Genetics Quiz 1
Why is primase required for replication?
Primase is a DNA-dependent RNA polymerase. Primase synthesizes the short RNA molecules, or primers, that have a free 3'-OH to which DNA polymerase can attach deoxyribonucleotides in replication initiation. The DNA polymerases require a free 3'-OH to which they add nucleotides, and therefore they cannot themselves initiate replication. Primase does not have this requirement.
List the different proteins and enzymes taking part in bacterial replication. Give the function of each in the replication process.
(1) DNA polymerase III is the primary replication polymerase. It elongates a new nucleotide strand from the 3'-OH of the primer .(2) DNA polymerase I removes the RNA nucleotides of the primers and replaces them with DNA nucleotides. (3) DNA ligase connects Okazaki fragments by sealing nicks in the sugar phosphate (4) DNA primase synthesizes the RNA primers that provide the 3'-OH group needed for DNA polymerase III to initiate DNA synthesis. (5) DNA helicase unwinds the double helix by breaking the hydrogen bonding between the two strands at the replication fork. (6) DNA gyrase reduces DNA supercoiling and torsional strain that is created ahead of the replication fork by making double-stranded breaks in the DNA and passing another segment of the helix through the break before resealing it. Gyrase is also called topoisomerase II. (7) Initiator proteins bind to the replication origin and unwind short regions of DNA. (8) Single-stranded binding protein (SSB protein) stabilizes single-stranded DNA prior to replication by binding to it, thus preventing the DNA from pairing with complementary sequences.
What are the three basic stages of transcription? Describe what happens at each stage.
(1) Initiation: Transcription proteins assemble at the promoter to form the basal transcription apparatus and begin synthesis of RNA. (2) Elongation: RNA polymerase moves along the DNA template in a 3' to 5' direction, unwinding the DNA and synthesizing RNA in a 5' to 3' direction. (3) Termination: Synthesis of RNA is terminated, and the RNA molecule separates from the DNA template.
What are the three principal elements in mRNA sequences in bacterial cells?
(1) The 5′ untranslated region, which contains the Shine-Dalgarno sequence (2) The protein-encoding region, which begins with the start codon and ends with the stop codon (3) The 3′ untranslated region
The following sequence of nucleotides is found in a single-stranded DNA template: ATTGCCAGATCATCCCAATAGATAssume that RNA polymerase proceeds along this template from left to right. Give the sequence and label the 5' and 3' ends of the RNA copied from this template.
5′-UAACGGUCUAGUAGGGUUAUCUA-3′
Poly(A) tail
A poly(A) tail is added to the 3' end of the pre-mRNA. It affects mRNA stability.
Sense codon
A sense codon is a group of three nucleotides that code for an amino acid. In a standard genetic code there are 61 sense codons that code for the 20 amino acids commonly found in proteins.
Describe two types of alternative processing pathways. How do they lead to the production of multiple proteins from a single gene?
Alternative processing of pre-mRNA can take the form of either alternative splicing of pre-mRNA introns or the alternative cleavage of 3′ cleavage sites in a pre-mRNA molecule containing two or more cleavage sites for polyadenylation. Alternative splicing results in different exons of the pre-mRNA being ligated to form mature mRNA. Each mRNA formed by an alternative splicing process will yield a different protein. In pre-mRNA molecules with multiple 3′ cleavage sites, cleavage at the different sites will generate mRNA molecules that differ in size. Each alternatively cleaved RNA potentially could code for a different protein depending on the location of the alternative 3′ cleavage site. A single pre-mRNA transcript can undergo both alternative processing steps, thus potentially producing multiple proteins.
Initiation codon
An initiation codon establishes the appropriate reading frame and specifies the first amino acid of the protein chain. Typically, the initiation codon is AUG; however, GUG and UUG can also serve as initiation codons although rarely.
For the ovalbumin gene shown in Figure 14.3, where would the 5′ untranslated region and 3′ untranslated regions be located in the DNA and in the RNA?
Assuming that alternative splicing is not occurring then the 5′ UTR would be located in exon 1 and the 3′ UTR would be located in exon 8 for both the DNA and RNA molecules.
How do the mRNAs of bacterial cells and the pre-mRNAs of eukaryotic cells differ? How do the mature mRNAs of bacterial and eukaryotic cells differ?
Bacterial mRNA is translated immediately upon being transcribed. Eukaryotic pre- mRNA must be processed and exported from the nucleus. Bacterial mRNA and eukaryotic pre-mRNA have similarities in structure. Each has a 5′ untranslated region as well as a 3′ untranslated region. Both also have protein-coding regions. However, the protein-coding region of the pre-mRNA is disrupted by introns. The eukaryotic pre- mRNA must be processed to produce the mature mRNA. Eukaryotic mRNA has a 5′ cap and a poly(A) tail, unlike bacterial mRNAs. Bacterial mRNA also contains the Shine-Dalgarno consensus sequence. Eukaryotic mRNA does not have the equivalent.
Would you expect to find more molecules of H2A or more molecules of H3? Explain your reasoning.
Because each nucleosome contains two molecules of H2A and two molecules of H3, eukaryotic cells should have equal amounts of these two histones.
In a typical eukaryotic cell, would you expect to find more molecules of the H1 histone or more molecules of the H2A histone? Explain your reasoning.
Because each nucleosome contains two molecules of histone H2A and only one molecule of histone H1 is associated with each nucleosome, eukaryotic cells will have more H2A than H1.
What is the function of the 5′ cap?
CAP binding proteins recognize the 5′ cap and stimulate binding of the ribosome to the 5′ cap and to the mRNA molecule. The 5′ cap may also increase mRNA stability in the cytoplasm. Finally, the 5′ cap is needed for efficient splicing of the intron that is nearest the 5′ end of the pre-mRNA molecule.
Gunter Korge examined several proteins that are secreted from the salivary glands of Drosophila melanogaster during larval development (G. Korge. 1975. Proceedings of the National Academy of Sciences of the United States of America72:4550-4554). One protein, called protein fraction 4, was encoded by a gene found by deletion mapping to be located on the X chromosome at position 3C. Korge observed that, about 5 hours before the first synthesis of protein fraction 4, an expanded and puffed-out region formed on the X chromosome at position 3C. This chromosome puff disappeared before the end of the third larval instar stage, when the synthesis of protein fraction 4 ceased. He observed that there was no puff at position 3C in a special strain of flies that lacked secretion of protein fraction 4. Explain these results. What is the chromosome puff at region 3 and why does its appearance and disappearance roughly coincide with the secretion of protein fraction 4?
Chromosomal puffs correspond to relaxation of chromatin structure and transcriptional activity at the locus. The puff at region 3C indicates active transcription of that region, including the gene for protein fraction 4.
How are transcription and replication similar and how are they different?
Common characteristics of transcription and replication: (1) Utilize a DNA template (2) Synthesize molecules in a 5' to 3' direction (3) Synthesize molecules that are antiparallel and complementary to the template (4) Use nucleotide triphosphates as substrates (5) Involve complexes of proteins and enzymes necessary for catalysis Unique characteristics of transcription: (1) Unidirectional synthesis of only a single strand of nucleic acid (2) Initiation does not require a primer (3) Subject to numerous regulatory mechanisms (4) Each gene is transcribed separately Unique characteristics of replication: (1) Bidirectional synthesis of two strands of nucleic acid (2) Initiates from replication origins
Describe in steps how the double helix of DNA, which is 2 nm in width, gives rise to a chromosome that is 700 nm in width.
DNA is first packaged into nucleosomes; the nucleosomes are packed to form a 30- nm fiber. The 30-nm fiber forms a series of loops that packs to form a 250-nm fiber, which in turn coils to form a 700-nm chromatid.
A diploid human cell contains approximately 6.4 billion base pairs of DNA. How many histone proteins are complexed to this DNA?
Each nucleosome contains two of each of the following histones: H2A, H2B, H3, and H4. A nucleosome plus one molecule of histone H1 constitute the chromatosome. Therefore, nine histone protein molecules occur for every nucleosome.3.2 × 107 nucleosomes × 9 histones = 2.9 × 108 molecules of histones are complexed to 6.4 billion bp of DNA.
What similarities and differences exist in the enzymatic activities of DNA polymerases I and III? What is the function of each DNA polymerase in bacterial cells?
Each of the five DNA polymerases has a 5' to 3' polymerase activity. They differ in their exonuclease activities. DNA polymerase I has a 3' to 5' as well as a 5' to 3' exonuclease activity. DNA polymerase III has only a 3' to 5' exonuclease activity. (1) DNA polymerase I carries out proofreading. It also removes and replaces the RNA primers used to initiate DNA synthesis. (2) DNA polymerase III is the primary replication enzyme and also has a proofreading function in replication.
What are epigenetic changes and how are they brought about?
Epigenetic changes are changes in gene expression that are passed on to cells or future generations but do not involve alteration of the nucleotide sequence. Epigenetic changes are brought about by altering DNA structure, such as methylation of the DNA, or altering chromatin structure by modifying histones.
Exons
Exons are transcribed regions that are not removed in intron processing. They include the 5′ UTR, coding regions that are translated into amino acid sequences, and the 3′ UTR.
How many different mRNA sequences can encode a polypeptide chain with the amino acid sequence Met-Leu-Arg? (Be sure to include the stop codon.)
From Figure 15.10, we can determine that leucine and arginine each have six different potential codons. There are also three potential stop codons. As for methionine, only one codon, AUG, is typically found as the initiation codon. (However, UUG and GUG have been shown to serve as start codons on occasion. For this problem, we will ignore these rare cases.) Therefore, the number of potential sequences is the product of the number of different potential codons for this tripeptide, which gives us a total of (1 × 6 × 6 × 3) = 108 different mRNA sequences that can code for the tripeptide Met-Leu- Arg.
A diploid human cell contains approximately 6.4 billion base pairs of DNA. How many nucleosomes are present in such a cell? (Assume that the linker DNA encompasses 40 bp.)
Given that each nucleosome contains about 140 bp of DNA tightly associated with the core histone octamer, another 20 bp associated with histone H1, and 40 bp in the linker region, then one nucleosome occurs for every 200 bp of DNA. 6.4 × 109 bp divided by 2 × 102 bp/nucleosome = 3.2 × 107 nucleosomes (32 million).
Overlapping code
If an overlapping code were present, then a single nucleotide would be expected to be included in more than one codon. The result for a sequence of nucleotides within a single gene would be to encode more than one type of polypeptide. However, because the genetic code is nonoverlapping, codons within the same gene do not overlap. Where overlap occurs is in overlapping genes in some viruses, where the same segment of the genome encodes multiple—two or even three (in case of HIV Env gene)—peptides. In such cases, the codons of different peptides are translated in different frames in nonoverlapping fashion.
Nonoverlapping code
In a nonoverlapping code, a single nucleotide is part of only one codon. This results in the production of a single type of polypeptide from one polynucleotide sequence.
Universal code
In a universal code, each codon specifies, or codes, for the same amino acid in all organisms. The genetic code is nearly universal, but not completely. Most of the exceptions occur in mitochondrial genomes.
What is semiconservative replication?
In semiconservative replication, the original two strands of the double helix serve as templates for new strands of DNA. When replication is complete, two double- stranded DNA molecules will be present. Each will consist of one original template strand and one newly synthesized strand that is complementary to the template.
Phosphorous is required to synthesize the deoxyribonucleoside triphosphates used in DNA replication. A geneticist grows some E. coli in a medium containing nonradioactive phosphorous for many generations. A sample of the bacteria is then transferred to a medium that contains a radioactive isotope of phosphorus (32P). Samples of the bacteria are removed immediately after the transfer and after one and two rounds of replication. Assume that newly synthesized DNA contains 32P and the original DNA contains nonradioactive phosphorous. What will be the distribution of radioactivity in the DNA of the bacteria in each sample? Will radioactivity be detected in neither, one, or both strands of the DNA?
In the initial sample removed immediately after transfer, noincorporated into the DNA because replication in the medium containing P has not yet occurred. After one round of replication in the 32P-containing medium, one strand of each newly synthesized DNA molecule will contain 32P, while the other strand will contain only nonradioactive phosphorous. After two rounds of replication in the 32P-containing medium, 50% of the DNA molecules will have 32P in both strands, while the remaining 50% will contain 32P in one strand and nonradioactive phosphorous in the other strand.
How is the poly(A) tail added to pre-mRNA? What is the purpose of the poly(A) tail?
Initially, a complex consisting of several proteins forms on the 3′ UTR of the pre- mRNA molecule. Proteins necessary for cleavage and polyadenylation bind to the AAUAAA consensus sequence, which is located upstream of the 3′ cleavage site, and to a site downstream of the cleavage site. Once the complex has formed, the pre-mRNA is cleaved. Other proteins add adenine nucleotides to the 3′ end creating the poly(A) tail. The presence of the poly(A) tail increases the stability of the mRNA molecule through the interaction of proteins at the poly(A) tail. The length of the poly(A) tail influences the time in which the transcript remains intact and available for translation. The poly(A) tail also assists with the binding of the ribosome to the mRNA and nuclear export.
How is the 5′ cap added to eukaryotic pre-mRNA?
Initially, the terminal phosphate of the three 5′ phosphates linked to the end of the mRNA molecule is removed. Subsequently, a guanine nucleotide is attached to the 5′ end of the mRNA using a 5′ to 5′ phosphate linkage. Next, a methyl group is attached to position 7 of the guanine base. Ribose sugars of adjacent nucleotides may also be methylated, but at the 2′-OH.
What are some characteristics of introns?
Introns are intervening sequences that typically do not encode proteins. Eukaryotic genes commonly contain introns. However, introns are rare in bacterial genes. The number of introns found in an organism's genome is typically related to complexity— more complex organisms possess more introns.
Introns
Introns are noncoding sequences of DNA that intervene within coding regions of a gene.
What are isoaccepting tRNAs?
Isoaccepting tRNAs are tRNA molecules that have different anticodon sequences but accept the same amino acids.
How did Meselson and Stahl demonstrate that replication in E. coli takes place in a semiconservative manner?
Meselson and Stahl demonstrated that replication in E. Coli takes place in a semiconservative manner by adding 15N, then switching to 14N and observing replication. 15N only present, but after one cycle 14N appeared, then separate.
Would you expect to see more or less acetylation in regions of DNA that are sensitive to digestion by DNase I? Why?
More acetylation. Regions of DNase I sensitivity are less condensed than DNA that is not sensitive to DNase I, the sensitive DNA is less tightly associated with nucleosomes, and it is in a more open state. Open states can represent areas of euchromatin which are undergoing DNA repair and/or transcription. Such a state is associated with acetylation of lysine residues in the N-terminal histone tails. Acetylation eliminates the positive charge of the lysine residue and reduces the affinity of the histone for the negatively charged phosphates of the DNA backbone, leading to a more relaxed or looser chromatin structure.
Nonuniversal codons
Most codons are universal (or nearly universal) in that they specify the same amino acids in almost all organisms. However, there are exceptions where a codon has different meanings in different organisms. Most of the known exceptions are the termination codons, which in some organisms do code for amino acids. Occasionally, a sense codon is substituted for another sense codon.
Nonsense codon
Nonsense codons or termination codons signal the end of translation. These codons do not code for amino acids.
What are polytene chromosomes and chromosomal puffs?
Polytene chromosomes are giant chromosomes formed by repeated rounds of DNA replication without nuclear division, found in the larval salivary glands and several other tissues of Drosophila and a few other species of flies. Each polytene chromosome can contain over a thousand copies of DNA laid side by side. Certain regions of polytene chromosomes can become less condensed, resulting in localized swelling, or chromosomal puffs, because of intense transcriptional activity at the site.
Compare and contrast prokaryotic and eukaryotic chromosomes. How are they alike and how do they differ?
Prokaryotic chromosomes are usually circular, whereas eukaryotic chromosomes are linear. Prokaryotic chromosomes generally contain the entire genome, whereas each eukaryotic chromosome has only a portion of the genome: The eukaryotic genome is divided into multiple chromosomes. Prokaryotic chromosomes are generally smaller and have only a single origin of DNA replication. Eukaryotic chromosomes are often many times larger than prokaryotic chromosomes and contain multiple origins of DNA replication. Prokaryotic chromosomes are typically condensed into nucleoids, which have loops of DNA compacted into a dense body. Eukaryotic chromosomes contain DNA packaged into nucleosomes, which are further coiled and packaged into successively higher-order structures. The condensation state of eukaryotic chromosomes varies with the cell cycle.
Draw an RNA nucleotide and a DNA nucleotide, highlighting the differences. How is the structure of RNA similar to that of DNA? How is it different?
RNA and DNA are polymers of nucleotides that are held together by phosphodiester bonds. An RNA nucleotide contains a ribose sugar, whereas a DNA nucleotide contains a deoxyribose sugar. Also, the pyrimidine base uracil is found in RNA, but thymine is not. DNA, however, contains thymine but not uracil. Finally, an RNA polynucleotide is typically single stranded even though RNA molecules can pair with other complementary sequences. DNA molecules are almost always double stranded.
The following sequence of nucleotides is found in a single-stranded DNA template: ATTGCCAGATCATCCCAATAGATAssume that RNA polymerase proceeds along this template from left to right. Which end of the DNA template is 5′ and which end is 3′?
RNA is synthesized in a 5′ to 3′ direction by RNA polymerase, which reads the DNA template in a 3′ to 5′ direction. So, if the polymerase is moving from left to right on the template then the 3′ end must be on the left and the 5′ end on the right. 3′-ATTGCCAGATCATCCCAATAGAT-5′
Summarize the different types of processing that can take place in pre-mRNA.
Several modifications to pre-mRNA take place to produce mature mRNA in eukaryotes. (1) Addition of the 5′ 7-methylguanosine cap to the 5′ end of the pre-mRNA (2) Cleavage at a site downstream of the AAUAAA consensus sequence at the 3′ end of the pre-mRNA (3) Addition of the poly(A) tail to the 3′ end of the mRNA immediately following cleavage (4) Removal of the introns (splicing)
Suppose a chemist develops a new drug that neutralizes the positive charges on the tails of histone proteins. What would be the most likely effect of this new drug on chromatin structure? Would this drug have any effect on gene expression? Explain your answers.
Such a drug would disrupt the ionic interactions between the histone tails and the phosphate backbone of DNA and thereby cause a loosening of the DNA from the nucleosome. The drug may mimic the effects of histone acetylation, which neutralizes the positively charged lysine residues. Changes in chromatin structure would result from the altered nucleosome-DNA packing and possible changes in interaction with other chromatin modifying enzymes and proteins. Changes in transcription would result because DNA may be more accessible to transcription factors.
How does supercoiling arise? What is the difference between positive and negative supercoiling?
Supercoiling arises from topoisomerases catalyzing the overwinding (positive supercoiling) or underwinding (negative supercoiling) of the DNA double helix. Supercoiling may occur: (1) when the DNA molecule does not have free ends, as in circular DNA molecules, or (2) when the DNA molecule is folded into loops that are bound to proteins that prevent the ends of the DNA strands from rotating about each other, as in eukaryotic chromosomes.
What functions does supercoiling serve for the cell?
Supercoiling compacts the DNA. Negative supercoiling helps to unwind the DNA duplex for replication and transcription.
What is the significance of the fact that many synonymous codons differ only in the third nucleotide position?
Synonymous codons code for the same amino acid, or, in other words, they have the same meaning. A nucleotide at the third position of a codon pairs with a nucleotide in the first position of the anticodon. Unlike the other nucleotide positions involved in the codon-anticodon pairing, this pairing is often weak, or "wobbles," and nonstandard pairings can occur. Because the "wobble," or nonstandard base pairing with the anticodons, affects the third nucleotide position, the redundancy of codons ensures that the correct amino acid is inserted in the protein when nonstandard pairing occurs.
For the RNA molecule shown in Figure 13.1a, write out the sequence of bases on the template and nontemplate strands of DNA from which this RNA is transcribed. Label the 5′ and 3′ ends of each strand.
Template strand: 5′—GTCA—3′, Nontemplate strand: 5′—TGAC—3′
3′ untranslated region
The 3′ untranslated region is a sequence of nucleotides at the 3′ end of the mRNA that is not translated into proteins. However, it does affect the translation of the mRNA molecule as well as the stability of the mRNA.
Are the 5′ untranslated regions (5′ UTR) of eukaryotic mRNAs encoded by sequences in the promoter, exon, or intron of the gene? Explain your answer.
The 5′ UTR is located in the first exon of the gene. It is not part of the promoter because promoters for RNA polymerase II (which transcribes pre-mRNA) are not normally transcribed, and the 5′ UTR is in the mRNA. It is not located in an intron because introns are removed during processing of pre-mRNA, and the 5′ UTR is part of the mRNA.
5′ cap
The 5′ cap functions in the initiation of translation and mRNA stability.
What is the 5′ cap?
The 5′ end of eukaryotic mRNA is modified by the addition of the 5′ cap. The cap consists of an extra guanine nucleotide linked 5′ to 5′ to the mRNA molecule. This nucleotide is methylated at position 7 of the base. The ribose sugars of adjacent bases may be methylated at the 2′ -OH.
5′ untranslated region
The 5′ untranslated region lies upstream of the translation start site. The eukaryotic ribosome binds at the 5′ cap of the mRNA molecule and scans to the first methionine codon (AUG). The region 5′ of this start codon is the 5′ UTR.
AAUAAA consensus sequence
The AAUAAA consensus sequence lies near the 3′ end of the pre-mRNA. It determines the location of the 3′ cleavage and poly(A) tail addition to the pre- mRNA molecule.
The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit. The bottom strand is the template strand. Give the sequence found on the RNA molecule transcribed from this DNA and label the 5′ and 3′ ends of the RNA. 5′-ATAGGCGATGCCA-3′ 3′-TATCCGCTACGGT-5′Template strand
The RNA molecule would be complementary to the template strand, contain uracil, and be synthesized in an antiparallel fashion. The sequence would be:5′-A U A G G C G A U G C C A-3′.The RNA strand contains the same sequence as the nontemplate DNA strand except that the RNA strand contains uracil in place of thymine.
What would be the most likely effect of moving the AAUAAA consensus sequence shown in Figure 14.7 10 nucleotides upstream?
The cleavage site would also be moved 10 nucleotides upstream, likely resulting in a shorter mRNA.
Write the consensus sequence for the following set of nucleotide sequences. A GG AG TT A GC T AT T T G C A AT A A C G AA AA T C C T AA T T G C A AT T
The consensus sequence is identified by determining which nucleotide is used most frequently at each position. For the two nucleotides that occur at an equal frequency at the first position, both are listed at that position in the sequence and identified by a slash mark: T/AGAATT
How is the reading frame of a nucleotide sequence set?
The initiation codon on the mRNA sets the reading frame.
Describe the composition and structure of the nucleosome.
The nucleosome core particle contains two molecules each of histones H2A, H2B, H3, and H4, which form a protein core with 145-147 bp of DNA wound around the core. For each nucleosome, one molecule of a fifth histone, H1, binds to DNA entering and exiting the nucleosome core to clamp the DNA around the nucleosome.
A geneticist discovers that two different proteins are encoded by the same gene. One protein has 56 amino acids and the other 82 amino acids. Provide a possible explanation for how the same gene could encode both of these proteins.
The pre-mRNA molecules transcribed from the gene are likely processed by alternative processing pathways. Two possible mechanisms that could have produced the two different proteins from the same pre-mRNA are alternative splicing or multiple 3′ cleavage sites in the pre-mRNA. The cleavage of the pre-mRNA molecule at different 3′ cleavage sites would produce alternatively processed mRNA molecules that differ in size. Translation from each of the alternative mRNAs would produce proteins containing different numbers of amino acids.Alternative splicing of the pre-mRNA could produce different mature mRNAs, each containing a different number of exons and thus the mRNAs differ in size. Again, translation from each alternatively spliced mRNA would generate proteins that differ in the number of amino acids contained.
Promoter
The promoter is the DNA sequence that the transcription apparatus recognizes and binds to initiate transcription.
Reading frame
The reading frame refers to how the nucleotides in a nucleic acid molecule are grouped into codons containing three nucleotides. Each sequence of nucleotides has three possible sets of codons, or reading frames.
Termination codon
The termination codon signals the termination or end of translation and the end of the protein molecule. There are three termination codons—UAA, UAG, and UGA—which can also be referred to as stop codons or nonsense codons. These codons do not code for amino acids.
Transcription start site
The transcription start site is the location of the first transcribed nucleotide of the mRNA and is located 25 to 30 nucleotides downstream of the TATA box.
Draw a typical bacterial promoter and identify any common consensus sequences.
The typical bacterial promoter consists of the -35 (TTGACA) and -10 (TATAAT) consensus sequences. An upstream element rich in AT sequences is found only in some bacterial promoters and is located upstream of the -35 consensus sequence typically around -40 or -60.
A nontemplate strand on bacterial DNA has the following base sequence. What amino acid sequence will be encoded by this sequence?5 ́-ATGATACTAAGGCCC-3 ́
To determine the amino acid sequence, we need to know the mRNA sequence and the codons present. The nontemplate strand of the DNA has the same sequence as the mRNA, except that thymine-containing nucleotides are substituted for the uracil- containing nucleotides. So the mRNA sequence would be as follows: 5'-AUGAUACUAAGGCCC-3'.Assuming that the AUG indicates a start codon, then the amino acid sequence would be starting from the amino end of the peptide and ending with the carboxyl end: fMet-Ile- Leu-Arg-Pro.
What would be the effect on DNA replication of mutations that destroyed each of the following activities in DNA polymerase I?
a) 3' => 5' exonuclease activity is important for proofreading newly synthesized DNA. If the activity is nonfunctional, then the fidelity of replication by DNA polymerase I will decrease, resulting in more misincorporated bases in the DNA. b) 5' => 3' exonuclease activity would result in the RNA primers used to initiate replication not being removed by DNA polymerase I. c) 5' => 3' polymerase activity DNA polymerase I would be unable to synthesize new DNA strands if the 5' => 3' polymerase activity was destroyed. RNA primers could be removed by DNA polymerase I using the 5' => 3' exonuclease activity, but new DNA sequences could not be added in their place by DNA polymerase I.
Referring to the genetic code presented in Figure 15.10, give the amino acids specified by the following bacterial mRNA sequences. a. 5 ́-AUGUUUAAAUUUAAAUUUUGA-3 ́ b. 5 ́AGGGAAAUCAGAUGUAUAUAUAUA UAUGA-3 ́ c.5 ́-UUUGGAUUGAGUGAAACGAUG GAUGAAAGAUUUCUCGCUUGA-3 d. 5 ́-GUACUAAGGAGGUUGUAUGG GUUAGGGGACAUCAUUUUGA-3 ́
a) Amino fMet-Phe-Lys-Phe-Lys-Phe Carboxyl b) Amino Arg-Glu-Ile-Arg-Cys-Ile-Tyr-Ile-Tyr-(Asp or Glu)----carboxylFor the last potential readable codon only 2 bases are shown and could potentially code for either Asp (GAU or GAC) or Glu (GAA or GAG) c) Amino Phe-Gly-Leu-Ser-Glu-Thr-Met-Asp-Glu-Arg-Phe-Leu-Ala Carboxyl d) Val—Leu-Arg- Arg-Leu-Tyr-Gly-Leu-Gly-Asp-Ile-Ile-Leu-(Ile, Met, Thr, Asn, Lys, Ser, or Arg)----Carboxyl For the last potential readable codon only 1base is shown and could potentially be part of a codon that encodes for Ile (AUU, AUG, or AUA), Met (AUG), Thr (ACU, ACC, ACA, or ACG), Asn (AAU OR AAC), Lys (AAA or AAG), Ser (AGU or AGC), or Arg (AGA or AGG). Each of the mRNA sequences begins with the three nucleotides AUG. This indicates the start point for translation and allows for a reading frame to be set. In bacteria, the AUG initiation codon codes for N-formylmethionine. Also, for each of these mRNA sequences, a stop codon is present either at the end of the sequence or within the interior of the sequence. The amino terminal refers to the end of the protein with a free amino group and will be the first peptide in the chain. The carboxyl terminal refers to the end of the protein with a free carboxyl group and is the last amino acid in the chain. For the following peptide chains reading from left to right, the first amino acid is located at the amino end, while the last amino acid is located at the carboxyl end.
Which of the following amino acid changes could result from a mutation that changed a single base? For each change that could result from the alteration of a single base, determine which position of the codon (first, second, or third nucleotide) in the mRNA must be altered for the change to result a) leu to gln b) phe to ser c) phe to ile d) pro to ala e) asn to lys f) ile to asn
a) Of the six codons that encode for Leu, only two could be mutated by the alteration of a single base to produce the codons for Gln: CUA (Leu)—Change the second position to A to produce CAA (Gln). CUG (Leu)—Change the second position to A to produce CAG (Gln). b) Both Phe codons (UUU and UUC) could be mutated at the second position to produce Ser codons: UUU (Phe)—Change the second position to C to produce UCU (Ser). UUC (Phe)—Change the second postion to C to produce UCC (Ser). c) Both Phe codons (UUU and UUC) could be mutated at the first position to produce Ile codons: UUU (Phe)—Change the first position to A to produce AUU (Ile). UUC (Phe)—Change the first position to A to produce AUC (Ile). d) All four codons for Pro can be mutated at the first position to produce Ala codons: CCU (Pro)—Change the first position to G to produce GCU (Ala). CCC (Pro)—Change the first position to G to produce GCC (Ala). CCA (Pro)—Change the first position to G to produce GCA (Ala). CCG (Pro)—Change the first position to G to produce GCG (Ala). e) Both codons for Asn can be mutated at a single position to produce Lys codons: AAU (Asn)—Change the third position to A to produce AAA (Lys). AAU (Asn)—Change the third position to G to produce AAG (Lys). AAC (Asn)—Change the third postion to A to produce AAA (Lys). AAC (Asn)—Change the third position to G to produce AAG (Lys). f) Only two of the three Ile codons can be mutated at a single position to produce Asn codons: AUU (Ile)—Change the second position to A to produce AAU (Asn). AUC (Ile)—Change the second position to A to produce AAC (Asn).
Based on the DNA sensitivity to DNase I illustrated in Figure 11.7, which type of chicken hemoglobin (embryonic or adult) is likely produced in highest quantity at the following tissues and developmental stages?
a. Erythroblasts during the first 24 hours None—neither the embryonic nor adult hemoglobin genes show DNase I sensitivity in erythroblasts at this time. b. Erythroblasts at day 5 Embryonic—the embryonic hemoglobin gene, but not the adult genes, shows DNase I sensitivity, indicating an open chromatin conformation that is conducive for transcription. c. Erythroblasts at day 14 Adult—now the adult hemoglobin genes show DNase I sensitivity, but the embryonic gene is DNase I insensitive. d. Brain cells throughout development None—neither embryonic nor adult hemoglobin genes show DNase I sensitivity in brain cells at any time of development.
