Answers to Review Questions
1.1 Theme: Dimensions in Cell Biology
General comment: 1 m = 103 mm = 106 μm = 109 nm.
1. E. 2,000 nm = 2 μm. Most bacteria are in the range 1 μm to 2 μm.
2. F. 20,000 nm = 20 μm. Most eukaryotic cells are in the range 5 μm to 100 μm.
3. I. 1,000,000,000 nm = 1 m. Nerve cells supplying the fingers and toes are about this length.
4. D. 250 nm. The wavelength of green light is 500 nm; light microscopes can achieve a resolution of about half this.
5. B. 0.2 nm. This allows the electron microscope to reveal structures that are invisible in the light microscope.
1. C = fibroblast. Fibroblasts secrete collagen and other components of the connective tissue extracellular matrix.
2. E = glial cell. Glial cells and nerve cells are the two main types found in nervous tissue.
3. B = epithelial cell. An epithelium is a sheet of cells.
4. D = macrophage. Macrophage means “big eater” in Latin; macrophages take up unwanted material and digest it.
5. A = bacterium. Prokaryotic cells have no nuclear envelope; the genetic material is free in the cytosol.
6. F = skeletal muscle cell. Many precursor cells, each with one nucleus, fuse to form the large cells that form our muscles.
7. G = stem cell.
1.3 Theme: Some Basic Components of the Eukaryotic Cell
1. Cytosol = D.
2. Internal membranes = C.
3. Mitochondrion = F.
4. Nucleus = B.
5. Plasma membrane = A.
2.1 Theme: Types of Organic Chemicals
1. A disaccharide = F, lactose. Lactose is a disaccharide found, as the name suggests, in milk.
2. A fatty acid = H, oleic acid.
3. A nucleoside = B, adenosine.
4. A nucleotide = C, adenosine triphosphate. A nucleotide is a phosphorylated nucleoside.
5. A pentose sugar = I, ribose. Ribose, one of the building blocks of nucleic acids, is the most important pentose sugar in the body.
2.2 Theme: Chemical Groups and Bonds
General comments: molecule (i) is an acetylated sucrose molecule. You should have recognized the oxygen-
containing ring structures as sugar residues, and hence the link between them is a glycosidic bond. On the right, attached to the sucrose by an ester bond, is an acetate residue. Singly acetylated sugars like this can easily be made but have no major role either in nature or industry. However sucrose with many acetate and butyrate residues attached by ester bonds, a molecule called sucrose acetate isobutyrate, is a food additive—it is referred to as E444 in Europe.
Molecule (ii) is a dipeptide of cysteine on the left and serine on the right, connected by a peptide bond. The side chain of serine has been phosphorylated, a modification we will meet again in Chapter 9.
Diagram (iii) shows part of an RNA molecule hydrogen bonded to a base on another nucleic acid, which could be either DNA or RNA. We will cover RNA in depth in Chapter 6.
1. amino group = D.
2. carboxyl group = G.
3. hydroxyl group = A. Note that OH groups within a carboxyl group, such as the group at G, are not called hydroxyl groups.
4. phosphate group = F. It would not be unreasonable to answer I, but since the phosphate atom at I is connected to the rest of the molecule by two bonds, we would not normally refer to the group at I as a phosphate group.
5. ester bond = C. Adding the elements of H2O to this bond would regenerate a hydroxyl group on the sugar and a carboxyl group on the acetate, so this is an ester bond between a hydroxyl group and an organic acid.
6. glycosidic bond = B.
7. hydrogen bond = H.
8. peptide bond = E.
9. phosphodiester link = I.
1. Considering the acetate, both the protonated form CH3COOH and the deprotonated form CH3COO− are present at significant concentrations = B. pH 5. When the pH is close to the pKa of the reaction CH3COOH ↔ CH3COO− + H+ then both the protonated and unprotonated forms will be present.
2. Considering the ammonium, both the protonated form NH4+ and the unprotonated form NH3 are present at significant concentrations = D. pH 9. When the pH is close to the pKa of the reaction NH3 + H+ ↔ NH4+ then both the protonated and unprotonated forms will be present.
3. The vast majority of both the acetate and the ammonium are in their protonated forms, CH3COOH and NH4+ respectively = A. pH 3. At this pH the H+ concentration is very high, driving protons onto all potential acceptors.
4. The vast majority of both the acetate and the ammonium are in their unprotonated forms, CH3COO− and NH3 respectively = E. pH 11. At this pH the H+ concentration is very low, so that even ammonium ions give up their H+.
5. The vast majority of both the acetate and the ammonium are in their ionic forms, CH3COO− and NH4+ respectively = C. pH 7. This pH is more alkaline than the pKa of the reaction CH3COOH↔CH3COO− + H+, so the acetic acid gives up its H+ to become acetate. However, the pH is more acid than the pKa of the reaction NH3 + H+ ↔ NH4+, so ammonium remains in its protonated form, NH4+.
6. The vast majority of both the acetate and the ammonium are in their uncharged forms, CH3COOH and NH3 respectively = F. This situation is impossible. For acetate to be in its uncharged, protonated form CH3COOH, the pH must be much more acid than the pKa of the reaction CH3COOH↔CH3COO− + H+, that is, more acid than pH 4.8. However for ammonia to be in its unchanged, unprotonated form the pH must be much more alkaline than the pKa of the reaction NH3 + H+ ↔ NH4+, that is, more alkaline than pH 9.2.
1. D. Mitochondria, the nucleus, and in plant cells, chloroplasts, are surrounded by a double membrane envelope.
2. A.
3. B.
4. F.
5. A.
3.2 Theme: Organelles in Eukaryotic Cells
1. a site of protein synthesis = A, endoplasmic reticulum.
2. contains many powerful digestive enzymes = C, lysosome.
3. contains small circular chromosomes = D, mitochondrion. Only the nucleus, mitochondria, and (in plants) chloroplasts contain DNA.
4. contains the enzyme catalase = F, peroxisome.
5. filled with chromatin = E, nucleus. Chromatin, so called because it stains strongly with colored dyes, is a complex of DNA and histones.
6. made up of flattened sacks called cisternae = B, Golgi apparatus.
7. most of the cell's ATP is made here = D, mitochondrion.
8. usually found at the cell centre = B, Golgi apparatus. The cell center is a specific region, close to the nuclear envelope, where the Golgi complex and the centrosome are found.
3.3 Theme: Transport Across Membranes
1. An RNA molecule of Mr = 10,000 = C. RNA molecules are hydrophilic because their building blocks are hydrophilic molecules such as sugars and phosphate. They cannot therefore cross lipid bilayers by simple diffusion. Most are also too large to go through gap junctions. The Mr of this particular RNA was 10,000, and gap junctions do not allow molecules of Mr > 1000 to pass.
2. Inositol trisphosphate (Mr = 649) = B. Inositol trisphosphate is highly hydrophilic since it bears six negative charges. It cannot therefore cross lipid bilayers by simple diffusion. However, inositol trisphosphate is small enough to pass through gap junctions. This is thought to be an important mechanism of cell-cell signaling; we will learn some of the signaling functions of inositol trisphosphate in Chapter 15.
3. K+ (atomic weight = 39) = B. As a small ion, K+ is highly hydrophilic and cannot cross lipid bilayers by simple diffusion. However K+ is readily permeable through gap junctions. Passage of electrical current, by the movement of K+ and other ions, allows heart cells to contract in concert.
4. Nitric oxide = NO (Mr = 30) = A. NO is a small uncharged molecule and has sufficient solubility in hydrophobic solutes to be able to cross lipid bilayers by simple diffusion. In Chapter 16 we will describe how the movement of NO from one cell to its neighbor in this way is critical in allowing the blood supply to respond to the needs of the tissues.
1. A change of a U to a A in the 6th codon of the sequence, generating the sequence 5′ACU AUC UGU AUU AUG UAA CAC CCA3′ = C, nonsense mutation. UAA is a stop codon, so translation into a polypeptide chain will terminate prematurely.
2. A change of a U to a C in the 6th codon of the sequence, generating the sequence 5′ACU AUC UGU AUU AUG CUA CAC CCA3′ = D, none of the above. CUA, like the original UUA, codes for leucine, so the encoded polypeptide sequence is unchanged. This type of mutation is known as a synonymous mutation.
3. A change of a U to a G in the 2nd codon of the sequence, generating the sequence 5′ACU AGC UGU AUU AUG UUA CAC CCA3′ = B, missense mutation. The AGC now codes for serine in place of the original isoleucine.
4. Deletion of a U in the 3rd codon, generating the sequence 5′ACU AUC UGA UUA UGU UAC ACC CA3′ = C, nonsense mutation. The deletion has generated the stop codon UGA, so translation into a polypeptide chain will terminate prematurely.
5. Deletion of an A in the 4th codon, generating the sequence 5′ACU AUC UGU UUA UGU UAC ACC CA3′ = A, frameshift mutation. From the deletion onwards, the sequence will be read using the wrong reading frame, creating a different polypeptide sequence.
4.2 Theme: Bases and Amino Acids
1. A nitrogen-rich base that is not a component of DNA = J, uracil. Uracil replaces thymine in RNA.
2. A positively charged amino acid that is found in large amounts in chromatin, where it neutralizes the negative charge on the phosphodiester links of DNA = C, arginine. Both positively charged amino acids, arginine and lysine, are present in large amounts in histones, the proteins around which DNA wraps to form the nucleosome.
3. A protein is described as having the mutation G5E. Which amino acid is present in this protein in place of the amino acid present in the normal protein? = F, glutamate. The notation G5E means that the fifth amino acid in the protein is usually glycine (G) but is glutamate (E) in the mutant.
4. The base that pairs with guanine in double stranded DNA = E, cytosine.
5. The base that pairs with thymine in double stranded DNA = A, adenine.
4.3 Theme: Structures Associated with DNA
1. A highly compacted, darkly staining substance comprising DNA and protein found at the nuclear periphery = E, heterochromatin. Heterochromatin is the form adopted by DNA that is not being transcribed into RNA. In Chapter 3 we described how heterochromatin is located at the edge of the nucleus.
2. A mass of DNA and associated proteins lying free in the cytoplasm = F, nucleoid. This is the organization found in prokaryotes.
3. A structure formed when a 146 base pair length of DNA winds around a complex of histone proteins = G, nucleosome.
4. The form adopted by those parts of chromosomes that are being transcribed into RNA = C, euchromatin.
5.1 Theme: Synthesis on a DNA Template
1. the sequence that is generated from 3′GCGAAGTCGTA 5′ during the process of transcription = F. Because the strand is being transcribed (i.e., RNA synthesis) a U is inserted into the RNA strand when an A occurs in the template strand.
2. the sequence that is generated from 3′GCGAAGTCGTA 5′ during the process of replication = G. Replication is the process of DNA synthesis. Sequence G is complementary to the template strand shown.
3. the sequence that is generated from 5′ATGCTGAAGCG3′ during the process of transcription = F. Sequence F is complementary to the template strand shown. Because the strand is being transcribed (i.e., RNA synthesis) a U is inserted into the RNA strand when an A occurs in the template strand.
4. the sequence that is generated from 5′ATGCTGAAGCG3′ during the process of replication = G. Replication is the process of DNA synthesis. Sequence G is complementary to the template strand shown.
1. DNA ligase = B. DNA ligase connects adjacent deoxyribonucleotides within an otherwise complete double-stranded DNA molecule, both during normal DNA synthesis, as here, and during DNA repair.
2. DNA polymerase I = F. DNA polymerase I is responsible for building the DNA chain in those regions previously occupied by the RNA primer.
3. DNA polymerase III = E. DNA polymerase III runs nonstop to create the leading strand. It also synthesizes most of the lagging strand, but is incapable of running through the sections already occupied by RNA primers; DNA polymerase I is required in those sections.
4. exonuclease I = D. We have described how during mismatch repair exonuclease I runs in a 3′ to 5′ direction, destroying the DNA strand.
5. exonuclease III = D. We have described how during mismatch repair exonuclease VII runs in a 5′ to 3′ direction, destroying the DNA strand.
6. helicase = A. As the first step in DNA replication, helicase splits the DNA double helix into two single strands.
7. primase = C. This occurs repeatedly during synthesis of the lagging strand. As we will describe in Chapter 6, other enzymes (the RNA polymerases) synthesize RNA during transcription, but here we are concerned with DNA replication.
8. ribonuclease H = G. In prokaryotes DNA polymerase performs this function and then goes on to synthesize the complimentary DNA strand; however in eukaryotes there is a specialized stand-alone enzyme to do the job.
5.3 Theme: Regions within Eukaryotic Chromosomes
1. A section of DNA that, read in triplets of bases, encodes successive amino acids in a polypeptide chain = A. In eukaryotic genes, only the exons encode the polypeptide sequence of a protein.
2. A section of DNA with a sequence similar to a working gene but which no longer encodes a functional protein = E, pseudogene.
3. A section of DNA within a gene that is transcribed into RNA but which does not encode amino acids and which must be removed from the RNA before it leaves the nucleus = C, intron.
4. A series of identical or almost identical genes all of which are transcribed so as to generate identical or almost identical RNA products and, in the case of protein coding genes, identical proteins = G. Examples are the genes that encode ribosomal RNA, transfer RNA, and histone proteins. Note that the answer “gene family” is wrong—members of a gene family encode significantly different, albeit strongly similar, proteins.
5. A type of DNA with a presumed structural role that makes up much of the chromosome in the centromere region = F, satellite DNA.
6. An extragenic DNA sequence that is repeated more than a million times = F, satellite DNA. Note that the answer “long interspersed nuclear element” is wrong—the copy number of LINEs is in the thousands, not in the millions.
6.1 Theme: Codes within the Base Sequence
1. A stretch of DNA rich in adenine and thymine, called the TATA box = E, transcription initiation in eukaryotes. TATA binding protein, a component of transcription factor IID, binds to the TATA box on the DNA, allowing recruitment of the other transcription factors and then RNA polymerase II.
2. A stretch of DNA rich in guanine and cytosine, followed by a string of adenines = H, transcription termination in prokaryotes. The resulting RNA strand forms a hairpin loop in the GC-rich region, reducing the size of the transcription bubble, while the attachment of the following uracils to the string of adenines is weak since there are only two hydrogen bonds in each UA pair.
3. The DNA sequence GGGGCGGGGC, called the GC box = E, transcription initiation in eukaryotes. This is an alternative transcription initiator used in many eukaryotic genes. A protein called Sp1 binds to the GC box and recruits TATA binding protein, allowing recruitment of the other transcription factors and then RNA polymerase II.
4. The DNA sequence TATATT, called the −10 or Pribnow box = F, transcription initiation in prokaryotes. The −10 box is part of the prokaryotic promoter sequence, which recruits the σ factor, allowing recruitment of the other subunits of RNA polymerase.
5. The RNA motif GU … . AG, where the … . indicates a long sequence of bases = D, removal of introns in eukaryotic mRNA. All eukaryotic introns conform to the sequence GU … . AG, and although the process is incompletely understood, these bases are involved in the process by which the spliceosome recognizes the length of RNA to be removed.
6. The RNA sequence AAUAAA = C, polyadenylation of eukaryotic mRNA. This sequence is found close to the 3′ end of most mRNAs and is thought to be a signal for poly-A polymerase to attach and add a string of A residues to the 3′ end of the mRNA.
6.2 Theme: The Control of Transcription
1. Catabolite activator protein = C. In the presence of cyclic AMP, catabolite activator protein binds to a regulatory site of the lac operon of E. coli and increases transcription. In turn, cAMP is only high if glucose is in low supply.
2. Glucocorticoid hormone receptor = C. In the presence of glucocorticoid hormone, the glucocorticoid hormone receptor binds to an enhancer site of various mammalian genes and increases transcription.
3. Lac repressor protein = B. If the lac repressor protein binds a β-galactoside sugar, it adopts a shape that can no longer bind to the operator region of the operon. Thus the enzymes that allow utilization of β-galactoside sugars are only made if these sugars are present.
4. Trp aporepressor protein = D. When the aporepressor protein binds tryptophan, it can then bind to the operator region of the trp operon and inhibit transcription. Thus the enzymes that synthesize tryptophan are only made when tryptophan is lacking.
6.3 Theme: Events that Occur after Transcription in Eukaryotes
1. A chemical modification of the 3′ end of the RNA molecule = D, polyadenylation. A long stretch of adenosine residues is added to the 3′ end of the mRNA transcript.
2. A chemical modification of the 5′ end of the RNA molecule = A, capping. The methylated guanosine is added by a 5′ to 5′ phosphodiester link, unlike the 3′ to 5′ links formed by RNA polymerase.
3. A process that allows two or more polypeptide chains of different amino acid sequence to be generated from the same mRNA transcript = E, RNA splicing. Alternative splicing allows one primary mRNA transcript to give rise to two processed mRNAs that share some exons but differ in others. It is distinct from the phenomenon of polycistronic mRNA seen in prokaryotes, where sequential lengths of mRNA on the same molecule are translated to generate completely different proteins.
4. A process that reduces, often dramatically, the length of the RNA molecule prior to its subsequent translation into protein = E, RNA splicing. Splicing removes the
introns to leave only exonic, coding RNA. Answer B, digestion by nucleases, is wrong because after digestion the mRNA cannot be translated into protein.
7.1 Theme: A Mammalian Expression Plasmid
1. DNA encoding cytochrome c is inserted into the plasmid. Which element in the plasmid makes this possible = D, the multiple cloning site. This comprises the recognition sites of a number of restriction enzymes, allowing the experimenters to choose a site also present at the appropriate places on the DNA encoding the cytochrome c, cut both pieces of DNA with the same enzyme, and then assemble the recombinant plasmid.
2. To generate large amounts of the recombinant plasmid, the plasmid is grown up in bacteria. The plasmid is used to transform competent E. coli which are then cultured in such a way that only bacteria containing the plasmid survive. Which element of the plasmid allows the survival of host bacteria when sister bacteria are dying = E, antibiotic resistance gene. The bacteria are grown on agar containing the toxic antibiotic. Only bacteria containing the resistance gene survive.
3. The transformed bacteria divide repeatedly, producing colonies each derived from a single transformed progenitor. What element in the plasmid allows it to be copied in parallel with the host bacterium's DNA = A, origin of replication. This is the point where the host enzyme DnaA attaches, allowing formation of the open replication complex.
4. Clones containing the recombinant plasmid are grown up and then lysed, allowing purification of large amounts of recombinant plasmid. The purified plasmid is then used to transfect human cells, which synthesize the GFP:cytochrome c chimera. Which element in the plasmid allows the chimaeric protein to be expressed in the HeLa cells, even though it was not expressed in the bacteria = B, cytomegalovirus promoter. This is a powerful promoter used by the cytomegalovirus to drive transcription of its genes in preference to those of the host, and is very commonly used by experimenters to drive expression from plasmids introduced into mammalian cells. This promoter is not recognized by the bacterial RNA polymerase and therefore the chimeric mRNA (and hence protein) will not be made in bacterial cells.
7.2 Theme: Choosing an Oligonucleotide for a Specific Task
1. In the polymerase chain reaction: indicate the oligonucleotide that should be used together with the oligonucleotide 5′TACGGATCCCTTTGCAGGAT3′ to
amplify the double stranded DNA molecule shown at the top. = B. The oligonucleotide 5′TGCCTACTGCAGCGTCTGCA3′ can be used to copy the top strand of the sequence shown in the 5′ to 3′ direction.
2. If you wished to clone the amplified product into the EcoR1 site of a plasmid how would you modify the oligonucleotide 5′TACGGATCCCTTTGCAGGAT3′ = E. The sequence the enzyme will recognize is simply added to the 5′ end of the oligonucleotide generating 5′GAATTCTACGGATCCCTTTGCAGGAT3′.
3. An oligonucleotide that could be used to prime the synthesis of DNA from most of the mRNAs present in a tissue, in order to generate a cDNA library = A. Most eukaryotic mRNAs have at their 3′ end a poly-A tail. The primer 5′TTTTTTTTTTTTTTT3′ will hydrogen bond with the tail and can then be extended by reverse transcriptase.
4. An oligonucleotide that could be used in a Southern blot to identify carriers of sickle cell anemia. Note that this disease is caused because an A in the sequence 5′GTGCATCTGACTCCTGAGGAGAAGTCT3′ in the normal β globin gene is mutated to a T to generate the sequence 5′GTGCATCTGACTCCTGTGGAGAAGTCT3′. = F. Both strands of DNA are present on a Southern blot so the sequence 5′GTGCATCTGACTCCTGTGGAGAAGTCT3′ will hydrogen bond with its complementary sequence.
5. An oligonucleotide that could be used for northern blotting to detect mRNA containing the sequence 5′GUCAGCUUACGAUGGCAGUC3′ = G, 5′GACTGCCATCGTAAGCTGAC3′. The oligonucleotide must be complementary in sequence to the mRNA. Oligonucleotides are made as DNA and therefore contain Ts instead of Us.
7.3 Theme: Uses of cDNA Clones
1. Amplifying a known or partially known DNA sequence using the polymerase chain reaction = A. The strands are separated at high temperature and then the primers (which have been chosen such that once attached, the 3′ ends point towards each other) allowed to anneal to their respective strands. Thermostable DNA polymerase then generates the complementary strands so that for each cycle of operation there is a replication of the DNA between the primer two attachment points. At present polymerase chain reaction amplification of lengths of up to 4 kb are routine; some users amplify lengths of up to 8 kb.
2. Automated DNA sequencing by the dideoxy chain termination method = D. DNA polymerase will attach to the 3′ end of the annealed oligonucleotide and then run along copying the remainder of the DNA sequence until a dideoxynucleotide is incorporated, at which point replication of that molecule stops. Note that this is automated sequencing, in which detection of the various products is by fluorescence of labelled dideoxynucleotides. The oligonucleotide does not need to be radiolabelled (and both cost and safety considerations mean that a radioactive reagent would never be used when a nonradioactive one will suffice).
3. Detection of specific DNA sequences by Southern blotting, for example to differentiate DNA from two human subjects = C. The oligonucleotide will hybridize to the specific DNA sequence and can then be detected by autoradiography.
4. Investigation, by northern blotting, of the degree to which a gene of interest is transcribed in a particular tissue = C. The oligonucleotide will hybridize to the specific RNA sequence and can then be detected by autoradiography.
8.1 Theme: Translation Initiation
1. In the early stages of translation initiation in prokaryotes, the small ribosomal subunit attaches by complimentary base pairing to this sequence at the 5′ end of the mRNA, sometimes called the Shine-Dalgano sequence = H, 5′GGAGG3′.
2. In contrast in the early stages of translation initiation in eukaryotes, the small ribosomal subunit attaches to this group at the extreme 5′ end of the mRNA molecule = M, 7-methyl guanosine cap.
3. The eukaryotic small subunit then slides along the mRNA until it encounters this sequence, known as the Kozak sequence = E, 5′CCACC3′.
4. The subsequent steps are similar in prokaryotes and eukaryotes. The small subunit slides a few more bases along until it encounters this sequence, the start codon for translation = D, 5′AUG 3′.
5. Initiation factors then act to catalyze the assembly of the complete ribosome. The first tRNA, with its formyl methionine (in prokaryotes) or methionine (in eukaryotes) attached, locates in one of the three tRNA binding sites on the ribosome: state which one = C, P site.
8.2 Theme: Translation Elongation and Termination
1. Following translocation of the ribosome three bases along the mRNA, this site on the ribosome is empty and can be occupied by a charged tRNA whose anticodon is complimentary to the corresponding codon on the mRNA = A, A site.
2. Peptidyl transferase now catalyzes the formation of a peptide bond between the new amino acid and the existing polypeptide chain. Immediately following this, the polypeptide chain is only attached to the mRNA via the tRNA in which of the three sites on the ribosome = A, A site.
3. The next step is translocation, the physical movement of the ribosome three bases along the mRNA. Energy from this is provided by the hydrolysis of GTP by which enzyme, which occupies the A site on the ribosome = D, EF-G. EF-G is a GTPase, one of a large family of fundamentally similar proteins that guide and drive biological processes through the binding and hydrolysis of GTP. GTPases will be described in more detail in Chapter 10.
4. As a result of translocation the uncharged tRNA, which gave up its amino acid during the formation of the peptide bond, moves to this site on the ribosome, from where it is released = B, E site.
5. When translocation brings a stop codon UGA, UAA or UAG into the position facing the A site on the ribosome, the A site becomes occupied not by a normal charged tRNA molecule but rather by this molecule = I, release factor 1 or 2.
6. Finally the ribosome splits into two subunits as a result of energy released in GTP hydrolysis by EF-G. The released small ribosomal subunit already has one initiation factor attached, ready to accept the other initiation factors and the first charged amino acid to initiate synthesis of a new polypeptide. Name this pre-attached initiation factor = F, IF3.
1. Methionine = B, 5′CAU3′. If you answered H = 5′UAC3′ you forgot that nucleic acid strands pair up in an antiparallel fashion.
2. Asparagine = D, 5′GUU3′. The wobble phenomenon is operating: the G at the 5′ end of the anticodon can pair with either U or C at the 3′ end of the codon.
3. Phenylalanine = C, 5′GAA3′. Once again the wobble phenomenon is operating: the G at the 5′ end of the anticodon can pair with either U or C at the 3′ end of the codon.
4. Isoleucine = E, 5′IAU3′. The inosine at the 5′ end of the anticodon can pair with any of U, C or A at the 3′ end of the codon.
1. Can be phosphorylated = E, glutamate.
2. Has a strongly acidic side chain = E, glutamate.
3. Has a strongly basic side chain = R, arginine.
4. Is an imino acid, not an amino acid, and therefore imposes greater constraints on the shape of the polypeptide chain = P, proline.
5. Two can form a disulphide bond = C, cysteine.
9.2 Theme: Terms Used to Describe Proteins
1. A covalent bond between the side chains of two cysteine residues = D, disulfide bond.
2. A protein secondary structure where the backbone coils in a right-handed helix with hydrogen bonds between the amide and carboxyl groups of the peptide bonds with the hydrogen bonds running parallel with the direction of the helix = A, α helix.
3. A separately folded region of a single polypeptide chain = E, domain.
4. Loss of all of the higher levels of structure with accompanying loss of biological activity of a protein = C, denaturation.
5. The tendency for hydrophobic molecules to cluster together away from water = F, hydrophobic effect.
9.3 Theme: Specific Binding Partners
1. Calmodulin = C, calcium ions.
2. Catabolite activator protein = A, a specific base sequence on DNA. Catabolite activator protein also binds cAMP, but does not directly bind either glucose or β-galactoside sugars. Rather, low glucose concentrations cause the concentration of cAMP to rise, and it is cAMP that activates catabolite activator protein so that it can bind to its specific base sequence on DNA.
3. Connexin 43 = D, connexin 43. Connexins on one cell bind a compatible connexin on a neighboring cell to form a gap junction channel.
4. Glucocorticoid receptor = A, a specific base sequence on DNA. The glucocorticoid receptor also binds glucocorticoid hormones, and can only adopt a shape that binds to the DNA when it has the steroid bound.
10.1 Theme: The Three Modes of Intracellular Transport
1. β globin = B. β globin is used to assemble hemoglobin, which remains in the cytosol.
2. Catalase (required inside peroxisomes) = C. All other organelles bound by a single membrane receive the majority of their proteins by vesicular trafficking, but proteins destined for peroxisomes are synthesized on cytosolic ribosomes and only then imported into peroxisomes.
3. Glucocorticoid hormone receptor = A. If the glucocorticoid hormone receptor binds its steroid hormone, it is imported into the nucleus.
4. NFAT (Nuclear Factor of Activated T cells) = A. Dephosphorylated NFAT reveals a nuclear localization sequence and is imported into the nucleus.
5. Platelet-derived growth factor receptor = D. All integral proteins of the membrane reach this location via the endoplasmic reticulum and Golgi apparatus.
6. Pyruvate dehydrogenase (required in the mitochondrial matrix) = C. Proteins destined for mitochondria are synthesized on cytosolic ribosomes, and only then imported into mitochondria.
10.2 Theme: Trafficking Processes
1. Arf = D, transport between Golgi cisternae.
2. Dynamin = A, endocytosis.
3. Rab = B, exocytosis. Rab family members control fusion events, including the fusion of vesicles with the plasma membrane.
4. Ran = C, traffic through the nuclear pore.
5. Signal recognition particle = G, transport into the endoplasmic reticulum.
6. TAP (transporter associated with antigen processing) = G, transport into the endoplasmic reticulum.
1. GTPases have a binding site for a nucleotide. Identify the nucleotide present in the pocket when the GTPase is in its off state (e.g. Arf in the state that cannot associate with membranes) = D, GDP.
2. GTPases are activated when the nucleotide in the binding pocket is replaced by a nucleotide present at higher concentration in the cytosol. Give the general name for the protein partner that catalyzes this switch = H, guanine nucleotide exchange factor, GEF.
3. GTPases turn off when the nucleotide in the binding pocket is hydrolysed. Identify the product of this hydrolysis = D, GDP.
4. Hydrolysis of the nucleotide in the binding pocket is activated by a protein partner. Give the general name for the protein partner that accelerates the hydrolysis = G, GTPase activating protein, GAP.
General comment: The term GTPase activating protein is somewhat confusing. These proteins activate the GTPase catalytic activity of the GTPase, and therefore speed the process whereby the GTPase turns itself off.
11.1 Theme: Changing the Preferred Shape of a Protein
1. A fall of pH from 7.5 to 6.5 = B, favours the closed configuration. At the lower pH the histidine will spend a greater fraction of time protonated and therefore positively charged. While protonated it will be attracted to the negative charge on the glutamate.
2. Phosphorylation of the serine = B, favours the closed configuration. The negative phosphate group will be attracted to the positive charge on the lysine.
3. Phosphorylation of both threonines = A, favours the open configuration. The two negative phosphate groups will repel.
1. kcat (catalytic rate constant) = E, the number of moles of product formed per mole of enzyme per unit time.
2. KM (Michaelis constant) = B, that substrate concentration that gives an initial velocity equal to half the maximum velocity (Vm).
3. v0 (initial velocity) = G, the rate of product formation at the start of an enzyme reaction.
4. Vm (maximum velocity) = C, the maximum initial velocity possible when an enzyme is fully saturated with its substrate.
1. F. The catalytic rate constant… can be determined if both Vmax and the total enzyme concentration are known.
2. G. When an enzyme can work on two substrates the best substrate is the one that… gives the highest specifity constant (ratio of kcat over KM).
3. E. If you double the amount of an enzyme (keeping other conditions constant) you will… increase Vm by twofold.
4. B. A sigmoid curve obtained when v0 is plotted against substrate concentration shows… that the enzyme binds its substrate cooperatively, that it is an allosteric enzyme.
5. C. A substrate with KM of 5 × 10−3 mol L−1 binds to the enzyme more… weakly than one with KM = 5 × 10−4 mol L−1.
12.1 Theme: Cell Spaces and Regions in Energy Trading
1. ADP/ATP exchanger = B.
2. ATP synthase = B.
3. Coenzyme Q = B.
4. Cytochrome c = C.
5. [Na+] > 100 mmole liter−1 = G.
6. porin = D.
7. sodium/potassium ATPase = F.
8. The electron transport chain = B.
12.2 Theme: The Electron Transport Chain and ATP Synthase
1. Is not a carrier = B.
2. Oxidises coenzyme Q = C.
3. Oxidises NADH = A.
4. Oxidises succinate = B.
5. Oxidises the reduced form of cytochrome c = D.
6. Reduces molecular oxygen to water = D.
7. Will reverse direction if the H+ electrochemical gradient across the membrane is dissipated after application of an uncoupler = E.
1. Contains a pyrimidine residue = F. UTP is a nucleotide comprising the pyrimidine uracil coupled to triphosphorylated ribose. In contrast NADH, ATP and GTP all contain purine residues: adenine, adenine, and guanine respectively. NADH also contains a nicotinamide residue.
2. Generated by the action of an integral membrane protein of the plasma membrane = E. The sodium ion electrochemical gradient across the plasma membrane is created by the action of the sodium/potassium ATPase.
3. Has no energy content under anaerobic conditions = C. Under anaerobic conditions NADH cannot pass hydrogens to molecular oxygen, and so cannot be used to drive H+ up its electrochemical gradient out of the mitochondrial matrix.
4. In a well oxygenated cell, this is the most energy rich of the energy currencies = C.
5. This currency is directly depleted by the action of uncouplers such as 2,4 dinitrophenol = D. Uncouplers allow H+ to cross the inner mitochondrial membrane, dissipating the H+ electrochemical gradient.
13.1 Theme: Reactions and Pathways
1. Fatty acid synthesis = I, stearic acid. Stearic is one of many fatty acids synthesized by the body. Double bonds and longer-chain fatty acids are made by other systems after fatty acid sythesis has made stearic acid.
2. Glucose-6-phosphate dehydrogenase = G, NADPH. This is the first step in the pentose phosphate pathway, generating reducing power as NADPH.
3. Gluconeogenesis = E, glucose 6-phosphate. Liver cells can then dephosphorylate this compound to generate glucose itself for export from the cell.
4. Lactate dehydrogenase = F, NADH. Lactate dehydrogenase oxidizes lactate to pyruvate, reducing NAD+ in the process.
5. Phosphofructokinase = C, fructose-1,6-bisphosphate. This reaction commits the sugar to being broken down and used to provide energy rather than used for other purposes.
13.2 Theme: Pathways and Enzymes
1. Converts fatty acids to acetyl CoA = B, β oxidation.
2. Converts pyruvate to acetyl CoA = G, pyruvate dehydrogenase.
3. The only way the red blood cell has to make ATP = F, glycolysis.
4. The source of 5-carbon sugars and a source of NADPH for biosynthesis = I, the pentose phosphate pathway.
5. Uses UDP-glucose = E, glycogen synthesis.
1. B. Essential amino acids are… amino acids that an organism cannot make and so must be present in the diet.
2. A. Basic amino acids are… amino acids with a side chain that can be protonated.
3. G. The enzymes of the Krebs cycle are… located mainly in the mitochondrial matrix with one in the inner mitochondrial membrane.
4. F. Pyruvate carboxylase makes… oxalacetate. This is used to top up oxalacetate for the Krebs cycle and is an important step in gluconeogenesis.
5. D. It necessary to convert pyruvate to lactate when there is little oxygen available because… the reaction regenerates the NAD+ used up during glycolysis (during the oxidation of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate).
14.1 Theme: Cytosolic and Extracellular Concentrations of Important Ions
1. Cytosolic calcium = I. A typical value for cytosolic calcium is 100 nmoles liter−1.
2. Extracellular calcium = E.
3. Cytosolic chloride = D.
4. Extracellular chloride = A.
5. Cytosolic H+ (H3O+) = I. A typical value for cytosolic H+ is 60 nmoles liter−1, so the response “less than or equal to 100 nmoles liter−1” is correct.
6. Extracellular H+ (H3O+) = I. A typical value for extracellular H+ is 40 nmoles liter−1, so the response “less than or equal to 100 nmoles liter−1” is correct.
7. Cytosolic potassium = A. A typical value for cytosolic potassium is 140 mmoles liter−1, so the response “greater than or equal to 100 mmoles liter−1” is correct.
8. Extracellular potassium = D.
9. Cytosolic sodium = C.
10. Extracellular sodium = A. A typical value for extracellular sodium is 150 mmoles liter−1, so the response “greater than or equal to 100 mmoles liter−1” is correct.
14.2 Theme: Pathways for Solute Movement Across the Plasma Membrane
1. An enzyme that carries out a hydrolytic reaction = A. The calcium ATPase hydrolyses ATP to ADP + Pi; this hydrolysis provides the energy that drives a calcium ion out of the cytosol.
2. A protein that carries one solute in one direction and a second solute in the opposite direction = E.
3. A channel that can pass glucose = B. The molecular weight of glucose is 180, well below the 1000 or so limit for a connexon. Note that C is not correct because the glucose carrier is not a channel.
4. A protein whose expression in almost all human cells is responsible for the fact that their cytosol is at a negative voltage with respect to the extracellular medium = D.
5. A channel expressed in pain receptor neurons which opens at damagingly high temperatures = F.
6. A channel expressed in almost all neurons, even those that are not sensitive to painful stimuli, but which is not expressed in most non-neuronal human cells (it is not found, for example, in liver or blood cells) = G.
14.3 Theme: Ion Fluxes in a Nerve Cell
1. Although gradients of this ion play a crucial role at other cell membranes, the concentration gradient of this ion across the plasma membrane is small: the concentration in the cytosol is the same, within a factor of two, as the concentration in the extracellular medium = B. H+ has a crucial role in mitochondria, but the concentration gradient across the plasma membrane is small, with the concentration in the cytosol being about 1.5 times that in the extracellular medium.
2. In a resting nerve cell, this ion is constantly leaking into the cell, and must be removed by the action of an ATP-consuming carrier = D.
3. In a resting nerve cell, this ion is constantly leaking out of the cell, and must be pumped back in by the action of an ATP-consuming carrier = C.
4. The more a nerve cell is depolarized, the greater the electrochemical gradient favouring entry of this ion into the cell = A. When the cytosol becomes less negative, the voltage force pulling the positive ions H+, K+ and Na+ will get smaller. Only a negative ion such as Cl− will experience a greater inward electrochemical gradient when the cell is depolarized.
5. In a resting nerve cell, this ion is at equilibrium across the plasma membrane = A. Chloride is at equilibrium across the plasma membranes of most resting animal cells.
15.1 Theme: Processes Downstream of Receptor Tyrosine Kinases
1. A domain comprising a pocket with a positively charged arginine at the base. Proteins with this domain are recruited to phosphorylated tyrosine residues = F.
2. A guanine nucleotide exchange factor for Ras = G.
3. A hydrolytic enzyme that is activated when phosphorylated by receptor tyrosine kinases = D.
4. A kinase that is activated when phosphorylated by receptor tyrosine kinases = B. A kinase is an enzyme that transfers the γ phosphate of ATP to another molecule;
most of the kinases described in this book are protein kinases that phosphorylate serine, threonine, or tyrosine residues, but phosphoinositide 3-kinase phosphorylates the lipid PIP2. Note that protein kinase B is not a correct answer because this kinase is activated by being phosphorylated on serine and threonine residues; this phosphorylation is not carried out by receptor tyrosine kinases, which only phosphorylate tyrosine residues.
5. An enzyme that is activated by Ras:GTP = B. As well as activating MAPKKK, Ras can also activate PI 3-kinase, as shown in Figure 15.16.
15.2 Theme: Proteins Activated by Nucleotides
1. A protein kinase activated by cAMP = G.
2. A protein responsible for generating electrical signals in photoreceptors = C.
3. A protein that activates a phospholipase C when in the GTP bound state = D.
4. A protein that activates adenylate cyclase when in the GTP bound state = E.
5. A protein that activates MAP kinase kinase kinase when in the GTP bound state = J.
15.3 Theme: Inositol Compounds
1. A ligand that binds to and opens a calcium channel = C.
2. A lipid that recruits protein kinase B to the plasma membrane = G.
3. A substrate for phosphoinositide 3-kinase (PI3K) = F.
4. A substrate for phospholipases C (PLC) = F.
5. The product of phosphoinositide 3-kinase (PI3K) = G.
6. The product of phospholipases C (PLC) = C.
General comment: Unphosphorylated inositol, IP2, and IP4 play important roles in mammalian cells but are not covered in this book. IP6 is found in plants but is relatively unimportant in animals.
1. A receptor that signals to the trimeric G protein Gq = A.
2. A receptor that signals to the trimeric G protein Gs = B.
3. A receptor tyrosine kinase = F. Note that the interleukin 2 receptor is not a receptor tyrosine kinase; its downstream effects require the assistance of JAK tyrosine kinase which is associated with, but not part of, the receptor.
4. An intracellular receptor that is always cytosolic = D.
5. An intracellular receptor that moves to the nucleus upon binding transmitter = C.
6. An ionotropic receptor = G.
1. A hormone = A. Adrenaline is released by the adrenal gland and is distributed throughout the body by the blood system.
2. A paracrine transmitter = F. Noradrenaline is released by the axons of vasoconstrictor and other nerve cells and diffuses through the tissue, causing smooth muscle cell contraction and other responses.
3. A transmitter that acts to cause release of calcium from the endoplasmic reticulum of many cells including smooth muscle = F.
4. A transmitter that is a γ amino acid = B.
5. A transmitter that is an α amino acid = C.
6. An excitatory synaptic transmitter = C.
7. An inhibitory synaptic transmitter = B.
1. A burst of activity in a presynaptic GABAergic neurone for a postsynaptic cell that is receiving steady excitatory input from a number of presynaptic glutaminergic neurones = D. Since the postsynaptic cell is already depolarized, the opening of GABA receptors will cause chloride ions to move in, carrying negative change and reducing the depolarization of the postsysnaptic cell.
2. A rapid burst of action potentials in a presynaptic pain receptor neurone caused by a painful stimulus such as a pin prick = B. Note that the stimulus was painful. For the stimulus to be perceived as painful by the subject, the pain relay cell must have been depolarized to threshold.
3. A single action potential in a motoneurone = B. The synapse between motoneurons and skeletal muscle cells is unusual in that one presynaptic action potential evokes a postsynaptic depolarization large enough to evoke an action potential.
4. A single action potential in a presynaptic GABAergic neurone, at a time when no other synapses onto the postsynaptic cell are active = E. Chloride is at equilibrium across the plasma membrane of an otherwise unperturbed cell.
5. A single action potential in a presynaptic glutaminergic neurone, at a time when no other synapses onto the postsynaptic cell are active = A.
17.1 Theme: Cytoskeletal Structures
1. Cilia = C, tubulin.
2. Fingernails = B, intermediate filament proteins. Fingernails are formed of keratin.
3. Flagella = C, tubulin.
4. Microfilaments = A, actin.
5. Microtubules = C, tubulin.
6. Microvilli = A, actin.
7. Stress fibers = A, actin.
17.2 Theme: Proteins of the Cytoskeleton
1. D. Keratin forms the intermediate filaments in skin and structures formed from skin such as hair, fingernails, horns and hooves.
2. A. Microfilaments are made of actin monomers.
3. B. Microtubules are formed from α- and β-tubulin.
4. F. Myosin is found in all cells, and in skeletal muscle cells forms part of the contractile apparatus. It is a motor molecule that interacts with actin filaments (microfilaments)
5. E. Kinesin and dynein are the two molecular motors that act on microtubules
6. G.
7. C. γ-tubulin does not form microtubules, but resides at the centrosome as part of the microtubule organizing center.
1. Amoeboid locomotion is powered by ATP hydrolysis performed by this ATPase = E, myosin.
2. The contraction of muscle is powered by ATP hydrolysis performed by this ATPase = E, myosin.
3. The rowing motion of a cilium is powered by ATP hydrolysis performed by this ATPase = C, dynein.
4. The transport of vesicles and organelles from the tips of nerve cell axons to the cell body is powered by ATP hydrolysis performed by this ATPase = C, dynein.
5. The wriggling motion of sperm tails is powered by ATP hydrolysis performed by this ATPase = C, dynein.
General comment: Do not confuse dynein, an ATPase, with dynamin. The latter uses energy released by GTP hydrolysis to power the pinching off of vesicles from budding membranes .
1. chromosome condensation occurs in the nucleus and spindle formation begins in the cytoplasm = E, prophase.
2. the nuclear envelope breaks down and chromosomes become associated with the spindle = D, prometaphase.
3. the chromosomes are aligned on the spindle and no longer make individual excursions towards and away from the spindle poles = C, metaphase.
4. paired chromatids separate and begin to move toward the spindle poles = A, anaphase.
5. chromosomes decondense and the nuclear envelope reforms = F, telophase.
6. physical separation into two cells = B, cytokinesis.
18.2 Theme: Checkpoints in the Cell Cycle
1. During this period of interphase, the cell replicates its DNA = H, S phase.
2. For animal cells to begin DNA replication, many conditions must be met. One is that the transcription factor E2F must be released from a ligand that holds it in an inactive dimer. What is this ligand = K, Rb.
3. If the DNA is damaged, it must be repaired before it is replicated. DNA damage activates two kinases, Chk1 and Cds1. These phosphorylate a transcription factor that acts to upregulate cyclin dependent kinase inhibitors, CKIs. Phosphorylation of the transcription factor allows its concentration to increase in cells. What is the identity of this anti-division trancription factor = I, P53.
4. Once the DNA is replicated and the cell is large enough, it can enter mitosis. Passing the G2/M checkpoint requires activity of cyclin dependent kinase 1, Cdk1. Cdk1 is only active when dimerized with a protein partner; identify this essential partner = C, cyclin B.
5. Cdk1 is also regulated by posttranslational modification. Which reaction must be performed on Cdk1 in order for it to be active = D, dephosphorylation at T14 and Y15.
6. During prometaphase the chromosomes line up on the metaphase plate. The sister chromatids are joined together at the kinetochores by which protein A, = cohesin.
7. When all the kinetochores are under tension the anaphase promoting complex targets securin for destruction, allowing activation of an enzyme that digests the link between the chromatids, allowing the separation of the chromatids in anaphase. What is that link-destroying enzyme = M, separase.
1. In animals, cells are kept alive by the activation of other cells that supply growth factors. Growth factor receptors activate PI 3-kinase, which generates PIP3 at the plasma membrane. PIP3 recruits a critical survival-promoting kinase to the plasma membrane; name that kinase = H, protein kinase B.
2. If PIP3 dissapears from the plasma membrane and the kinase described above becomes inactive, BAX is activated, allowing a protein to escape from the mitochondria. Name the released mitochondrial protein = D, cytochrome c.
3. White blood cells can kill target cells by activating this cell surface death domain receptor = E, Fas. Many other mechanisms by which white blood cells kill targets are described in Chapter 19.
4. Cells also die if their DNA is damaged so badly that it cannot be repaired in a reasonable time. What is the transcription factor whose concentration increases after DNA damage and which upregulates synthesis of BAX = F, p53.
5. All the death initiation pathways described above converge on the activation of this family of cytosolic proteases = B, caspases.
19.1 Theme: Immune System Cells
1. A cell generated by a process that includes somatic hypermutation = A. Somatic hypermutation only occurs in the B cell lineage. It occurs in no other lineage, not even that leading to T cells.
2. A cell that attacks any body cell that does not express major histocompatibility complex protein on its cell surface = F.
3. A cell that attacks any body cell whose major histocompatibility complex proteins are presenting novel peptides = C.
4. A cell that makes antibodies = A.
5. A cell that migrates from the tissues to lymph nodes, where it presents antigens to the lymphocytes = D.
6. A cell that upon stimulation expresses CD40L, a protein that in turn activates B cells = B.
7. An important phagocytic cell of body tissue which differentiates from blood monocytes = E.
8. The commonest phagocytic cell of the blood = G.
19.2 Theme: The Antibody Heavy Chain Locus
1. A component that experiences somatic hypermutation = A. Only the V sections of the locus experience somatic hypermutation, in the process that generates antibodies that bind the antigen with even higher affinity than did the original, genetically encoded version.
2. A component that includes a sequence encoding a transmembrane domain = C. At all stages before conversion into a plasma cell, antibodies are integral membrane proteins. Their heavy chain crosses the membrane once in a polypeptide chain encoded by a length of DNA that is given a Greek letter corresponding to the Latin letter of the corresponding immunoglobulin class, in this case, γ for IgG.
3. Although chromosome 14 in every other cell of the body contains 6 of these arranged sequentially, the DNA of mature B cells contains a linear sequence comprising any number from 1 to 6 = E. The DNA splicing that occurs during B cell maturation leaves only one V and one D section. In contrast, between zero and five of the six J sections are removed. The primary RNA transcript contains all the remaining J sections, with all except the first remaining one being removed during RNA splicing.
4. Genomic DNA contains more than 20, but less than 100, slightly different versions of this section arranged sequentially along the chromosome = D. The best current estimate is that there are 23 D sections in the human genome.
5. The first protein coding region that would be encountered when reading along the locus from the 5′ end = A. The heavy chain locus begins with approximately 100 slightly different copies of a V section.
19.3 Theme: T Cells and Their Interaction with Other Cells
1. A death domain receptor that, when activated, causes the proteolysis and activation of caspase 8 and hence cell death by apoptosis = E.
2. A mitogen secreted by CD4+ T cells = G. CD40L on the surface of CD4+ T cells induces mitosis in B cells, but it is an integral membrane protein—it is not secreted.
3. A protease secreted by CD8+ T cells = F.
4. A protein complex that cuts cytosolic proteins into short lengths of peptide = J.
5. A protein found in the cytosol of resting T cells which translocates to the nucleus upon cell stimulation = I. NFAT is named for this behavior: “Nuclear Factor of Activated T cells.”
6. A protein whose expression allows scientists and clinicians to recognise the cell as a T killer cell, that is, a member of that population of T cells that attacks body cells infected by viruses or other pathogens = C.
7. A receptor expressed by mature B cells that, when activated, causes differentiation and proliferation = B.
8. A transmembrane protein that can, if the match is correct, bind to a major histocompatability complex protein that is presenting a foreign peptide = K.
9. A transmembrane protein that presents short lengths of peptide at the cell surface = H.
1. This condition is likely to be a nonsense mutation and therefore may be treatable by drugs that cause the ribosome to read through a stop codon = B. Only a premature stop codon can generate a truncated protein. If the stop codon was created by a base change, then there will be no frameshift problems. Thus if the ribosome can be caused to read through the stop, the protein generated will either be completely normal or (more likely) differ in one amino acid only.
2. This condition is likely to be caused by a mutation in a promoter or enhancer region of the CFTR gene = F. Mutations that change a promoter or enhancer sequence, and in turn decrease how effectively RNA polymerase II can bind, will decrease the rate of transcription and less mRNA will be made. Some mutations can of course change a promoter or enhancer sequence to one that makes RNA polymerase II bind more efficiently. In this case more mRNA protein will be made. Once the mRNA is made it will be translated into a normal protein.
3. This condition may in the future be treatable by drugs that favour the open configuration of the CFTR channel = D. If CFTR is present at the plasma membrane and would allow chloride ions to pass if it opened, but is aberrant in the residues that allow it to be switched into the open state, then a drug that artificially opens the channel is likely to be of use.
4. This condition may in the future be treatable by drugs that inhibit or modify the function of the proteasome = C. The proteasome is the protein complex that destroys misfolded proteins, and by inhibiting its function it may be possible to retain partially misfolded CFTR that will nevertheless function, at least to some extent, if it reaches the plasma membrane.
20.2 Theme: The CFTR as a Permeability Pathway
Answer: H. If protein kinase A is activated then it will phosphorylate CFTR, causing the channels to open. Chloride is at equilibrium across the plasma membrane of most cells, so opening CFTR channels will not cause a net flow of chloride in either direction, and will therefore not change the membrane voltage. However, the open CFTR channels are a pathway for electrical current flow, so the resistance of the plasma membrane will decrease.
20.3 Theme: A Medly of Sentences about CF
1. C. Because CF is recessive the disease must… be inherited from both parents.
2. H. Reading through a stop codon may help some CF patients because… some of the many CF mutations have a mutation that introduces a premature stop. The truncated protein does not function properly.
3. B. A zoo blot is useful to… identify genes that have been conserved in several species. Such genes are likely to code for important proteins.
4. D. CFTR protein was definitively proven to be a chloride channel by… a lipid bilayer voltage clamp experiment. CFTR protein was inserted into a lipid bilayer and chloride ions were shown to move through the channel.
5. E. Gene therapy for CF is very difficult because… the healthy gene needs to be inserted into a huge number of cells in the lungs and these cells are relatively short-lived. In addition the modified virus vectors used seem to cause side effects.