Genetics Exam 2

A double-stranded DNA molecule with the sequence shown here produces a polypeptide that is five amino acids long (not including the stop codon), beginning with a starting methionine and ending with a stop codon. TACATGATCATTTCACGGAATTTCTAGCATGTA ATGTACTAGTAAAGTGCCTTAAAGATCGTACAT A) Which strand of DNA is transcribed and in which direction? (2.5 points) B) Label the 5' and 3' end of each strand. (2.5 points) C) Provide the mRNA sequence. (2.5 points) D) Provide the amino acid sequence of the polypeptide produced.

5' TACATGATCATTTCACGGAATTTCTAGCATGTA 3' template 3' ATGTACTAGTAAAGTGCCTTAAAGATCGTACAT 5' non-template Top strand serves as the template with transcription moving from right to left. mRNA 5' AUG CUA GAA AUU CCG UGA peptide N- Met Leu Glu Ile Pro Stop -C

The DNA polymerases are positioned over the following DNA segment (which is part of a much larger molecule) and moving from right to left. If we assume that an Okazaki fragment is made from this segment, what will be the fragment's sequence? Label its 5′ and 3′ ends. 5′... CCTTAAGACTAACTACTTACTGGGATC....3′ 3′... GGAATTCTGATTGATGAATGACCCTAG....5′

5′ . . . .CCTTAAGACTAACTACTTACTGGGATC. . . . 3′

The DNA sense (coding) strand for a particular amino acid is 5′-CAT-3′. What RNA sequence would be transcribed for this codon, what tRNA anticodon would recognize it, and what amino acid would be added in response to this codon?

5′-CAU-3′ (RNA sense); 3′-GUA-5′ (tRNA anticodon); histidine

How many nucleotides would be expected for a gene coding for a protein with 300 amino acids?

900 300 amino acids require 300 codons. A codon is 3 base pairs, therefore 900 nucleotides are required.

A certain nonsense suppressor corrects a nongrowing mutant to a state that is near, but not exactly, wild type (it has abnormal growth). Suggest a possible reason why the reversion is not a full correction.

A nonsense suppressor is a mutation in a tRNA such that its anticodon can base-pair with a stop codon. In this way, a mutant stop codon (nonsense mutation) can be read through and the polypeptide can be fully synthesized. However, the mutant tRNA may be for an amino acid that was not encoded in that position in the original gene. For example, the codon UCG (serine) is instead UAG in the nonsense mutant. The suppressor mutation could be in tRNA for tryptophan such that its anticodon now recognizes UAG instead of UGG. During translation in the double mutant, the machinery puts tryptophan into the location of the mutant stop codon. This allows translation to continue but does place tryptophan into a position that was serine in the wild-type gene. This may create a protein that is not as active and a cell that is "not exactly wild type." Another explanation is that translation of the mutant gene is not as efficient and that premature termination still occurs some of the time. This would lead to less product and, again, a state that is "not exactly wild type."

Which statement below best describes the situation between nucleotides on opposite strands in a DNA molecule?

A-T bonding between opposite strands involves two hydrogen bonds, whereas G-C bonding between strands involves three hydrogen bonds

If you extract the DNA of the coliphage φX174, you will find that its composition is 25 percent A, 33 percent T, 24 percent G, and 18 percent C. Does this composition make sense in regard to Chargaff's rules? How would you interpret this result? How might such a phage replicate its DNA?

Chargaff's rules are that A = T and G = C. Because these equalities are not observed, the most likely interpretation is that the DNA is single stranded. The phage would first have to synthesize a complementary strand before it could begin to make multiple copies of itself.

The protein encoded by the cystic fibrosis gene is 1480 amino acids long, yet the gene spans 250 kb. How is this difference possible?

Conservatively, the amount of DNA necessary to encode this protein of 445 amino acids is 445 3 = 1335 base pairs. When compared with the actual amount of DNA used, 60 kb, the gene appears to be roughly 45 times larger than necessary. This "extra" DNA mostly represents the introns that must be correctly spliced out of the primary transcript during RNA processing for correct translation. (There are also comparatively very small amounts of both 5´ and 3´ untranslated regions of the final mRNA that are necessary for correct translation encoded by this 60-kb of DNA.)

Pyrimidines

Cytosine and Thymine

Gel electrophoresis can be used to separate DNA or RNA molecules based on their:

Gel electrophoresis separates molecules exclusively based on their size and not charge because the charge per unit weight is constant (longer DNA has proportionately greater charge). This means all molecules have the same acceleration. The faster migration of smaller fragments results not from greater acceleration but lesser retardation by the gel matrix.

If the GC content of a DNA molecule is 48 percent, what are the percentages of the four bases (A, T, G, and C) in this molecule?

If the DNA is double stranded, G = C = 24% and A = T = 26%.

We know that DNA and RNA (each with four nucleotide components) both use a three nucleotide genetic code and 64 codons (43 = 64). Knowing that the minimum number of codons for the genetic code is 21 (20 amino acids, 1 stop codon), what codon sizewould be required if only three nucleotides were present in the genome?

If the genome consisted of only three different nucleotides, 2 nucleotides per codon will result in 32=9 different codons or amino acids, which is insufficient to encode 20 amino acids. 3 nucleotides per codon will result in 33=27 different codons, which are sufficient to encode 20 amino acids and one stop codon.

Which of the following statements about PCR is/are FALSE?

It is incorrect to say that the amount of DNA increases in a linear manner since the amount doubles every cycle (grow exponentially).

If after 6 cycles of PCR, we have 64 picomoles of amplified DNA, then the starting amount of template DNA must be:

Let x be the starting amount of DNA template. Since the amount doubles every cycle, after 6 cycles we will have 26x amount of DNA. Since we have 64 picomoles after cycles 26x = 64 picomoles 64x = 64 picomoles Therefore, x=1 picomole.

A mutant has no activity for the enzyme isocitrate lyase. Does this result prove that the mutation is in the gene encoding isocitrate lyase?

No. The enzyme may require posttranslational modification to be active. Mutations in the enzymes required for these modifications would not map to the isocitrate lyase gene.

In Northern blotting, electrophoresis is used to resolve which biological molecules? What type of probe is used to identify the target molecule(s)?

Northern blotting is used to resolve RNA (Ribonucleic Acid) molecules. Northern blotting is done to determine quantity and identity of a RNA molecule (mRNA). Radioactive or non isotopically labeled DNA/RNA/oligonucleotide molecule, with complementary base pairs to the target sequence (if known). Probes are of variable length depending on the type of mRNA investigated.

In what cellular compartment are introns removed from pre-mRNA to make mature mRNA?

Nucleus

A molecule of composition 5′-AAAAAAAAAAA-3′ 3′-TTTTTTTTTTTTT-5′ is replicated in a solution containing unlabeled (not radioactive) GTP, CTP, and TTP plus adenine nucleoside triphosphate with all its phosphorus atoms in the form of the radioactive isotope 32P. Will both daughter molecules be radioactive? Explain. Then repeat the question for the molecule 5′-ATATATATATATAT-3′ 3′-TATATATATATATA-5′

Only the DNA molecule that used the poly-T strand as a template would be radioactive. The other daughter molecule would not be radioactive because it would not have required any dATP for its replication. Because each strand of the second molecule contains T, both daughter molecules would require dATP for replication, so each would be radioactive.

Why is it necessary to use a special DNA polymerase ( Taq polymerase) in PCR?

PCR requires a special DNA polymerase, such as the Taq polymerase, because PCR needs the denatured DNA to be replicated at a high temprature (75-80 degree Celcius) and at a high speed, and both of the conditions are met by Taq polymerase.

Below is the structure of a parental strand with a primer attached. By numbering the carbons on the ribose sugars, determine which end of the parent is 5' and which is 3'.

Position 1 is 5' and position 8 is 3' on the parental strand.

The "RNA World" theory suggests that:

RNA was the genetic material in the first living cells

In a chromosome, which of the following is TRUE?

RNAs of different genes can be transcribed off either DNA strand, but always 5' to 3'

Tazswana has one normal sized b-globin mRNA. Explain why she has severe symptoms and from whom she inherited the intron mutation (you may wish to refer to the pedigree below).

She has severe symptoms because she has two bad alleles, one that produces mRNA that is abnormally spliced and one that produces a protein with an amino acid substitution (a different amino acid than normal) She inherited the intron splicing mutation from her dad.

If the DNA template 5′-ATGCATGC-3′ were transcribed to RNA, the RNA would read:

Since this is the template strand, the RNA will be the reverse complement: Template: 5′-ATGCATGC-3′ RNA: 3'-UACGUACG-5' Or writing it 5'-to-3': 5'-GCAUGCAU-3'

In the Sanger sequencing method, the use of dideoxy adenosine triphosphate stops nucleotide polymerization opposite:

T's in the template strand since adenine base pairs with thymine and dideoxy nucleotides lack the 3' OH and prevent further elongation of the chain.

The functional equivalent of the TATA binding protein in prokaryotes is:

TATA binding protein finds the TATA box, that is, eukaryotic promoter. In prokaryotes, the promoter is located by sigma subunit of the RNA polymerase holoenzyme complex

Formation of the peptide bond could be described as:

The -OH group on the carboxyl end of one amino acid and a hydrogen on the amino end of the other amino acid are removed during the formation of the peptide bond. The reaction therefore releases water and is a condensation or dehydration reaction. Peptide bonds do not form spontaneously and require the activity of the ribosome and therefore are catalyzed by the ribosome.

If the coding strand has sequence 5′-ATGCATGC-3′ then RNA sequence would be:

The RNA has the same sequence and orientation as the coding strand, except the T's are replaced with U's. The RNA will be: 5'-AUGCAUGC-3'.

Assume the b-globin mRNA in Lane 3 is from Tazswana's grandmother. How is it different from the normal mRNA shown in Lane 4? Discuss the cause of this difference in terms of the mutation and why Tazswana's grandmother is symptom free.

The band in lane 3 is less intense than the band in lane 4 because grandma produces half as much mRNA as a person with two normal alleles (only one of her two alleles is expressed). Grandma is symptom free because one normal allele produces enough protein to have a normal phenotype

The autoradiogram shown was generated by the dideoxy sequencing method. What is the sequence of the first 5 bases of the template strand in the 5'-to-3' direction. Note that the DNA fragments migrate top to bottom in the gel.

The bands correspond to the bases in the daughter strand. The sequence of the daughter strand in the 3'-to-5' direction can be read top to bottom: 3'-ATCTA-5'. Reverse complementing to get the sequence of the template strand, we obtain 5'- TAGAT-3'.

Which of the following is/are role(s) of the 5′ cap of eukaryotic mRNAs?

The cap acts as a binding site for the ribosome and protects the RNA from degradation

the following segment of DNA, which is part of a much longer molecule constituting a chromosome: 5'....ATTCGTACGATCGACTGACTGACAGTC....3' 3'....TAAGCATGCTAGCTGACTGACTGTCAG....5' If the replication fork moves from right to left, a. which will be the template for the leading strand?

The top strand will be leading: 5'....ATTCGTACGATCGACTGACTGACAGTC....3'

A researcher was mutating prokaryotic cells by inserting segments of DNA. In this way, she made the following mutation: Original TTGACAT 15 to 17 bp TATAAT Mutant TATAAT 15 to 17 bp TTGACAT What does this sequence represent? What do you predict will be the effect of such a mutation? Explain.

a. The original sequence represents the -35 and -10 consensus sequences (with the correct number of intervening spaces) of a bacterial promoter. The σ factor, as part of the RNA polymerase holoenzyme, recognizes and binds to these sequences. b. The mutated (transposed) sequences will not be a binding site for the σ factor. The orientation of the two regions with respect to each other is not correct; therefore, they will not be recognized as a promoter.

A certain Drosophila protein-encoding gene has one intron. If a large sample of null alleles of this gene is examined, will any of the mutant sites be expected a. in the exons? b. in the intron? c.in the promoter? d. in the intron-exon boundary?

a. Yes. The exons encode the protein, so null mutations would be expected to map within exons. b. Possibly. There are sequences near the boundaries of and within introns that are necessary for correct splicing. If these are altered by mutation, correct splicing will be disrupted. Although transcribed, it is likely that translation will not occur. c. Yes. If the promoter is deleted or altered such that GTFs cannot bind, transcription will be disrupted. d. Yes. There are sequences near the boundaries of and within introns that are necessary for correct splicing.

The intermediate temperature cycle (72°C) in the polymerase chain reaction enables:

activity of the thermostable DNA polymerase enzyme.

Initiation of replication occurs at an "origin of replication" site that typically includes an AT-rich region. Initiation benefits from these AT-rich regions because:

adenine-thymine pairs are held together by two H-bonds, making them easier to separate during unwinding

The "wobble" base is less important than the other two nucleotides in a codon and is found:

at the 5′ end of the tRNA anticodon.

The anticodon on the tRNA molecule:

binds to the mRNA in a complementary fashion.

Which of the following is NOT true of a cDNA clone?

cDNA clones lack exons, as introns are spliced together before cloning. The statement is incorrect because cDNA is reverse transcribed from mature mRNA, which includes exons and excludes introns.

The cDNA for a eukaryotic gene B is 900 nucleotide pairs long. A cDNA clone is used to isolate a genomic clone of gene B, and the gene is sequenced. From start to stop codon, the gene is found to be 1800 nucleotide pairs long. The most probable reason for the discrepancy is that:

cDNA does not contain introns since it is made from mature mRNA. The genomic DNA will include the introns and this means 900 bp of the genomic DNA must be introns.

What type of library would be most valuable in the isolation of promoter sequences found in front of a gene's transcriptional start site?

cDNA libraries are made from RNA and therefore lack promoter sequences. In order to find the promoter sequence, then, one has to use genomic libraries.

Transgenic plants can be generated using T-DNA plasmid carrying a gene of interest. To get the DNA into the plant cells, the researchers:

co-cultivate bacteria with T-DNA and plant cells, resulting in DNA transfer.

Embryonic stem (ES) cells have the unique ability to:

combine with other stem cells to generate a chimeric mouse pup.

Which of the following is not a key property of hereditary material? a. It must be capable of being copied accurately. b. It must encode the information necessary to form proteins and complex structures. c. It must occasionally mutate. d. It must be able to adapt itself to each of the body's tissues.

d. It must be able to adapt itself to each of the body's tissues

In 1959, Arthur Kornberg isolated DNA polymerase activity from Escherichia colicells, revealing its ability to copy DNA by mixing ___________________ in a tube and measuring synthesis of a DNA strand.

dATP, dGTP, dTTP, dCTP, template DNA, and pure DNA polymerase enzyme

Perhaps the most common use of plasmid vectors is to:

enable the amplification of cloned DNA within a bacterial host cell.

For protein-encoding genes, six general transcription factors (GTFs) function to:

identify a gene's promoter, facilitating RNA polymerase II binding

Examine Figure 7-16, recalling that DNA synthesis by DNA polymerases always occurs in the 5′to 3′direction. The predominant location of small Okazaki fragments during DNA replication occurs at the:

lagging strand of replication

A linear DNA molecule has n target sites for restriction enzyme EcoRI. How many fragments will be produced after complete digestion?

n+1

Of the three key building blocks of DNA, which type(s) of building block stabilizes the structure via weak van der Waals interactions (through stacking) and regular hydrogen bonds?

nitrogenous bases

When replicating the end of a chromosome, the lagging strand cannot copy the last ~10 nucleotides at the end of the chromosome. As a result, chromosomes contain telomere sequences at their ends, which are defined as:

noncoding, repetitive sequences that can be copied independent of the replisome

RNA synthesis is always 5' to 3' because:

nucleotides can only be added to an available 3'-OH group on the transcript terminus

During the initiation of prokaryotic translation, a defect in IF3 function would generate:

premature association of the 30S and 50S subunits. Initiation factor 3 (IF3) prevents premature assembly of the ribosome. Therefore, if IF3 is defective, it will lead to the premature association of the 30S and 50S subunits.

Cell-cycle progress enables the initiation of genome replication by

providing Cdc6 and Cdt1, which aid assembly of initiation components at the origin of replication

The -10 and the -35 boxes found in bacterial gene promoters function to:

recruit and orient RNA polymerase at a gene's transcription start site

The role of most microRNAs within a eukaryotic cell is to

repress the expression of genes by destroying mRNAs

In bacteria, the Shine-Dalgarno sequence is found on the mRNA and is recognized by the ________________________ to reveal __________________________.

the 30S subunit; the translation START codon

The complexity of lagging strand replication is necessary because

the polymerase elongates in a direction opposite to the replication fork

Assuming that the cellular splicing machinery will "cut" at the GU 5' splice site and the very next 3' splice site, CAG, predict how Tazswana's pre-mRNA will be processed into mRNA. That is, write down the sequence of the abnormal mRNA given the wildtype sequence in panel A below. (Note: "N" stands for any nucleotide.)

the resulting mRNA would read 5' AGNNNNG-3' because the two mutations create two new splice sites at CAG/NNN and NNN/GU.

If Escherichia coli, grown for a period of time in 15N, is transferred to 14N for one generation of DNA replication, the resulting DNA should have 14N added to all "new" DNA. If conservative replication is occurring, the 14N-containing "new" DNA will compose:

1 chromosome consisting of both parental (15N) strands and 1 chromosome consisting of both daughter (14N) strands.

E. coli is widely used in laboratories to produce proteins from other organisms. A.) You have isolated a yeast gene that encodes a metabolic enzyme and want to produce this enzyme in E. coli. You suspect that the yeast promoter will not work in E. coli. Why? B.) After replacing the yeast promoter with an E. coli promoter, you are pleased to detect RNA from the yeast gene but are confused because the RNA is almost twice the length of the mRNA from this gene isolated from yeast. Explain why this result might have occurred.

A. Eukaryotic promoters bind to several different transcription factors, which attract the RNA polymerase complex. Prokaryotic promoters bind to sigma factors, which attract the RNA polymerase complex. The sites for these factors differ, so the eukaryotic promoter would not attract the prokaryotic sigma factors. B. If it magically does transcribe the gene, it still won't work. Eukaryotic genes contain introns, 'junk' DNA excised during eukaryotic transcription. Prokaryotic genes DON'T have introns, so they transcribe the 'good' DNA with the 'junk' DNA.

In vitro (in a test tube) translation of a synthetic RNA with the repeating sequence (AGA)20 could produce which of the following?

AGAAGAAGAAGAAGA... Could be read as: A. AGA AGA AGA ... (polypeptide containing only arginines) B. GAA GAA GAA ... (polypeptide containing only glutamic acid) C. AAG AAG AAG ... (polypeptide containing only lysine) Depending on the reading frame. Therefore, depending on the placement of the start codon and the reading frame chosen, any one of the polypeptides could be produced.

purine

Adenine and Guanine

The following data were obtained from three organisms : an RNA virus, a DNA virus, and a wombat (an Australian mammal). Which is likely the sample from the RNA virus? Sample Adenine Cytosine Guanine Thymine Uracil (a) 28.0 22.0 22.0 0.0 28.0 (b) 21.0 29.0 29.0 21.0 0.0 (c) 27.0 24.0 26.0 23.0 0.0

DNA does not contain Uracil, while RNA does. Sample (a) is the only one with Uracil and is therefore the correct choice. All organisms have dsDNA which obeys Chagraff's rules that A=T, C=G, and A+G=C+T (amount of purines is the same as the amount of pyrimidines). Therefore sample (b) is the wombat. DNA viruses can be single stranded and don't have to obey Chagraff's rules. Therefore sample (c) is the DNA virus.

. You have two DNA molecules: A) 5' --------------------------------------GGATCC-3' 3'---------------------------------------CCTAGG-5' and B) 5'-AGATCT----------------------------------------------3' 3'-TCTAGA----------------------------------------------5' Suppose you digest DNA "A" with BamH1: 5'-G | GATCC-3' 3'-CCTAG | G-5' and DNA "B" with BglII: 5'-A | GATCT-3' 3'-TCTAG | A-5' ( | indicates where the phosphodiester bond is cleaved). Can these two DNA pieces be ligated together? Can they be separated again by digesting with either BamH1 or BglII?

Digesting DNA "A" with BamHI, we get: 5' --------------------------------------G 3'---------------------------------------CCTAG Digesting DNA "B" with BglII, we get: GATCT----------------------------------------------3' A----------------------------------------------5' If we mix these two DNAs together, the GATC overhang ("sticky end") in "B" will base pair with the CTAG overhang in "A". Adding ligase will join the two molecules with phosphodiester bonds and give: 5' --------------------------------------GGATCT-----------------------------3' 3'---------------------------------------CCTAGA-----------------------------5' Therefore the two molecules can be ligated together. The sequence GGATCT does not correspond to the restriction site of either BamHI or BglII and therefore neither enzyme will be able to separate the two molecules (assuming of course that the molecules don't have BamHI or BglII elsewhere.

A radioactive probe is generated using the actin gene from yeast. This probe is used in a Southern blot analysis of EcoRI digested genomic DNA from the ciliated protozoan Tetrahymena thermophila. The autoradiogram shows a single-labeled band of 4 kb in size. This means that the Tetrahymenaactin gene is:

Digesting genomic DNA with a restriction enzyme such as EcoRI results in hundreds of thousands of fragments of varying size. A single band in the Southern blot implies that there is a unique restriction fragment containing the sequence of the actin gene. This means that actin is present in one copy in the genome. If there were multiple copies the gene, multiple restriction fragments would have hybridized (bound) the actin probe and resulted in multiple bands.

Deduce what the six wild-type codons may have been in the mutants that led Brenner to infer the nature of the UAG codon.

The codon is UAG which is a stop (Amber) codon as it doesn't code for any amino acid. To identify six codons, we need 6 amino acids which will be there in the mutant protein. Following amino acids need to be inserted to continue wild-type chain & their codons: lysine => AAA, AAG glutamine => CAA, CAG glutamate => GAA, GAG tyrosine => UAU, UAC tryptophan => UGG SERINE => AGU, AGC, UCU, UCC, UCA, UCG (In each case, the the Underlined codon will require a single base change to give UAG.)

Why was cDNA and not genomic DNA used in the commercial cloning of the human insulin gene?

The commercial cloning of insulin was into bacteria. Bacteria are not capable of processing introns. Genomic DNA would include the introns, while cDNA is a copy of processed (and thus intron-free) mRNA.

Tazswana's mRNA sample (extracted from her reticulocytes) has been run in Lane 2. a. Compare Lane 2 to Lane 4. In terms of pre-mRNA processing, explain why Tazswana has an extra band. Include drawings of the mRNAs shown in Lane 4 versus those in Lane 2 (symbolize two exons as rectangles, and the one intron as a single line).

The extra band on the gel in lane 2 is due to abnormal splicing of mRNA from one allele, which produces a longer mRNA molecule than normal (i.e, all or part of an intron has been retained). Hence, the mRNA migrates slower through the gel.

Describe two possible mechanisms by which the size(s) of her mRNA bands could affect the size and amino acid content of her b-globin protein. In each case, will her b-globin protein be functional or not?

There are several possible consequences. The longer mRNA might code for more amino acids and produce a protein that is longer than normal. However, the retained intron would likely shift the reading frame. This would result in a new set of amino acids that are not consistent with the ß-globin protein sequence. It is also possible that a stop codon could be introduced because of this frameshift and therefore result in a smaller than normal ß-globin protein. Regardless, due to the intron sequence the protein will not likely be functional.

The following structure was identified at the 3′ end of a transcribed RNA in prokaryotes. Choose the option that best describes this structure and its function

This is a hairpin loop that is involved in the termination of prokaryotic transcription

Degeneracy in the genetic code is best illustrated by:

Threonine Degeneracy of the genetic code means that many amino acids are encoded by multiple codons. Both methionine and tryptophan are encoded by a single codon, AUG and UGG respectively, while threonine is encoded by 4 codons: ACU, ACC, ACA, and ACG. Therefore, threonine best illustrates degeneracy.

2. Shown below is the structure of a Drosophilagene, divided into 10 segments, designated A-J. The gene contains three exons, two introns, a promoter, and a site in I for poly(A) addition. What segment or segments of the gene will be found in the completely processed transcription

Transcription begins downstream of the promoter, so the promoter will not be part of the transcript. The introns will be removed. Segment I is the poly(A) signal and will be part of the transcript. The transcript is terminated ~20bp downstream of the poly(A) signal and therefore J will not be part of the transcript. The transcript will contain: D,F,H,I.