have white fur. We can see that any block in signal due to absence of extracellular signal or defective GPCR lead to mice with white fur. Thus, with defective G-protein the signal will be blocked (since it attaches to GTP but cannot hydrolyze it). If the signal is not regulated (as in absence of phosphodiesterase) the synthesis of the some pigment protein becomes constitutive (Non stop gene expression) and that results in dark brown fur.
Chapter 17
1)
Do you agree or disagree with this statement? Explain your answer.
Minus end-directed microtubule motors (like dyneins) deliver their
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cargo to the periphery of the cell, whereas plus end-directed
microtubule motors (like kinesins) deliver their cargo to the
interior of the cell.
Disagree. The plus ends of microtubules usually point toward the cell periphery,
whereas the minus ends point toward the cell center. This is because the γ-tubulin
in the centrosome serves to nucleate microtubule growth. Because the
centrosomes are near the center of the cell, the minus ends of microtubules are
located there. Therefore, a minus-end-directed microtubule motor would direct its
cargo toward the center of the cell, and a plus-end-directed microtubule motor
would direct its cargo toward the cell periphery.
2)
Do you agree or disagree with the following statement? Explain your answer.
When skeletal muscle receives a signal from the nervous system to
contract, the signal from the motor neuron triggers the opening of a
voltage-sensitive Ca2+ channel in the muscle cells' plasma
membrane, allowing Ca2+ to flow into the cell.
Disagree. The increase in intracellular Ca2+ during muscle contraction comes from an intracellular source. The Ca2+ is released from the lumen of the sarcoplasmic reticulum, which is a specialized region of endoplasmic reticulum inside a muscle cell. The signal from the nerve terminal triggers an action potential in the muscle cell plasma membrane, which causes a voltage-sensitive transmembrane protein in the membranous transverse tubules to open a Ca2+
release channel in the membrane of the sarcoplasmic reticulum
Chapter 20
1)
Indicate the direction in which the plant cell shown in Figure Q20-4 is most likely to grow. The black lines indicate the direction of the cellulose microfibrils around
the cell. Explain your answer.
Vertically, Cellulose fibers are highly resistant to stretching,
so a plant cell tends to grow, under the stimulus of turgor pressure, in a direction
perpendicular to the orientation of the fibers in the cell wall.
2)
Indicate whether the following molecules are found in plants, animals, or both.
A. intermediate filaments
B. cell walls
C. microtubules
D. cellulose
E. collagen
A. animals
B. plants
C. both
D. plants
E. animals
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3)
What are the main structures providing tensile strength in the following?
A. animal connective tissue
B. animal epidermis
C. plant cell walls
A. collagen fibers
B. intermediate filaments
C. cellulose fibers
4)
Do you agree or disagree with the following statement? Explain your answer.
Like many other extracellular proteins, newly synthesized collagen
molecules undergo post-translational processing inside the cell to
convert them into their mature form; they are then secreted and
self-assemble into fibrils in the extracellular space.
Disagree. The cell secretes newly synthesized collagen molecules in an immature form as procollagen, and the peptides at the ends of the procollagen molecules then have to be cleaved off in the extracellular space before fibril assembly can
occur. This process ensures that collagen fibrils will not assemble prematurely inside the cell.
5)
Fill IN the blank: Desmosomes join the intermediate filaments in one cell to those in the
neighboring cell. Hemidesmosomes anchor intermediate filaments in a cell to the extracellular matrix. Adherens junctions involve cadherin connections between
neighboring cells and are anchorage sites for actin filaments. Gap junctions
permit the passage of small molecules from one cell to its adjacent cell. Tight junctions prevent the leakage of molecules between adjacent cells.
6)
Name the three key mechanisms important for maintaining the organization of
cells into tissues.
1. cell communication
2. selective cell-cell adhesion
3. cell memory
7)
Place the following in order of their replacement times, from shortest to longest.
A. epidermal cell
B. nerve cell
C. bone matrix
D. red blood cell
E. cell lining the gut
cell lining the gut (few days) < epidermal cell (1 or 2 months) < red blood cell
(4 months) < bone matrix (10 years) < nerve cell (lifetime)
8)
A stem cell divides into two daughter cells. One of the daughter cells goes on to become a terminally differentiated cell. What is the typical fate of the other daughter cell?
The other daughter cell typically remains a stem cell.
9)
Your friend is a pioneer in ES cell research. In her research, she uses an ES cell line that
originated from an inbred strain of laboratory mice called FG426. She has just figured
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out methods that allow her to grow an entire liver from an ES cell and has successfully grown 10 livers. She demonstrates that the newly grown
livers are functional by successfully transplanting one of the new livers into a FG426 laboratory mouse
You are particularly excited about this, because you have a sick pet mouse,
Squeaky. You are very attached to Squeaky, as you found him when you were out
camping in New Hampshire. Unfortunately, Squeaky has developed liver disease
and will not live much longer without a liver transplant. After you see your friend
on TV talking about her new method for growing mouse livers, you immediately
grab your cell phone to ask her whether Squeaky could have one of the newly
grown livers. Just as you are about to dial your friend, you remember something
you learned in cell biology and realize that instead, you should ask your friend
about possibly using therapeutic cloning for Squeaky's benefit.
A. Why do you think that one of the newly grown livers may not work in
Squeaky?
B. Explain why therapeutic cloning would solve this problem.
A. For organ transplantation to be successful, the donor and the recipient
should be as close a genetic match as possible to minimize the risk of
immunological rejection. Because you found Squeaky in the fields of New
Hampshire, Squeaky is likely to have genetic differences from the FG426
inbred laboratory mice.
B. Therapeutic cloning may help save Squeaky because it is a method for
generating ES cells from Squeaky himself—cells that will be genetically
identical to Squeaky. You hope that your friend will be successful when
applying her methods to the ES cell line derived from Squeaky and can
grow a functional liver for him.
10)
Cancer is a disease of enhanced proliferation and cell survival. DNA repair mechanisms are normally important for cell survival. When a cell senses DNA damage, the cell cycle is inhibited until the damage is fixed. Given the importance of DNA repair mechanisms, how can their failure can lead to the production of cancer cells with a competitive advantage over normal cells?
A cell with a defect in its DNA repair mechanisms will have an increased mutation rate, thus increasing its chances to acquire further mutations that give it
(and its progeny) a competitive growth advantage.
is normal and one copy that is mutated—that is, people who are heterozygous for Rb—have a greatly increased risk of cancer.
Given this information, do you agree or disagree with the following statement?
Explain your answer.
The Rb mutation must have a dominant effect, which means that it
must result in an increase in Rb function. Thus, Rb in its mutant
form must be an oncogene.
Disagree. It is true that the Rb mutation is dominant, in the sense that a person
who is heterozygous (inherits one normal copy and one mutant copy of the gene)
is likely to show the mutant trait (that is, will be cancer-prone). However, this
does not mean that the mutation in Rb causes an increase in Rb gene function; in
fact, the opposite is true—the propensity for cancer arises from a loss of Rb gene
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function. Therefore, Rb should be classified as a tumor suppressor gene and not as
an oncogene. Most people have two functional Rb genes in each of their cells.
Thus, for one of their cells to turn cancerous by losing Rb function, both copies of
the gene in that cell must be inactivated or lost, a two-step process. However, in a
person born lacking one copy of the Rb gene, each cell is only one step away from
a complete loss of Rb function. Consequently, such a person has a high risk that at
least one of the cells in the body will undergo a mutation that precipitates cancer.
In this way, at the level of the whole person, the Rb loss-of-function mutation is
dominant, even though at the level of the individual cell it is recessive.
11)
Figure Q20-47 shows a sequence of mutations that might underlie the
development of colorectal cancer. Explain why the loss of p53 is advantageous to
cancerous cells.
When DNA is damaged, the protein p53 is activated and stabilized. When p53 is
active, it stops the cell cycle to give the cell time to repair its damaged DNA, or, if
that is not feasible, it causes the cell to commit suicide by apoptosis. Cells lacking
p53 will continue to replicate their DNA, will avoid suicide, and will go through
cell division even when the DNA has been damaged, producing mutant daughter
cells. Repeated rounds of cell division under these circumstances perpetuate the
genetic damage and allow still more mutations to occur. Some of these newly
accumulated mutations will give the cell an increased ability to survive,
proliferate, and metastasize, resulting in invasive cancer.
12)
Drugs that block the function of oncogenic proteins hold great promise in the
fight against cancer. Should cancer researchers also be attempting to design drugs
that will interfere with the products of tumor suppressor genes? Explain.
Oncogenic proteins lead toward cancer, because they have excessive or
unregulated activity in comparison with the corresponding normal proteins.
Blocking this activity with a drug molecule that simply clogs the active site of the
oncogenic protein removes the danger. For a tumor suppressor gene, the danger
lies in a loss of function, and there is generally no simple way for a drug molecule
to restore a protein function that has been lost. It is therefore hard to see how we
could achieve any useful effect on cancer by means of drugs that interfere with
tumor suppressor gene products. A drug that simply inhibited their function would
be expected to promote, not cure, cancer.
Random: 1)
Receipt of extracellular signals can change cell behavior quickly (for example, in seconds or less) or much more slowly (for example, in hours).
A. What kind of molecular changes could cause quick changes in cell behavior?
B. What kind of molecular changes could cause slow changes in cell behavior?
C. Explain why the response you named in A results in a quick change, whereas the response you named in B results in a slow change.
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A. Any answer that involves the modification of existing cell components is correct. Protein phosphorylation, protein dephosphorylation, protein ubiquitylation, lipid phosphorylation, and lipid cleavage are all examples of correct answers.
B. Responses that involve altercations in gene expression occur slowly.
C. Modification of existing cell components can happen quickly, whereas responses that depend
on changes in gene expression that take much longer, because the genes will need to be transcribed the mRNAs will need to be translated, and the proteins need to accumulate to high-
enough level to instigate change.
2)
Phosphorylation of nuclear lamins regulates their assembly and disassembly during mitosis. You add a drug to cells undergoing mitosis that inhibits the
activity of an enzyme that dephosphorylates nuclear lamins. What do you predict will happen to these cells? Why?
Key points are; relationship between phosphorylation of lamins, and nuclear envelop formation and phosphorylation of lamins promote disassembly.
-Phosphorylation of lamins by M-cdk required for nuclear envelope breakdown during mitosis.
-Dephosphorylation of lamins after metaphase in mitosis is required for reassembly of nuclear envelope at telophase.
-Therefore, this drug would cause defect in reassembly of nuclear envelope in telophase.
3)
When cells need to eliminate the mitochondria, they use the autophagy mechanism to destroy mitochondria. From our discussions in class, why do think cells need to use that mechanism?
Key points are; relationship between mitochondria and intrinsic apoptosis, and key molecule (cytochrome c), and how autophagy can prevent initiation of apoptosis.
1. Initial step: Leaking of cytochrome C from mitochondria
2. Cells need to prevent cytochrome C leakage to prevent unwanted apoptosis
3. Autophagy surrounds mitochondria with membrane to separate it then destroy it.
4. This way cytochrome C won't contact cytoplasmic components and form apoptosome
4)
You studied signal transduction with the technology called "single cell stimulation". This technology allows you to apply a ligand to a single cell in a group of cells in contact with each other. You find that group of cells surrounding the stimulated cell showed the same response but cells in different area of the dish, which are not in direct contact with the stimulated cell, did not show any response. From these results, 1) what kind of mechanism do you expect the cell is using in this particular signal transduction pathway?
2) what do you expect regarding the cell/cell junction of responding cells? (6pt)
Key points are; Utilizing "Second messenger" molecules for signal transduction. And, GAP junction to share that messenger molecules among the cells in direct contact.
For instance, if you indicated adherens junction, you have to mention the signal need to use actin filament as an intra-cellular signaling that could affect the junction then make contacted
cells might respond to it.
1. Cells connected by GAP junction can share small biomolecules
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2. Second messenger molecules can be shared by all cells attached by GAP junction (Ca+2 or cAMP)
3. SO, all responding cells should be connected via Gap junction
5)
In the intestine, cells on top of the _
villus
_ undergo _
apoptotic
_ cell death continuously. Lost cells are supplied by newly divided cells at the _
crypt
__. All newly divided cells are provided by the cell division of __
stem
__ cells by __
asymmetrical
__ cell division that produce different type of daughter cells. Some of the newly divided cells
are differentiated into _
goblet
__ cells, which produce mucus.
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