05-01-2008, 08:45 PM
THE ANSWERS BY ONE SHOT(81-90)
81.
The answer is b.
The structures of glycine and alanine are quite similar, with the .H side group of glycine being replaced with the .CH3 side group of alanine. Consequently, a mutation causing such a change is unlikely to produce a dysfunctional protein. In contrast, a mutation changÂing a splice site could result in either the abnormal exclusion of an exon or the inclusion of an intron in a protein, which would drastically change its properties. Likewise, frame-shift mutations caused by deletion or addition of one or two bases completely distort the remainder of the protein. The closer a frame shift is to the amino terminal that is synthesized first, the more garbled the protein. All nonsense mutations involve inserting a stop codon in place of whatever amino acid would have been coded. In the case given, only half the protein would be synthesized.
82.
The answer is e.
Chain termination is determined by three codons: UAA, UAG, and UGA. Aside from chain termination codons, each group of three bases in a sequence codes for an amino acid. The next three bases specify another amino acid. Thus, the genetic code is nonoverlapÂ
ping. The triplet genetic code is degenerate, which is to say that for most amino acids there is more than one code word. The triplets of bases (codons) that specify the same amino acid usually differ only in the last base of the triplet.
83. The answer is b.
The following mutations were shown in the DNA changes:
a.
Missense (cys to leu)
b.
Nonsense (tyr to stop)
c.
Missense (ser to phe)
d.
Harmless (tyr to tyr)
e.
Missense (leu to phe)
f.
No mutation
g. Missense (phe to tyr) Most of the mutations shown result in a missense effect, with a different amino acid being incorporated into the same site in a protein. This may or may not have an effect depending upon its location. Some single-base mutations are harmless because of the degeneracy of the genetic code, whereby more than one triplet code exists for all amino acids except trypÂtophan and methionine. Choice a contains two mutations, one degenerate and the other missense.
DNA coding: 3.-ACGACGACG-5. to 3.-ACAAACACG-5. mRNA: 5.-UGCUGCUGC-3. to 5.-UGUUUGUGC-3. Protein: cys-cys-cys to cys-leu-cys
Nonsense mutations occur when the reading of the normal termination sigÂnal is changed. This can occur by mutation to a stop signal as in choice b, by deletions near a stop codon, or by insertions.
84. The answer is d.
Several operons in E. coli, including the lac operon, are subject to catabolite repression. In the presence of glucose, there is decreased manufacture of cyclic AMP (cAMP) by adenylate cyclase. Low glucose levels increase production of cAMP, which binds to the catabolite activator protein (CAP). The cAMP-CAP complex binds to the promoters of several responsive operons at catabolite activator protein (CAP) binding sites, greatly enhancing transcription of operon RNA. This positive control stimulates use of more exotic metabolites when glucose is not available and conserves energy when glucose is plentiful. High levels of glucose lower cAMP levels and direct metabolism toward constitutive gluÂcose pathways such as glycolysis.
85.
The answer is b.
The lactose (lac) operon is a classic model for understanding gene regulation. It is negatively controlled through two regulatory genes”the lac I gene that constitutively (always) expresses a repressor protein and the operator (o) region to which the repressor binds. The lac operon is inducible by lactose and lactose analogues, inactivating the repressor and uncovering the operon and its neighboring promoter (p) sequence. RNA polymerase then transcribes the inducible, structural genes .-galactosidase (z), permease (y), and transacetylase (a). The RNA transcript is polycistronic, so that one regulatory site allows transcription of all three genes needed for the metabolism of lactose. The bacterial riboÂsomes immediately attach to the nascent RNA transcript, allowing for simultaneous transcription and translation. When all the lactose is metabÂolized, the repressor returns to its native conformation, binds to the operÂator, and shuts down lac operon transcription (see the figure below).
86.
The answer is c.
Certain regulatory elements act on genes on the same chromosome (œcis), while others can regulate genes on the oppoÂsite chromosome (œtrans). The terminology makes analogy to carbon-carbon double bonds where two modifying groups may both be above or below the bond (cis) or opposite it (trans). Cis regulatory elements like the lac operator and promoter or mammalian enhancers are usually DNA sequences (regulaÂtory sequences) adjoining or within the regulated gene. Trans regulatory eleÂments like the lac repressor protein or mammalian transcription factors are usually diffusible proteins (regulatory factors) that can interact with adjoinÂing target genes or with target genes on other chromosomes. Classification of bacterial elements as cis or trans requires mating experiments where portions of a second chromosome are introduced by transfection (with bacteriophage) or conjugation (with other bacteria). The distinction between cis and trans is fundamental for understanding how regulators work.
A. Regulation of lac operon in E.coli showing repression in the absence of inducer. B. Induction of gene expression with inducer and no glucose. Glucose prevents cAMP and its cAMP-binding protein (CAP) from binding to the promoter to facilitate transcription by RNA polymerase (catabolite repression).
(Reproduced,with permission,from Murray RK,Granner DK,Mayes PA,Rodwell VW: Harper™s Biochemistry, 25/e.New York,McGraw-Hill,2000:471.)
87.
The answer is d.
Mammalian regulatory factors are much more diverse than those of bacteria, possessing several types of structural domains. Activators of transcription, such as steroid hormones, may enter the cell and bind to regulatory factors at specific sites called ligand-binding domains; these intracellular œreceptors are analogous to G protein“linked membrane receptors that extend into the extracellular space. Response eleÂments are not regulatory factors but DNA sequences near the transcription site for certain types of genes (e.g., steroid-responsive and heat shock“ responsive genes). Regulatory factors interact with specific DNA sequences through their DNA-binding domains, and with other regulatory factors through transcription-activating domains. Some regulatory factors have antirepressor domains that counteract the inhibitory effects of chromatin proteins (histones and nonhistones).
88.
The answer is d.
The POMC gene provides a mammalian example in which several proteins are derived from the same RNA tranÂscript. Unlike the polycistronic mRNA of the bacterial lactose operon, mammalian cells generate several mRNAs or proteins from the same gene by variable protein processing or by alternative splicing. Variable protein processing preserves the peptide products of some gene regions but degrades those from others. Alternative splicing would often produce proÂteins composed of different exon combinations with the same terminal exon and carboxy-terminal peptide, but could remove the terminal exon in some proteins and produce different C-terminal peptides. Different tranÂscription factors or enhancers in different brain regions could regulate the total amounts of POMC gene transcript but not the types of protein proÂduced. Elongation of protein synthesis involves GTP cleavage but is not differentially regulated in mammalian tissues.
89.
The answer is a.
Imbalance of globin chain synthesis occurs in the thalassemias. Deficiency of .-globin chains (. thalassemia) is comÂmon in Asian populations and may be associated with abnormal hemogloÂbins composed of four .-globin chains (hemoglobin H) or (in fetuses and newborns) of four .-globin chains (hemoglobin Bart™s). Mutation in a tranÂscription factor necessary for expression of .-globin could ablate .-globin
expression, since the same factor could act in trans on all four copies of the .-globin genes (two .-globin loci). Mutation of a regulatory sequence eleÂment that acts in cis would inactivate only one .-globin gene, leaving others to produce .-globin in reduced amounts (mild . thalassemia). Deletions of one .-globin would produce a similar mild phenotype, and deficiencies of transcription factors regulating .- and .-globin genes would not produce chain imbalance.
90.
The answer is a.
) Antibody classes, called isotypes, are deterÂmined by the constant region of heavy chains. There are five isotypes that include IgM, IgD, IgG, IgE, and IgA. During B cell maturation, DNA rearrangements produce light chains with unique V-J segments and heavy chains with unique V-D-J segments. After activation, B cells can change their preferred DNA recombination to join the V-D-J segment to a different constant © segment. Different constant region exons are clustered downÂstream of the Cµ exon used for initial IgM synthesis by the activated B cell. Recombination at designated switch sites places another constant chain exon (e.g., C. for IgG) next to the V-D-J segment and allows the activated B cell to secrete a different antibody isotype (heavy chain class switching). Alternative splicing allows activated B cells to switch from membrane-bound to secreted IgM.
81.
The answer is b.
The structures of glycine and alanine are quite similar, with the .H side group of glycine being replaced with the .CH3 side group of alanine. Consequently, a mutation causing such a change is unlikely to produce a dysfunctional protein. In contrast, a mutation changÂing a splice site could result in either the abnormal exclusion of an exon or the inclusion of an intron in a protein, which would drastically change its properties. Likewise, frame-shift mutations caused by deletion or addition of one or two bases completely distort the remainder of the protein. The closer a frame shift is to the amino terminal that is synthesized first, the more garbled the protein. All nonsense mutations involve inserting a stop codon in place of whatever amino acid would have been coded. In the case given, only half the protein would be synthesized.
82.
The answer is e.
Chain termination is determined by three codons: UAA, UAG, and UGA. Aside from chain termination codons, each group of three bases in a sequence codes for an amino acid. The next three bases specify another amino acid. Thus, the genetic code is nonoverlapÂ
ping. The triplet genetic code is degenerate, which is to say that for most amino acids there is more than one code word. The triplets of bases (codons) that specify the same amino acid usually differ only in the last base of the triplet.
83. The answer is b.
The following mutations were shown in the DNA changes:
a.
Missense (cys to leu)
b.
Nonsense (tyr to stop)
c.
Missense (ser to phe)
d.
Harmless (tyr to tyr)
e.
Missense (leu to phe)
f.
No mutation
g. Missense (phe to tyr) Most of the mutations shown result in a missense effect, with a different amino acid being incorporated into the same site in a protein. This may or may not have an effect depending upon its location. Some single-base mutations are harmless because of the degeneracy of the genetic code, whereby more than one triplet code exists for all amino acids except trypÂtophan and methionine. Choice a contains two mutations, one degenerate and the other missense.
DNA coding: 3.-ACGACGACG-5. to 3.-ACAAACACG-5. mRNA: 5.-UGCUGCUGC-3. to 5.-UGUUUGUGC-3. Protein: cys-cys-cys to cys-leu-cys
Nonsense mutations occur when the reading of the normal termination sigÂnal is changed. This can occur by mutation to a stop signal as in choice b, by deletions near a stop codon, or by insertions.
84. The answer is d.
Several operons in E. coli, including the lac operon, are subject to catabolite repression. In the presence of glucose, there is decreased manufacture of cyclic AMP (cAMP) by adenylate cyclase. Low glucose levels increase production of cAMP, which binds to the catabolite activator protein (CAP). The cAMP-CAP complex binds to the promoters of several responsive operons at catabolite activator protein (CAP) binding sites, greatly enhancing transcription of operon RNA. This positive control stimulates use of more exotic metabolites when glucose is not available and conserves energy when glucose is plentiful. High levels of glucose lower cAMP levels and direct metabolism toward constitutive gluÂcose pathways such as glycolysis.
85.
The answer is b.
The lactose (lac) operon is a classic model for understanding gene regulation. It is negatively controlled through two regulatory genes”the lac I gene that constitutively (always) expresses a repressor protein and the operator (o) region to which the repressor binds. The lac operon is inducible by lactose and lactose analogues, inactivating the repressor and uncovering the operon and its neighboring promoter (p) sequence. RNA polymerase then transcribes the inducible, structural genes .-galactosidase (z), permease (y), and transacetylase (a). The RNA transcript is polycistronic, so that one regulatory site allows transcription of all three genes needed for the metabolism of lactose. The bacterial riboÂsomes immediately attach to the nascent RNA transcript, allowing for simultaneous transcription and translation. When all the lactose is metabÂolized, the repressor returns to its native conformation, binds to the operÂator, and shuts down lac operon transcription (see the figure below).
86.
The answer is c.
Certain regulatory elements act on genes on the same chromosome (œcis), while others can regulate genes on the oppoÂsite chromosome (œtrans). The terminology makes analogy to carbon-carbon double bonds where two modifying groups may both be above or below the bond (cis) or opposite it (trans). Cis regulatory elements like the lac operator and promoter or mammalian enhancers are usually DNA sequences (regulaÂtory sequences) adjoining or within the regulated gene. Trans regulatory eleÂments like the lac repressor protein or mammalian transcription factors are usually diffusible proteins (regulatory factors) that can interact with adjoinÂing target genes or with target genes on other chromosomes. Classification of bacterial elements as cis or trans requires mating experiments where portions of a second chromosome are introduced by transfection (with bacteriophage) or conjugation (with other bacteria). The distinction between cis and trans is fundamental for understanding how regulators work.
A. Regulation of lac operon in E.coli showing repression in the absence of inducer. B. Induction of gene expression with inducer and no glucose. Glucose prevents cAMP and its cAMP-binding protein (CAP) from binding to the promoter to facilitate transcription by RNA polymerase (catabolite repression).
(Reproduced,with permission,from Murray RK,Granner DK,Mayes PA,Rodwell VW: Harper™s Biochemistry, 25/e.New York,McGraw-Hill,2000:471.)
87.
The answer is d.
Mammalian regulatory factors are much more diverse than those of bacteria, possessing several types of structural domains. Activators of transcription, such as steroid hormones, may enter the cell and bind to regulatory factors at specific sites called ligand-binding domains; these intracellular œreceptors are analogous to G protein“linked membrane receptors that extend into the extracellular space. Response eleÂments are not regulatory factors but DNA sequences near the transcription site for certain types of genes (e.g., steroid-responsive and heat shock“ responsive genes). Regulatory factors interact with specific DNA sequences through their DNA-binding domains, and with other regulatory factors through transcription-activating domains. Some regulatory factors have antirepressor domains that counteract the inhibitory effects of chromatin proteins (histones and nonhistones).
88.
The answer is d.
The POMC gene provides a mammalian example in which several proteins are derived from the same RNA tranÂscript. Unlike the polycistronic mRNA of the bacterial lactose operon, mammalian cells generate several mRNAs or proteins from the same gene by variable protein processing or by alternative splicing. Variable protein processing preserves the peptide products of some gene regions but degrades those from others. Alternative splicing would often produce proÂteins composed of different exon combinations with the same terminal exon and carboxy-terminal peptide, but could remove the terminal exon in some proteins and produce different C-terminal peptides. Different tranÂscription factors or enhancers in different brain regions could regulate the total amounts of POMC gene transcript but not the types of protein proÂduced. Elongation of protein synthesis involves GTP cleavage but is not differentially regulated in mammalian tissues.
89.
The answer is a.
Imbalance of globin chain synthesis occurs in the thalassemias. Deficiency of .-globin chains (. thalassemia) is comÂmon in Asian populations and may be associated with abnormal hemogloÂbins composed of four .-globin chains (hemoglobin H) or (in fetuses and newborns) of four .-globin chains (hemoglobin Bart™s). Mutation in a tranÂscription factor necessary for expression of .-globin could ablate .-globin
expression, since the same factor could act in trans on all four copies of the .-globin genes (two .-globin loci). Mutation of a regulatory sequence eleÂment that acts in cis would inactivate only one .-globin gene, leaving others to produce .-globin in reduced amounts (mild . thalassemia). Deletions of one .-globin would produce a similar mild phenotype, and deficiencies of transcription factors regulating .- and .-globin genes would not produce chain imbalance.
90.
The answer is a.
) Antibody classes, called isotypes, are deterÂmined by the constant region of heavy chains. There are five isotypes that include IgM, IgD, IgG, IgE, and IgA. During B cell maturation, DNA rearrangements produce light chains with unique V-J segments and heavy chains with unique V-D-J segments. After activation, B cells can change their preferred DNA recombination to join the V-D-J segment to a different constant © segment. Different constant region exons are clustered downÂstream of the Cµ exon used for initial IgM synthesis by the activated B cell. Recombination at designated switch sites places another constant chain exon (e.g., C. for IgG) next to the V-D-J segment and allows the activated B cell to secrete a different antibody isotype (heavy chain class switching). Alternative splicing allows activated B cells to switch from membrane-bound to secreted IgM.