BSCI222 Exam 3 Textbook Problems (Topics 9-13) and gss practice problems

Ch.18 #15 List at least three different types of DNA repair and briefly explain how each is carried out.

(1) Mismatch Repair- Replication errors that are the result of base-pair mismatches are repaired. Mismatch-repair enzymes recognize distortions in the DNA structure due to mispairing and detect the newly synthesized strand by the lack of methylation on the new strand. The bulge is excised and DNA polymerase fills the gap and DNA ligase seals the repair. (2) Direct Repair- DNA damage is repaired by directly changing the damaged nucleotide back to its original structure as, for example, the repair done by a photolyase or methyltransferase. (3) Base-Excision Repair- After the damaged base is removed by glycosylases, the phosphodiester bond is excised by AP endonucleases and other enzymes remove the deoxyribose sugar; then the entire nucleotide is replaced by DNA polymerase and the nick is sealed by DNA ligase. (4) Nucleotide-Excision Repair- Repair enzymes recognize distortions of the DNA double helix. Damaged regions are excised by enzymes (nucleases and helicases), which separate the strands of DNA and cut phosphodiester bonds on either side of the damaged region. The gap generated by the excision step is filled in by DNA polymerase and sealed by DNA ligase.

The AST gene is a maternally imprinted gene. Joanna is heterozygous at AST; she carries one wildtype allele AST+and a rare nonfunctional allele, AST-. Why would Joanna express the phenotype of the AST- allele in her somatic cells? Could her children inherit her AST- allele? If they inherit it, could the children express the phenotype of the AST- allele? For each part, explain your answer. Assume that additional mutations do not play a role in this scenario.

--Why would Joanna express it? She would express the AST- allele if it was inherited from her dad. [That is because in a maternally imprinted gene, the allele from mom is methylated but the allele from dad is not.] --Could the kids inherit and/or express it? Yes they can inherit, the allele still exists in the genome. Because the gene is maternally imprinted, her kids would not express the AST- allele (it would be methylated in her eggs and thus in their somatic cells).

Ch.16 #5 Briefly describe the lac operon and how it controls the metabolism of lactose.

-The lac operon consists of three structural genes involved in lactose metabolism: lacZ, lacY, and lacA -lacZ codes for enzyme b-galactosidase, which breaks the disaccharide lactose into galactose and glucose, and converts lactose into allolactose -lacY is located downstream of the lacZ gene and codes for permease. Permease is necessary for the passage of lactose through the E.coli cell membrane. -lacA, located downstream of lacY, encodes the enzyme thiogalactoside transacetylase whose function in lactose metabolism has not yet been determined -all of these genes share a common overlapping promoter and operator region -upstream from the lactose operon is the lacI gene that encodes the lac operon repressor. The repressor binds at the operator region and inhibits transcription of the lac operon by preventing RNA polymerase from successfully initiating transcription. -when lactose is present in the cell, the enzyme b-gal converts some of it into allolactose. Allolactose binds to the lac repressor, altering its shape and reducing the repressor's affinity for the operator. Since this allolactose-bound repressor does not occupy the operator, RNA polymerase can initiate transcription of the lac structural genes from the lac promoter.

Ch.17 #1 How similar are the genomes of humans and chimpanzees? What genetic changes might be responsible for the large differences in the anatomy, physiology, and behavior of humans and chimpanzees?

-human and chimpanzees are identical at 96% of the genomes -humans and chimpanzees have different expression patterns for certain transcription factors -the differences in expression for transcription factors and changes in sequences that control gene expression could have a profound effect on the development process of both humans and chimpanzees -Because transcription factors may regulate multiple genes, a difference in a single transcription factor between humans and chimpanzees could impact the expression of many other genes, thus significantly affecting development within the two species -within brain tissues, transcriptional factor expression differences were more notable than in other tissues, which could explain the significant behavioral differences between humans and chimpanzees

A polyploid individual in a species with a haploid number of 10 would not have:

9 chromosomes

Ch.17 #9 What is an insulator?

An insulator or boundary element is a sequence of DNA that inhibits the action of regulatory elements called enhancers in a position dependent manner.

Ch.16 #8 What is antisense RNA? How does it control gene expression?

Antisense RNA molecules are small RNA molecules that are complementary to other DNA or RNA sequences and that form RNA-protein complexes. In bacterial cells, antisenseRNA molecules can bind to a complementary region in the 5' UTR of a mRNA molecule, blocking the attachment of the ribosome to the mRNA and stopping translation or they pair with specific regions of the mRNA and cleave the mRNA stopping translation.

Which of the following is a transversion mutation of the DNA codon CGA?

CCA

Ch.17 #22 (page 500) What would be the effect of moving the insulator (figure 17.8) to a position between enhancer II and promotor for gene B?

Enhancers I and II would now stimulate gene A. Neither enhancer would stimulate gene B.

Ch.18 #17 A codon that specifies the amino acid Gly undergoes a single-base substitution to become a nonsense mutation. Using the genetic code, is this mutation a transition or a transversion? At which position of the codon does the mutation occur?

Gly = GGU, GGC, GGA, GGG Stop = UAA, UAG, UGA GGA-->UGA transversion, occurs in the first position of the codon

Ch.8 #4 What is haploinsufficiency?

Haploinsufficiency is the term for a condition where having only one copy of a wild-type gene does not produce a wild-type phenotype in an otherwise diploid organism. For haploinsufficient genes, the gene dosage and relative amounts of gene products are important.

Ch.8 #9 How do translocations in which no genetic information is lost or gained produce phenotypic effects?

Like inversions, translocations can produce phenotypic effects if the translocation breakpoint disrupts a gene or if a gene near the breakpoint is altered in its expression because of relocation to a different chromosomal environment (a position effect)

Ch.21 #6 Briefly explain how patterns of DNA methylation are transmitted across cell division.

Methyltransferase enzymes recognize the hemimethylated state of CpG dinucleotides following replication and add methyl groups to the unmethylated cytosines, resulting in two new DNA molecules that are fully methylated.

Ch.18 #43 Which DNA-repair mechanism would most likely correct the error of a thymine on the original template strand base pairing with guanine through wobble, leading to an incorporated error.

Mismatch repair

What type of DNA repair mechanism is best suited to repair defects caused by errors in DNA replication? Why?

Mismatch repair - because it can determine which strand was the new strand and repairs the new one

Ch.18 #18 Use the genetic code to answer the following questions: a) If a single transition occurs in a codon that specifies Phe, what amino acids can be specified by the mutated sequence? b) If a single transversion occurs in a codon that specifies Phe, what amino acids can be specified by the mutated sequence?

Phe= UUU or UUC a) Transition: CUU, UCU, UUC, CUC, UCC, UUU. So, Phe --> Leu, Ser, Phe, Leu, Ser, Phe b) Transversion: AUU, GUU, UAU, UGU, UUA, UUG, AUC, GUC, UAC, UGC, UUA, UUG. So, Phe --> Ile, Val, Tyr, Cys, Leu, Leu, Ile, Val, Tyr, Cys, Leu, Leu

Ch.16 #4 What is the difference between positive and negative control? What is the difference between inducible and repressible operons?

Positive: requires an activator protein to stimulate transcription at the operon Negative: a repressor protein inhibits or turns off transcription at the operon Inducible: normally is not transcribed. It requires an inducer molecule to stimulate transcription either by inactivating a repressor protein in a negative inducible operon or by stimulating the activator protein in a positive inducible operon. Repressible: transcription normally occurs in a repressible operon. In a repressible operon, transcription is turned off either by the repressor becoming active in a negative repressible operon or by the activator becoming inactive in a positive repressible operon.

Ch.14 #16 What is the origin of small interfering RNAs (siRNA), microRNAs (miRNA), and Piwi-interacting RNAs (piRNA)? What do these RNA molecules do in the cell?

The siRNAs originate from the cleavage of mRNAs, RNA transposons, and RNA viruses by the enzyme Dicer. Dicer may produce multiple siRNAs from a single double-stranded RNA molecule. The double-stranded RNA molecule may occur due to the formation of hairpins or by duplexes between different RNA molecules. The miRNAs arise from the cleavage of individual RNA molecules that are distinct from other genes. The enzyme Dicer cleaves these RNA molecules that have formed small hairpins. A single miRNA is produced from a single RNA molecule. The piwi-interacting RNAs are typically 24-30 nucleotides in length and originate from a longer single-stranded precursor RNA molecule. Both siRNAs and miRNAs silence gene expression through a process called RNA interference. Both function by shutting off gene expression of a cell's own genes or to shut off expression of genes from the invading foreign genes of viruses or transposons. The microRNAs typically silence genes that are different from those from which the microRNAs are transcribes. However, the siRNAs usually silence genes from which they are transcribed. The piwi-interacting RNAs interact with Piwi proteins to silence gene expression of transposons. The silencing of transposon gene expression by the piwi-interacting RNAs occur in animal germ cells and can result in transposon mRNA degradation, chromatin rearrangement that inhibits transcription of the transposons gene, and inhibition of translation of transposon encoded proteins.

In a bacterial cell, you observe that translation of RNA transcribed from some channel-encoding genes is dependent on the presence of fluoride ions. The RNA is translated in the presence of fluoride ions and not translated in the absence of fluoride ions. A mutation study shows that the entire 5' UTR is necessary for this regulation to occur. What regulatory mechanism is this? Name the mechanism and describe how it is acting to affect expression when fluoride molecules are absent and when they are present.

This is a riboswitch. Binding of the fluoride ion signal makes the SD available by changing the secondary structure, resulting in translation. Without the signal, the secondary structure blocks the ribosome's binding because the SD is part of a double-stranded region, so translation does not occur. How did we get this answer? Clues from the question: In a bacteria Regulation is acting on translation (not transcription) 5' UTR is important

Ch.8 #36 Species I has 2n=16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? a) monosomic b) autotriploid c) autotetraploid d) trisomic e) double monosomic f) nullisomic g) autopentaploid h) tetrasomic

a) 15 b) 24 c) 32 d) 17 e) 14 f) 14 g) 40 h) 18

Ch.8 #19 Which types of chromosome mutations a) increase the amount of genetic material in a particular chromosome? b) increase the amount of genetic material in all chromosomes? c) decrease the amount of genetic material in a particular chromosome? d) change the position of DNA sequences in a single chromosome without changing the amount of genetic material? e) move DNA from one chromosome to a non-homologous chromosome?

a) duplications b) polyploidy c) deletions d) inversions e) translocations

Among the chemically-induced mutations covered in class are base analogs, alkylating agents, deaminating agents, and oxidative reactions. How is base analog different from these other chemically-induced mutations?

different: base analog is incorporated into a new DNA strand instead of a normal nucleotide during DNA replication whereas the others all cause a change to a nucleotide that is already part of the DNA strand. similar: all of these cause substitutions.

A uracil base found in DNA would have been created by: a. Depurination b. ionizing radiation c. DNA polymerase mis-incorporation d. UV light e. cytosine deamination

e. cytosine deamination

Mutant B OXT is found in new-world monkeys that form monogamous mating pairs. It binds the oxytocin receptor more tightly and may cause this stronger mate bonding. Mutant B in humans would be considered to be a) loss of function b) gain of function c) Neutral d) Silent e) recessive

gain of function

Ch.16 #1 Why is gene regulation important for bacterial cells?

gene regulation allows for biochemical and internal flexibility while maintaining energy efficiency by the bacterial cells

How does DNA methylation affect transcription?

it blocks the major groove of DNA

Neither LacI nor CAP bind to the DNA of the lac operon when:

lactose is high and glucose is high

In eukaryotes, each specific transcription factor recognizes its own

response element

Which of the following works only in eukaryotes?

siRNA

Ch.16 #26 At which level of gene regulation does attenuation occur? (Figure 16.1)

transcription

Ch.17 #13 Briefly list some ways in which siRNAs and miRNAs regulate genes.

(1) Through cleavage of mRNA sequences through "slicer activity": The binding of RISCs containing either siRNA or miRNA to complementary sequences in mRNA molecules stimulate cleavage of the mRNA through "slicer activity." This is followed by further degradation of the cleaved mRNA. (2) Through binding of complementary regions with the mRNA molecule by miRNAs to prevent translation: The miRNAs as part of RISC bind to complementary mRNA sequences preventing either translation initiation or elongation, which results in premature termination. (3) Through transcriptional silencing due to methylation of either histone proteins or DNA sequences: The siRNA binds to complementary DNA sequences within the nucleus and stimulated methylation of histone proteins. Methylated histones bind DNA more tightly preventing transcription factors from binding the DNA. The miRNA molecules bind to complementary DNA sequences and stimulate DNA methylases to directly methylate the DNA sequences, which results in transcriptional silencing. (4) Through slicer-independent mRNA degradation stimulated by miRNA binding to complementary regions in the 3'UTR of the mRNA: a miRNA binds to the AU rich element in the 3' UTR of the mRNA stimulating degradation using RISC and dicer.

Ch.21 #3 What three molecular mechanisms alter chromatin structure and are responsible for many epigenetic phenotypes?

(1) changes in DNA methylation (2) chemical modifications of histone proteins (3) RNA molecules that affect chromatin structure and gene expression

Match the following: 1. Adding methyl/ethyl groups 2. Caused by a mutagen 3. Oxidizes nucleotide bases 4. Amine groups are replaced by carbonyl groups 5. Inserting chemically modified bases 6. Mutagen is inserted in between double helix to distort it 7. Happens without outside external factors 8. Typically found in tandem repeat sequences 9. New nucleotide is inserted in between nucleotides 10. Creates Thymine dimers 11. Breaks strands a) Unequal Crossing Over b) Strand Slippage c) Depurination d) Spontaneous Change e) Induced Change f) Base Analog g) Alkylation h) Deaminating Agent i) Intercalating Agent j) UV Radiation k) X-ray Radiation

1. Adding methyl/ethyl groups (g) 2. Caused by a mutagen (e) 3. Oxidizes nucleotide bases (k) 4. Amine groups are replaced by carbonyl groups (h) 5. Inserting chemically modified bases (f) 6. Mutagen is inserted in between double helix to distort it (i) 7. Happens without outside external factors (d) 8. Typically found in tandem repeat sequences (a) 9. New nucleotide is inserted in between nucleotides (b) 10. Creates Thymine dimers (j) 11. Breaks strands (k)

Ch.8 #8 Explain why recombination is suppressed in individuals heterozygous for paracentric and pericentric inversions.

A crossover within a paracentric inversion produces a dicentric and an acentric recombinant chromatid. The acentric fragment is lost, and the dicentric fragment breaks, resulting in chromatids with large deletions that lead to nonviable gametes or embryonic lethality. A crossover within a pericentric inversion produces recombinant chromatids that have duplications or deletions. Again, gametes with these recombinant chromatids do not lead to viable progeny.

Ch.18 #3 What is the difference between a missense mutation and a nonsense mutation? Between a silent mutation and a neutral mutation?

A missense mutation includes a variant that changes the nucleotide sequence, while a nonsense mutation includes a variant that changes a codon in the amino acid sequence to a stop codon. A silent mutation is a change in the nucleotide sequence that doesn't affect the amino acid sequence. A neutral mutation does affect the amino acid sequence, but the phenotype expressed does not change.

Ch.8 #5 What is the difference between a paracentric and pericentric inversion?

A paracentric inversion does not include the centromere; a pericentric inversion includes the centromere

Ch.18 #1 What is the difference between a transition and a transversion? Which type of base substitution is more common?

A transition is a substitution between a purine and another purine or a pyrimidine and another pyrimidine. A transversion is a substitution between purines to pyrimidines. Transitions are more common because this error is easier to make during replication.

A mutation ocurs in one cell of a regenerating liver in an adult human. Cells in regenerating livers are undergoing mitosis. This mutation causes expression of an enzyme, trypsin, that does should not be expressed in the liver. Answer the following questions. (9 points) (A) Use a term to describe the phenotypic effects of this mutation: (B) Which cells in this individual's body will express this mutant allele? Should this individual worry that the mutation may be transmitted to their offspring? Explain your answer.

A) Gain of function (B) - the cell where the mutation occurred and all of it's decedents but no other cells -no concern that it will be passed to the offspring because it is a somatic mutation

For the following mutations, label the kind of mutation that has occurred. The original homologous chromosomes are ABCD/EFGH. (HINT: all mutations are different kinds of mutations). ABBCD/EFGH ACD/EFGH ABCBCD/EFGH ADCB/EFGH ABEF/CDGH ABC/DEFGH AE/BCDFGH ABCE/DFGH ABABC/DEFGH

ABBCD/EFGH: tandem duplication ACD/EFGH: deletion ABCBCD/EFGH: insertion ADCB/EFGH: inversion ABEF/CDGH: balanced reciprocal translocation ABC/DEFGH: unbalanced non-reciprocal translocation AE/BCDFGH: unbalanced reciprocal translocation ABCE/DFGH: balanced reciprocal translocation ABABC/DEFGH: insertion with a translocation

Label the following mutations as balanced or unbalanced (A pairs with E, B pairs with F, C pairs with G, and D pairs with H): ABCD/EFGH ABBCD/EFGH ACD/EFGH ABCBCD/EFGH ADCB/EFGH ABEF/CDGH ABC/DEFGH AE/BCDFGH ABCE/DFGH ABABC/DEFGH

ABCD/EFGH: balanced ABBCD/EFGH: unbalanced ACD/EFGH: unbalanced ABCBCD/EFGH: unbalanced ADCB/EFGH: balanced ABEF/CDGH: balanced ABC/DEFGH: unbalanced AE/BCDFGH: unbalanced ABCE/DFGH: balanced ABABC/DEFGH: unbalanced

You are creating a transgenic Drosophila where the gene EY is regulated by a conditional promoter under control of the LacI protein. You want the gene EY to be expressed during embryonic development but then not expressed in adults. What sequences would you need to add to the Drosophila genome to regulate EY expression with the LacI protein? How would you turn EY's expression on? How would you turn it off?

Add lacI gene under a consituitive promoter Add the operator that lacI recognizes in front of the EY gene. Turn on expression by feeding the fly IPTG (or allolactose) Turn off expression by feeding the fly a diet without IPTG (or allolactose)

Ch.16 #18 What would be the effect of a drug that altered the structure of allolactose so that it was unable to bind to the regulator protein?

Allolactose is produced when lactose is present; allolactose normally binds to the repressor protein and makes it inactive, allowing transcription to occur when lactose is present. If a drug altered the structure of allolactose, it would not bind to the repressor and the repressor would continue to bind to the operator, keeping transcription off. The result would be that transcription was repressed even in the presence of lactose; thus, no b-gal or permease would be produced

Match the appropriate ploidy to the correct mutation. Allotetraploid Autopolyploidy Monosomy Autotriploidy Tetrasomy Allopolyploidy 1. One chromosome lacks a homologous pair. 2. Four copies of a particular chromosome are present. 3. Created as a result of both chromosome sets of each parents being present in gametes 4. An individual has more than 2 sets of chromosomes 5. Having two or more sets of chromosomes derived from two different species. 6. Having a triploid set of chromosomes.

Allotetraploid: 5 Autopolyploidy: 3 Monosomy: 1 Autotriploidy: 6 Tetrasomy: 2 Allopolyploidy: 4

Ch.8 #6 How can inversions in which no genetic information is lost or gained cause phenotypic effects?

Although inversions do not result in loss or duplication of chromosome material, inversions can have phenotypic consequences if the inversion disrupts a gene at one of its breakpoints or if a gene near a breakpoint is altered in its expression because of a change in its chromosomal environment, such as relocation to a heterochromatic region. Such effects on gene expression are called position effect.

Ch.16 #7 What is attenuation? What are the mechanisms by which the attenuator forms when tryptophan levels are high and the anti-terminator forms when tryptophan levels are low?

Attenuation: the termination of transcription prior to the structural genes of an operon. It is a result of the formation of a termination hairpin structure or attenuator in the mRNA. -two types of secondary structures can be formed by the mRNA 5' UTR of the trp operon. If the 5' UTR forms two hairpin structures from the base pairing of region 1 with region 2 and the pairing of region 3 with region 4, then transcription of the structural genes will not occur. -The hairpin structure formed by the pairing of region 3 with region 4 results in a terminator being formed that stops transcription. -When region 2 pairs with region 3, the resulting hairpin acts as an anti-terminator, allowing for transcription to proceed -region 1 of the 5' UTR also encodes a small protein and has two adjacent tryptophan codons (UGG) -tryptophan levels affect transcription due to the coupling of translation with transcription in bacterial cells --> when tryptophan levels are high, the ribosome quickly moves through region 1 and into region 2, thus preventing region 2 from pairing with region 3. Therefore, region 3 is available to form the attenuator hairpin structure with region 4, stopping transcription -when tryptophan levels are low, the ribosome stalls or stutters at the adjacent tryptophan codons in region 1. Region 2 now becomes available to base pair with region 3, forming the antiterminator hairpin. transcription can now proceed through the structural genes

Ch.8 #17 Explain why autopolyploids are usually sterile, whereas allopolyploids are often fertile.

Autopolyploids arise from duplication of their own chromosomes. During meiosis, the presence of more than two homologous chromsomes results in faulty alignment of homologues in prophase I, and subsequent faulty segregation of the homologues in anaphase I. The resulting gametes have an uneven distribution of chromosomes and are genetically unbalanced. These gametes usually produce lethal chromosome imbalances in the zygote. Allopolyploids, however, have chromosomes from different species. As long as they have a diploid set of chromosomes from each species, as in an allotetraploid or even an allohexaploid, the homologous chromsome pairs from each species can align and segregate properly during meiosis. Their gametes will be balanced and will produce viable zygotes when fused with other gametes from the same type of allopolyploid individual.

Ch.16 #13 The blob operon produces enzymes that convert compound A into compound B. The operon is controlled by regulator gene S. Normally, the enzymes are synthesized only in the absence of compound B. If gene S is mutated, the enzymes are synthesized in the presence and in the absence if compound B. Does gene S produce a regulator protein that exhibits positive or negative control? Is this operon inducible or repressible?

Because the blob operon is transcriptionally inactive in the presence of B, gene S most likely codes for a repressor protein that requires compound B as a corepressor. The data suggests that the blob operon is repressible because it is inactive in the presence of compound B, but active when compound B is absent.

Ch.14 #20 Explain how some IncRNAs serve as molecular decoys for RNA-binding proteins and miRNAs

Binding sites for either RNA-binding proteins or miRNAs can be found on some IncRNAs. When these proteins or miRNAs bind to the IncRNAs, they are unavailable to bind to the RNA molecules that are their primary target. Essentially, these IncRNAs mimic the normal binding target of the RNA-binding proteins or miRNAs, thus providing a molecular decoy that regulated how many of certain RNA-binding proteins or miRNA molecules remain available to bind their primary target.

Ch.16 #14 A mutation prevents the catabolite activator protein (CAP) from binding to the promoter in the lac operon. What will the effect of this mutation be on the transcription of the operon?

Catabolite activator protein binds the CAP site of the lac operon and stimulates RNA polymerase to bind the lac promoter, thus resulting in increased levels of transcription from the lac operon. If a mutation prevents CAP from binding to the site, then RNA polymerase will bind to the lac promoter poorly. This will result in significantly lower levels of transcription of the lac structural genes.

Ch.17 #4 What changes take place in chromatin structure and what role do these changes play in eukaryotic gene regulation?

Changes in chromatin structure can result in repression or stimulation of gene expression. As genes become more transcriptionally active, DNA shows increased sensitivity to DNase I digestion, suggesting that the chromatin structure is more open. Acetylation of histone proteins by acetyltransferase proteins results in the destabilization of the nucleosome structure and increases transcription as well as hypersensitivity to DNase I. The reverse reaction by deacetylases stabililizes nucleosome structure and lessens DNase I sensitivity. Other transcription factors and reglatory proteins, called chromatin remodeling complexes, bind directly to the DNA-altering chromatin structure without acetylating histone proteins. The chromatin remodeling complexes allow for transcription to be initiated by increasing accessibility to the promoters by transcription factors. DNA methylation is also associated with decreased transcription. Methylated DNA sequences stimulate histone deacetylases to remove acetyl groups from the histone proteins, thus stabilizing the nucleosome and repressing transcription. Demethylation of DNA sequences is often followed by increased transcription, which may be related to the deacetylation of the histone proteins.

Ch.8 #1 List the three basic categories of chromosome mutations and define each one.

Chromosome Rearrangements: Deletion- loss of a portion of a chromosome Duplication- addition of an extra copy of a portion of a chromosome Inversion- a portion of the chromosome is reversed in orientation Translocation- a portion of one chromosome becomes incorporated into a different (non-homologous) chromosome Ploidy Changes: Aneupoloidy- loss or gain of one or more chromosomes so that the chromosome number deviates from 2n or the normal euploid complement Polyploidy- gain of entire sets of chromosomes so the chromosome number changes from 2n to 3n, 4n, and so on

The structure of DNA is not totally stable, particularly in aqueous solutions and acidic pH. Describe a spontaneous change that happens with DNA (before DNA replication) and can lead to a mutation if not repaired before the DNA is replicated.

Deamination -Loss of an amine group (NH2) -Cytosine looks like a uracil -In further replication will pair the U with an A -Adenine looks like a G Depurination - loss of a purine base from sugar-phosphate backbone -Hydrolysis of a purine base - loss of a base -During replication, because that base is empty, a random base can be put there → when this strand gets replicated it will pass down this random base

Ch.17 #8 What is an enhancer? How does it affect transcription of distant genes?

Enhancers are DNA sequences that are the binding sites of transcriptional activator proteins. Transcription at a distant gene is affected when the DNA sequence located between the gene's promoter and the enhancer is looped out, allowing for the interaction of the enhancer-bound proteins with proteins needed at the promoter, which in turn stimulates transcription. Additionally, the transcription of short enhancer (e)RNA molecules from an enhancer template may be involved in transcriptional activation, but a precise mechanism for such activation has not been determined.

Ch.8 #16 What is the difference between autopolyploidy and allopolypoloidy? How does each arise?

In autopolyploidy, all sets of chromosomes are from the same species. Autopolyploids typically arise from mitotic nondisjunction of all the chromosomes in early diploid (2n) embry, resulting in an autotetraploid, or from meiotic nondisjunction that results in a 2n gamete fusing with a 1n gamete to form an autotriploid. In allopolyploidy, the chromosomes of two different species are contained in one individual through the hybridization of two related species followed by mitotic nondisjunction. Fusion of gametes from two different (but usually related) species results in a hybrid with a haploid set of chromosomes from each parent. If an early embryonic cell then undergoes mitotic nondisjunction and doubles each chromosome, then a fertile 4n allotetraploid individual having two copies of each chromosome from each species may result.

Ch.17 #20 How do repressors that bind to silencers in eukaryotes differ from repressors that bind to operators in bacteria?

In bacteria, repressors that bind to the operator block RNA polymerase from binding to the promoter, and thus, directly block transcription. On the other hand, repressors that bind to silencers in eukaryotes block transcriptional activator proteins from binding at an activator site, thus eliminating transcriptional activation.

Ch.16 #6 What is catabolite repression? How does it allow a bacterial cell to use glucose in preference to other sugars?

In catabolite repression, the presence of glucose inhibits or represses the transcription of genes involved in the metabolism of other sugars. Because the gene expression necessary for utilizing other sugars is off, only enzymes involved in the metabolism of glucose will be synthesized. Operons that exhibit catabolite repression are under the positive control of catabolic activator protein (CAP). For CAP to be active, it must form a complex with cAMP. Glucose affects the level of cAMP. The levels of glucose and cAMP are inversely proportional—as glucose levels increase, the level of cAMP decreases. Thus, CAP is not activated.

Ch.21 #7 What types of histone modifications are responsible for epigenetic phenotypes?

Modifications of histone proteins especially in the positively charged tail are responsible for many epigenetic changes. The histone modifications include covalent modifications such as the addition or removal of chemical groups such as acetyl groups, methyl groups, phosphates, and ubiquitin to the positively charged tail. These modifications can change chromatin structure thus altering gene expression. Additionally, some modifications may change gene expression through their effect on the recruitment of transcription factors.

Ch.16 #28 Some mutations in the trp 5' UTR increase termination by the attenuator. Where might these mutations occur, and how might they affect the attenuator?

Mutations that disrupt the formation of the anti-terminator will increase termination by the attenuator. Could be caused by: - deletion in region 2 - mutations in region 1 if the mutation prevented the ribosome from stalling at the adjacent tryptophan codons within region 1 (any mutation that blocks translation initiation or stops translation early) - a mutation in region 1 that eliminates or replaces the two adjacent tryptophan codons in the small protein (ribosome cannot stall)

Ch.17 #21 (page 499) What would be the effect on transcription if a mutation occurred in the gene that encodes GAL3, so that no functional GAL3 was produced?

No transcription would take place when galactose was present and so the enzymes that are needed to break down galactose would not be produced. GAL80 normally prevents GAL4 from stimulating transcription. When galactose is present, it normally binds to GAL3, allowing GAL3 to bind to GAL80 and releasing the inhibition that GAL80 has on the interaction of GAL4 with the promoter. If GAL3 were mutated, so that it is nonfunctional, it would never bind to GAL80. GAL80 would always prevent GAL4 from stimulating transcription at the promoter.

Ch.8 #12 List four major types of aneuploidy.

Nullisomy: having no copies of a chromosome Monosomy: having only one copy of a chromosome Trisomy: having three copies of a chromosome Tetrasomy: having four copies of a chromosome

Why are partial diploid strains useful in the study of operon regulation

Partial diploid strains can be used to identify whether a regulator is cis- acting or trans- acting.

Label the following kinds of chromosomal mutations as balanced or unbalanced (in offspring): Pericentric Inversion Paracentric Inversion Paralog formation Ortholog formation Deletion Duplication Translocation Robertsonian Translocation

Pericentric Inversion: unbalanced Paracentric Inversion: unbalanced Paralog formation: unbalanced Ortholog formation: balanced Deletion: unbalanced (haploinsufficiency) Duplication: unbalanced Translocation: balanced (equally reciprocal) or unbalanced Robertsonian Translocation: balanced (equally reciprocal) or unbalanced

Ch.16 #9 What are riboswitches? How do they control gene expression? How do riboswitches differ rom RNA-mediated repression?

Riboswitches are regulatory sequences in RNA molecules. Most can fold into compact secondary structures consisting of a base stem and several branching hairpins. At riboswitches, regulatory molecules bind and influence gene expression by affecting the formation of secondary structures within the mRNA molecule. The binding of the regulatory molecule to a riboswitch sequence may result in repression or induction. Some regulatory molecules bind the riboswitch sequence and stabilize a terminator structure in the mRNA, which results in premature termination of the mRNA molecule. Other regulatory molecules bind riboswitch sequences resulting in the formation of secondary structures that block the ribosome binding sites of the mRNA molecules, thus preventing translation initiation. In induction, the regulatory molecule acts as an inducer, stimulating the formation of a secondary structure in the mRNA that allows for transcription or translation to occur. RNA-mediated repression occurs through the action of a ribosome. In RNA-mediated repression, an RNA sequence within the 5' UTR can act as a ribozyme that when stimulated by the presence of a regulatory molecule can induce self-cleavage of the mRNA molecule, which prevents translation of the molecule. When bound by a regulatory molecule, RNA mediated repression results in the self-cleavage of the mRNA molecule. When bound by a regulatory molecule, riboswitch sequences stimulate changes in the secondary structure of the mRNA molecule that affect gene expression.

Ch.14 #17 What are some similarities and differences between siRNAs and miRNAs?

Similarities: - both silence gene expression of foreign genes or a cell's own genes after combining with proteins to form a RNA-induced silencing complex (RISC) - both are dsRNA molecules 21 and 22 nucleotides in length -both are produced by the action of the enzyme Dicer on larger RNA molecules Differences: - the miRNAs originate from transcription products of distinct genes, while the siRNAs originate from mRNa, RNA transposons, or RNA viruses - for miRNAs, dicer cleaves single-stranded RNAs that form short hairpins, while for siRNAs, dicer cleaves single-stranded RNAs that form long hairpins or RNA duplexes - the miRNAs typically act by inhibiting translation although some do trigger the degradation of the mRNA, while some siRNAs work by stimulating degradation of the mRNA target molecule. Other siRNAs inhibit transcription - base-pairing between the miRNA and target is imperfect, while siRNA is usually perfect

Ch.18 #5 How do insertions and deletions arise?

Strand slippage that occurs during DNA replication and unequal crossover events due to misalignment to repetitive sequences have been shown to cause deletions and additions of nucleotides to DNA molecules. Strand slippage results from the formation of small loops on either the template or the newly synthesized strand. If the loop forms on the template strand, then a deletion occurs. Loops formed on the newly synthesized strands result in insertions. If, during crossing over, a misalignment of the two strands at repetitive sequence occurs, then the resolution of the crossover will result in one DNA molecule containing an insertion and the other molecule containing a deletion.

Ch.16 #29 Some of the mutations of the type mentioned in problem 28 have an interesting property: they prevent the formation of the anti-terminator that normally takes place when the tryptophan level is low. In one of these mutations, the AUG start codon for translation of the 5' UTR has been deleted. How might this mutation prevent anti-termination from taking place?

The AUG start codon is necessary for the translation initiation of the 5' UTR peptide. If translation does not initiate, then the mRNA 5'UTR region 1 will be available to pair with region 2. The resulting hairpin will prevent the formation of the anti-terminator.

Ch.14 #18 What role do CRISPR-Cas systems naturally play in bacteria?

The CRISPR-Cas system essentially serves as an adaptive RNA defense system or immune system for bacterial cells and protects cells against foreign DNA genomes from plasmids or infecting bacteriophages. The spacers between the palindromic repeat sequences consist of short nucleotide sequences captured from previous invading DNA molecules and can help to provide adaptive protection by targeting homologous invading foreign DNA molecules for cleavage by the Cas proteins.

Ch.14 #19 Outline the three stages of CRISPR-Cas action.

The action of the CRISPR-Cas system occurs through the three distinct stages of aquisition, expression, and interference. Aquisition: short DNA sequences from the destroyed foreign DNA molecules are inserted into a spacer region that is flanked by the palindromic repeat sequences within the CRISPR array Expression: a long precursor RNA is transcribed from the CRISPR array. The Cas proteins cleave and process the precursor RNA into shorter RNA fragments called crRNAs. Each of these crRNAs contains one of the spacer regions. These crRNAs then combine with the Cas proteins to form the effector complexes. Interference: the effector complexes bind to the homologous invading foreign DNA molecules allowing for their cleavage by the Cas proteins.

Ch.18 #44 A plant breeder wants to isolate mutants in tomatoes that are defective in DNA repair. However, this breeder does not have the expertise or equipment to study enzymes in DNA-repair systems. How can the breeder identify tomato plants that are deficient in DNA repair? What are the traits to look for?

The breeder should look for plants that have increased levels of mutations either in their germ-line or somatic tissues. Potentially mutant plants could be exposed to standard mutagens that damage DNA. If they are defective in DNA repair, they should have higher rates of mutation. For example, tomato plants with defective DNA repair systems should have an increased mutation rate when exposed to high levels of sunlight.

Ch.8 #2 Why do extra copies of genes sometimes cause drastic phenotypic effects?

The expression of some genes is balanced with the expression of other genes; the ratios of their gene products, usually proteins, must be maintained within a relatively narrow range for proper cell function. Extra copies of one of these genes cause that gene to be expressed at proportionally higher levels, thereby upsetting the balance of gene products.

Ch.16 #21 Explain why mutations in the lacI gene are trans in their effects, but mutations in the lacO gene are cis in their effects.

The lacI gene encodes the lac repressor protein, which can diffuse within the cell and attach to any operator. It can therefore affect the expression of genes on the same or different molecules of DNA. The lacO gene encodes the operator. The binding of the lac repressor to the operator affects the binding of RNA polymerase to the DNA, and therefore affects only the expression of genes on the same molecule of DNA.

Ch.8 #11 What is a Robertsonian translocation?

The long arms of two acrocentric chromosomes are joined to a common centromere through translocation, resulting in a large metacentric chromosome and a very small chromosome with two very short arms. The very small chromosome may be lost.

Ch.21 #4 What is the major form of DNA methylation that is seen in eukaryotes? At what type of DNA sequence is DNA methylation usually found?

The major form of DNA methylation seen in eukaryotes is 5-methylcytosine. In eukaryotes, DNA methylation is often seen at CpG dinucleotides. Additionally, in plants, DNA methylation is also frequently seen as CpNpG trinucleotides.

Ch.21 #5 How does DNA methylation repress transcription?

The methyl group of 5-methylcytosine sits within the major groove of the DNA and may inhibit the binding of transcription factors and other proteins required for transcription to occur; 5-methylcytosone attracts proteins that directly repress transcription. It also attracts histone deacteylase enzymes, which remove acetyl group histone proteins, altering the chromatin structure that represses transcription.

How does tryptophan affect the binding activity of the TrpR protein? Through what mechanism does this change occur?

The presence of tryptophan causes TrpR to bind the trp operon's DNA. This works because when tryptophan binds to the TrpR it changes the shape of the DNA binding domain so that it fits into the DNA structure.

Ch.16 #30 Several examples of antisense RNA regulating translation in bacterial cells have been discovered. Molecular geneticists have also used antisense RNA to artificially control transcription in both bacterial and eukaryotic genes. If you wanted to inhibit the transcription of a bacterial gene with antisense RNA, what sequences might the antisense RNA contain?

To block transcription, you will need to disrupt the action of RNA polymerase either directly or indirectly. Antisense RNA containing sequences complementary to the gene's promoter should inhibit the binding of RNA polymerase. If transcription initiation by RNA polymerase requires the assistance of an activator protein, then antisense RNA complementary to the activator protein-binding site of the gene could also disrupt transcription. By binding the activator site, the antisense RNA would block access to the site by the activator and prevent RNA polymerase from being assisted by the activator to initiate transcription.

Ch.17 #7 Briefly explain how transcriptional activator proteins and repressors affect the level of transcription of eukaryotic genes.

Transcriptional activator proteins stimulate transcription by binding DNA at specific base sequences such as an enhancer or regulatory promoter and attracting or stabilizing the basal transcription factor apparatus. Repressor proteins bind to silencer sequences or promoter regulator sequences. These proteins may inhibit transcription by either blocking access to the enhancer sequence by the activator protein, preventing the activator from interacting with the basal transcription apparatus, or preventing the basal transcription factor from being assembled.

A patient of yours is heterozygous for a chromosomal rearrangement that is unbalanced in his somatic cells. What does it mean that this chromosomal rearrangement is unbalanced? What chromosomal rearrangements are unbalanced in a heterozygote person's somatic cells?

Unbalanced means that there is a change in the amount of genetic information relative to the wildtype (either more or less) Unbalanced rearrangements in a heterozygote's somatic cells are duplications or deletions.

Ch.16 #17 A mutant strain of E.coli produces B-galactosidase in both the presence and the absence of lactose. Where in the operon might the mutation in this strain be located?

Within the operon, the operator region is the most probable location of the mutation. If the mutation prevents the lac repressor protein from binding to the operator, then transcription of the lac structural genes will not be inhibited. Expression will be constitutive. Outside of the operon, a mutation in the lacI gene that inactivates the repressor or keeps it from binding to the operator could also lead to constitutive expression of the structural genes.

Ch.8 #21 A chromosome initially has the following segments: AB * CDEFG Draw the chromosome, identifying its segments, that would result from each of the following mutations. a) tandem duplication of DEF b) displaced duplication of DEF c) deletion of FG d) paracentric inversion that includes DEFG e) pericentric inversion of BCDE

a) AB*CDEFDEFG b) DEFAB*CDEFG c) AB*CDE d) AB*CGFED e) AEDC*BFG

Ch.8 #38 Species I has 2n=8 chromosomes and species II has 2n=14 chromosomes. What would the expected chromosome numbers be in individuals with the following chromosome mutations? Give all possible answers. a) allotriploidy including species I and II b) autotetraploidy in species II c) trisomy in species I d) monosomy in species II e) tetrasomy in species I f) allotretraploidy including species I and II

a) Could have 1n from species I and 2n from species II for 3n=18 or they could have 2n from species I and 1n from species II for 3n=15 b) 4n=28 c) 2n+1=9 d) 2n-1=13 e) 2n+2=10 f) allotetraploids must have chromosomes from both species and total 4n. There are 3 possible combinations for such allotetraploids: 2n from each- 2(4)+2(7)=22 1n from species I+3n from species II- 1(4)+3(7)=25 3n from species I+1n from species II- 3(4)+1(7)=19

Ch.8 #22 The following diagram represents two nonhomologous chromosomes: AB * CDEFG RS * TUVWX What type of chromosome mutation would produce each of the following groups of chromosomes? a) AB * CD RS * TUVWXEFG b) AUVB * CDEFG RS * TWX c) AB * TUVFG RS * CDEWX d) AB * CWG RS * TUVDEFX

a) Nonreciprocal translocation of EFG b) Nonreciprocal translocation of UV c) Reciprocal translocation of CDE and TUV d) Reciprocal translocation of DEF and W

Ch.16 #27 Listed in parts a through g are some mutations that were found in the 5' UTR of the trp operon of E.coli. What will the most likely effect of each of these mutations be on the transcription of the trp structural genes? a) a mutation that prevents the binding of the ribosome to the 5' end of the mRNA 5' UTR b) a mutation that changes the Trp codons in region 1 of the mRNA 5' UTR into codons for alanine c) a mutation that creates a stop codon early in region 1 of the mRNA 5' UTR d) deletions in region 2 of the mRNA 5' UTR e) deletions in region 3 of the mRNA 5' UTR f) deletions in region 4 of the mRNA 5' UTR g) deletion of the string of adenine nucleotides that follows region 4 in the 5' UTR

a) Region 1 of the mRNA 5' UTR will be free to pair with region 2, thus preventing region 2 from pairing with region 3. Region 3 will be free to pair with region 4, forming the attenuator or termination hairpin. Transcription of the trp structural genes will be terminated. No gene expression will occur. b) If alanine codons replace trp codons, then under conditions of low alanine, the stalling of the ribosome will not occur (where regions 2 and 3 would normally form the anti-terminator hairpin). The attenuator will form, stopping transcription. The ribosome will stall when alanine is low, so transcription of the structural genes will occur only when alanine is low c) The ribosome will fall off region 1, allowing it to form a hairpin with region 2 and transcription will not occur bc regions 3 and 4 are free to form the attenuator d) If region 2 is deleted, the antiterminator cannot be formed. The attenuator will form and transcription will not occur e) The attenuator will not be able to form, transcription will be continuous. f) The attenuator will not be able to form, transcription will be continuous. g) For the attenuator to function, the presence of a string of uracil nucleotides following region 4 in the mRNA 5' UTR is required. So, no termination will occur and transcription will proceed.

Ch.16 #24 Ellis Engelsberg and his colleagues examined the regulation of genes taking part in the metabolism of arabinose, a sugar. Four structural genes encode enzymes that help metabolize arabinose (genes A,B,D, and E). An additional sequence C is linked to genes A,B, and D. These are in the order D−A−B−C. Gene E is distant from the other genes. Engelsberg and his colleagues isolated mutations at the C sequence that affected the expression of structural genes A,B,D, and E. In one set of experiments, they created various genotypes at A and C and determined whether arabinose isomerase (the enzyme encoded by gene A) was produced in the presence or absence of arabinose (the substrate of arabinose isomerase) by cells with these genotypes. Results from this experiment are shown in the following table, where a plus sign (+) indicates that the arabinose isomerase was synthesized and a minus sign (−) indicates that the enzyme was not synthesized. Genotype Arabinose Absent Arabinose Present 1. C+A+ — + 2. C-A+ — — 3. C-A+/C+A- — + 4. C^cA-/C-A+ + + a. On the basis of these results, is the C sequence an operator or a regulator gene? Explain your reasoning. b. Do these experiments suggest that the arabinose operon is negatively or positively controlled? Explain your reasoning. c. What type of mutation is Cc?

a) The C gene is a regulator gene. The C gene is trans-acting, thus it affects the expression of the A gene located on a different DNA molecule, which is typical of a gene encoding a regulatory protein. If the C gene was an operator, it would be cis-acting and only able to regulate the expression of the A gene found on the same DNA molecule, which is not the case as demonstrated from genotype 3. b) Positively controlled: the C gene appears to be a regulatory gene that is needed for the transcription of the A gene. For the A gene to be expressed, a functional C gene needs to be present within the cell. In the absence of a function C gene and arabinose, the A gene is not expressed. Both results would be explained if the C gene encodes a regulatory protein that is required to activate transcription of the A gene. c) the Cc mutation results in continuous activation of transcription from the A gene (constitutive expression)

Ch.16 #12 A mutation at the operator prevents the regulator protein from binding. What effect will this mutation have in the following types of operons? a) regulator protein is a repressor in a repressible operon b) regulator protein is a repressor in an inducible operon

a) The regulator protein-corepressor complex would normally bind to the operator and inhibit transcription. If a mutation prevented the repressor protein from binding at the operator, then the operon would never be turned off and transcription would occur all the time b) In an inducible operon, a mutation at the operator site that blocks binding of the repressor would result in constitutive expression and transcription would occur all the time

Ch.18 #27 The following nucleotide sequence is found in a short stretch of DNA: 5'-AG-3' 3'-TC-5' a) Give all the mutant sequences that can result from spontaneous depurination in this stretch of DNA b) Give all the mutant sequences that can result from spontaneous deamination in this stretch of DNA

a) The strand contains two purines, adenine and guanine. Because repair of depurination typically results in adenine being substituted for the missing purine, only the loss of the guanine by depurination will result in a mutant sequence. 5'-AG-3' --> 5'-AA-3' 3'-TC-5' 3'-TT-5' b) Deamination of guanine, cytosine and adenine can occur. However, the deamination of only cytosine and adenine are likely to result in mutant sequences because the deamination products can form improper base pairs. The deamination of guanine does not pair with thymine but can still form two hydrogen bonds with cytosine, thus no change will occur. 5'-AG-3' If A is 5'-GG-3' 3'-TC-5' deaminated 3'-CC-5' 5'-AG-3' If C is 5'-AA-3' 3'-TC-5' deaminated 3'-TT-3'

Ch.8 #37 Species I is diploid (2n=8) with chromosomes AABBCCDD; related species II is diploid (2n=8) with chromosomes MMNNOOPP. What types of chromosome mutations do individuals with the following sets of chromosomes have? a) AAABBCCDD b) MMNNOOOOPP c) AABBCDD d) AAABBBCCCDDD e) AAABBCCDDD f) AABBDD g) AABBCCDDMMNNOOPP h) AABBCCDDMNOP

a) Trisomy A b) Tetrasomy O c) Monosomy C d) Triploidy e) Ditrisomy A and D f) Nullisomy C g) Allotetraploidy h) Allotriploidy

Ch.17 #2 List some important differences between bacterial and eukaryotic cells that affect the way in which the genes are regulated.

a) bacterial genes are frequently organized into operons with coordinate regulation, and genes with operons can be transcribed as on a single long mRNA. Eukaryotic genes are not organized into operons and are singly transcribed from their own promotors. b) In eukaryotic cells, nucleosome structure of the DNA is remodeled prior to transcription occurring. Essentially, the chromatin must assume a more open configuration state, allowing for access by transcription-associated factors c) Activator and repressor molecules function in both eukaryotic and bacterial cells. However, in eukaryotic cells activators appear to be more common than in bacterial cells d) In bacteria, transcription and translation can occur concurrently. In eukaryotes, the nuclear membrane separates transcription from translation both physically and temporally. This separation results in a greater diversity of regulatory mechanisms that can occur at different points during gene expression.

Ch.16 #11 For each of the following types of transcriptional control, indicate whether the protein produced by the regulator gene will synthesized initially as an active repressor or as an inactive repressor. a) negative control in a repressible operon b) negative control in an inducible operon

a) inactive repressor b) active repressor

Ch.16 #20 Give all possible genotypes of a lac operon that produces, or fails to produce, beta-galactosidase and permease under the following conditions. Do not give partial-diploid genotypes. Lactose Absent Lactose Present b-gal permease b-gal permease a. - - + + b. - - - + c. - - + - d. + + + + e. - - - - f. + - + - g. - + - +

a) lacI+ lacP+ lacO+ lacZ+ lacY+ b) lacI+ lacP+ lacO+ lacZ- lacY+ c) lacI+ lacP+ lacO+ lacZ+ lacY- d) lacI+ lacP+ lacOc lacZ+ lacY+ e) lacIs lacP+ lacO+ lacZ+ lacY+ f) lacI+ lacP+ lacOc lacZ+ lacY- g) lacI+ lacP+ lacOc lacZ- lacY+

Ch.16 #19 a) lacI+ lacP+ lacO+ lacZ+ lacY+ b) lacI- lacP+ lacO+ lacZ+ lacY+ c) lacI+ lacP+ lacOc lacZ+ lacY+ d) lacI- lacP+ lacO+ lacZ+ lacY- e) lacI- lacP- lacO+ lacZ+ lacY+ f) lacI+ lacP+ lacO+ lacZ- lacY+/lacI- lacP+ lacO+ lacZ+ lacY- g) lacI- lacP+ lacOc lacZ+ lacY+/lacI+ lacP+ lacO+ lacZ- lacY- h) lacI- lacP+ lacO+ lacZ+ lacY-/lacI+ lacP- lacO+ lacZ- lacY+ i) lacI+ lacP- lacOc lacZ- lacY+/lacI- lacP+ lacO+ lacZ+ lacY- j) lacI+ lacP+ lacO+ lacZ+ lacY+/lacI+ lacP+ lacO+ lacZ+ lacY+ k) lacIs lacP+ lacO+ lacZ+ lacY-/lacI+ lacP+ lacO+ lacZ- lacY+ l) lacIs lacP- lacO+ lacZ- lacY+/lacI+ lacP+ lacO+ lacZ+ lacY+

a) lactose present b-gal: yes permease: yes lactose absent b-gal: no permease: no b) lactose present b-gal: yes permease: yes lactose absent b-gal: yes permease: yes c) lactose present b-gal: yes permease: yes lactose absent b-gal: yes permease: yes d) lactose present b-gal: yes permease: no lactose absent b-gal: yes permease: no e) lactose present b-gal: no permease: no lactose absent b-gal: no permease: no f) lactose present b-gal: yes permease: yes lactose absent b-gal: no permease: no g) lactose present b-gal: yes permease: yes lactose absent b-gal: yes permease: yes h) lactose present b-gal: yes permease: no lactose absent b-gal: no permease: no i) lactose present b-gal: yes permease: no lactose absent b-gal: no permease: no j) lactose present b-gal: yes permease: yes lactose absent b-gal: no permease: no k) lactose present b-gal: no permease: no lactose absent b-gal: no permease: no l) lactose present b-gal: no permease: no lactose absent b-gal: no permease: no

Ch.18 #22 A polypeptide has the following amino acid sequence: Met-Ser-Pro-Arg-Leu-Glu-Gly The amino acid sequence of this polypeptide was determined in a series of mutants listed in parts a through e. For each mutant, indicate the type of mutation that occurred in the DNA (single-base substitution, insertion, or deletion) and the phenotypic effect of the mutation (nonsense, missense, frameshift, etc.) a) Mutant 1: Met-Ser-Ser-Arg-Leu-Glu-Gly b) Mutant 2: Met-Ser-Pro c) Mutant 3: Met-Ser-Pro-Asp-Trp-Arg-Asp-Lys d) Mutant 4: Met-Ser-Pro-Glu-Gly e) Mutant 5: Met-Ser-Pro-Arg-Leu-Leu-Glu-Gly

a) missense single-base substitution b) nonsense single-base substitution c) framshift deletion in the first codon of Arg (CGA --> UGA) d) deletion e) insertion

Ch.16 #23 The mmm operon, which has sequences, A, B, C, and D (which may be structural genes or regulatory sequences), encodes enzymes 1 and 2. Mutations in sequences A, B, C, or D have the following effects, where a plus sign (+) indicates that the enzyme is synthesized and a minus sign (-) indicates that the enzyme is not synthesized. Mutation mmm absent mmm present Enz 1 Enz 2 Enz 1 Enz 2 no mut. + + - - A - + - - B + + + + C + - - - D - - - - a. Is the mmm operon inducible or repressible? b. Indicate which sequence (A, B, C, or D) is part of the following components of the operon (why?): - Regulator gene - Promoter - Structural gene for enzyme 1 - Structural gene for enzyme 2

a) repressible b) regulator gene: B (when sequence B is mutated, gene expression is not repressed by the presence of mmm) promoter: D (when sequence D is mutated, no gene expression occurs either in the presence or absence of mmm) enz 1 gene: A (when sequence A is mutated, enzyme 1 is not produced) enz 2 gene: C (when sequence C is mutated, enzyme 2 is not produced)

Ch.8 #20 A chromosome has the following segments, where * represents the centromere: AB * CDEFG What types of chromosome mutations are required to change this chromosome into each of the following chromosomes? (in some cases, more than one chromosome mutation may be required) a) ABAB*CDEFG b) AB*CDEABFG c) AB*CFEDG d) A*CDEFG e) AB*CDE f) AB*EDCFG g) C*BADEFG h) AB*CFEDFEDG i) AB*CDEFCDFEG

a) tandem duplication of AB b) displaced duplication of AB c) pericentric inversion of DEF d) deletion of B e) deletion of FG f) pericentric inversion of CDE g) pericentric inversion of ABC h) pericentric inversion of DEF, tandem duplication of DEF i) displaced duplication of CDEF, pericentric inversion of EF

Use the table below to identify the kind of mutation from this original protein sequence: Met-Ser-Phe-His-Gly (Hint: Phe is coded by UUU, and each of these is a different kind of mutation): a. Met-Ser-Phe b. Met-Ser-Phe-His-Gly-Trp-.... c. Met-Ser-Phe-Glu-His-Gly d. Met-Ser-Leu-His-Gly (Leu is coded by UUA) e. Met-Ser-Phe-His-Gly (Phe is coded by UUC) f. Met-Ser-Leu-His-Gly (Leu is coded by UUG) g. Met-Ser-Leu-Ile-Asp-...

a. nonsense b. loss of function c. insertion in frame mutation d. transversion point mutation e. silent transition point mutation f. Missense transversion point mutation g. deletion with a frameshift

Ch.16 #16 Under which of the following conditions would a lac operon produce the greatest amount of b-galactosidase? The least? Explain your reasoning. Lactose Present Glucose Present Condition 1 yes no Condition 2 no yes Condition 3 yes yes Condition 4 no no

condition 1 produces the most condition 2 produces the least

Ch.18 #20 The following nucleotide sequence is found on the template strand of DNA. First, determine the amino acids of the protein encoded by this sequence by using the genetic code. Then give the altered amino acid sequence of the protein that will be found in each of the following mutations: DNA template strand 3'- TAC TGG CCG TTA GTT GAT ATA ACT -5' 1 24 a) Mutant 1: A transition at nucleotide 11 b) Mutant 2: A transition at nucleotide 13 c) Mutant 3: A one-nucleotide deletion at 7 e) Mutant 5: An addition of TGG after 6

mRNA: 5'-AUG ACC GGC AAU CAA CUA UAU UGA-3' Amino Acid Sequence: Met- Thr - Gly - Asn- Gln- Leu - Tyr - Stop a) TTA -> TCA -> AGU, so Asn -> Ser b) GTT -> ATT -> UAA, so Gln -> stop c) TAC TGG CGT TAG TTG ATA TAA CT AUG ACC GCA AUC AAC UAU AUU start-thr-ala-lle-asn-tyr-lle-N e) Met-Thr-Thr-Gly-Asn-Gln-Leu-Tyr-Stop