Final Exam Practice
1. Assuming this sequence is from E. coli, in what region would you find the TATA-box? 2. Assuming this is from E. coli, in what regionwould you find the Shine-Dalgarno sequence? 3. In what region would you find the start codon?
1. Region 1 Hint: This question is testing your knowledge of what the TATA-box does. You should know it is a critical part of the promoter sequence. So where would the promoter sequence be? Feedback: The +1 of transcription is given, so that is where the RNA transcript begins to be formed. The promoter region has to be upstream of the +1. The TATA box is a sequence of T's and A's, found only in region 1. 2. Region 3 Hint: This question is testing your knowledge of what the S-D sequence does. Feedback: The SD sequence will be transcribed onto the mRNA for translation (the small subunit of the ribosome recognizes and binds to it in conjunction with the AUG start codon). The +1 of transcription is given, so that is where the RNA transcript begins to be formed. The SD has to be downstream of this. You were asked to memorize the E. coliSD consensus sequence (look this up if you have to), so just look for it, or something close to it, in regions 3 or 4. You'll find it in region 3 3. Region 4. Hint: If you located the SD in the above question, then you know the start codon (ATG in the DNA) shouldbe about 6-10 bases downstreamfor the ribosome to bind.
The sequence shown is part of an open reading frame (coding sequence). The tRNA anticodon sequence 5' to 3', for the first RNA codon in the open reading frame is:
5' CAU Hint: Antiparallel Feedback: Nucleic acids ALWAYS bind in an anti-parallel fashion, so in this case, the 5' end is the right end of each tRNA; 3' is on the left. Label the tRNA 5'and 3' ends. The start codon is 5' AUG, so the anticodon is 5' CAU.
A mutation that changes a T to a G in DNA, upon the next two rounds of DNA replication (without correction) will result in what kind of mutation? A) A transversion mutation B) A deletion mutation C) An C to a T substitution mutation D) A transition mutation
A Draw this out. It begins as a T:A basepair. What will be the pair become if the T becomes G?
Tyrosine is coded for by just two codons, UAU and UAC. The best explanation as to how this degeneracy occurs is: A) Wobble in the 5'position of the anticodon allows a single tRNA to bind to both codons B) Wobble in the 3'position of the anticodon allows a single tRNA to bind to both codons C) Tyrosine is carried by more than one tRNA D) Two different tRNAs are used
A Only one tRNA gene is needed.
If 35% of the bases in a region of the mouse genome are cytosine, what percentage in that region are adenine? A) 15% B) 30% C) 35% D) 20%
A Remember your base pairing rules: A pairs with T and C pairs with G.
The human BRCA-1 gene codes for a protein involved in DNA repair. Females who inherit one mutant BRCA-1 allele have a very high probability of developing breast and/or ovarian cancer, despite having one functional BRCA-1 allele. Thus, the mutant BRCA-1 appears as an autosomal dominant trait in heterozygotes. This is because: A) There is a strong likelihood that a mutation will eventually occur in the wild-type allele B) The mutant gene allows excessive proliferation of cells C) It induces mutations in other genes D) More than one of the answer options are correct
A This is an inherited situation very similar to inherited retinoblastoma.
During translation, after a peptide bond forms, what will happen next? A) The tRNA carrying the polypeptide will shift to the peptidyl (P) site B) The uncharged tRNA that transferred the polypeptide will shift to the peptidyl (P) site C) The uncharged tRNA that transferred the polypeptide will shift to the aminoacyl (A) site D) The tRNA carrying the polypeptide will shift to the aminoacyl (A) site
A This question is checking your knowledge on several aspects of translation. Do you know what a peptide bond is? Do you know what the different functional sites of the ribosome (E, P, A) are and what they do? Think about which direction the ribosome is travelling on the mRNA.
a) Draw all the elements you would expect to see in an operon encoding2 genes. b) Draw the resulting mRNA with all the needed elements. Assume this is in E. coli. c) If a frameshift mutation occurs near the 5' end of the coding region (ORF) of gene 1, will it affect gene 2/ORF 2? Explain why/why not? d) Why don't eukaryotes make polycistronic mRNA? Explain in molecular terms.
A/B in picture C)No it will not affect gene 2 because both genes are independently translated from the mRNA. They are not translated as a single combined protein D) Eukaryotes cannot read more than one ORF from a single mRNA because of the dependence for ribosome binding at the 5' cap. The small subunit binds at the cap and then scans the mRNA until it finds the first AUG codon and translation begins and ends at the first in-frame stop codon. Then the ribosome falls off. If there is a second ORF, the ribosome will never get to it.
What are eukaryotic ribosomes composed of? A) Protein and mRNA B) Protein and 18S rRNA C) Sigma factor and rRNA D) Protein and 16S rRNA
B All ribosomes are complexes of protein and RNA.
The function of aminoacyl-tRNA synthetases is to: A) Match tRNA anticodons and mRNA codons at the ribosome B) Attach amino acids to tRNAs C) Synthesize tRNA molecules D) Form the peptide bond between amino acids at the ribosome
B Charged tRNA
In the Meselson-Stahl experiment, E. coli cells that had been grown for many generations on 15N medium were transferred to 14N medium for two generations. The DNA was then extracted, and the resulting cesium chloride density gradient results showed: A) Two bands, one of 14N DNA and one of 15N DNA B) Two bands, one uniformly labeled with 14N DNA and one representing a mix of 14N and 15N DNA C) A single band of uniformly labeled 14N DNA D) A single band, representing a mix of 14N and 15N DNA
B DNA replication is semi-conservative.
A change of a codon from CGT to CGA, coding for the same amino acid, is best described as a ___________________ mutation. A) Transition B) Silent C) Nonsense D) Substitution
B Look at the 3rd nucleotide position. Would this likely code for the same or different amino acid?
What is found at the 3' OH end of a charged tRNA molecule? A) A codon B) An amino acid C) An anticodon D) A ribosome
B Protein building
With every round of DNA replication in most human somatic cells, what happens to telomeres? A) They are be repaired by telomerase B) They are progressively shortened with every round of DNA replication C) They are more subject to damage D) They are reduced by telomerase
B The key words here are MOST SOMATIC cells. Check your notes on this.
DNA synthesis always happens in the 5' to 3' direction because: A) Only short nucleic acid fragments can be made in the replication bubble B) DNA polymerase can only add nucleotides to a free 3' OH C) DNA polymerase can only move in one direction D) RNA primers are needed
B Think about what polymerases require.
Which of the following statements is correct regarding gene mutations in a coding sequence? A) In general, a deletion mutation near the 5' end is least likely to result in a mutant phenotype B) In general, a substitution mutation near the 3' end is least likely to result in a mutant phenotype C) In general, an insertion mutation near the 5' end is least likely to result in a mutant phenotype D) In general, a deletion mutation near the 3' end is least likely to result in a mutant phenotype
B Watch the context of this question; it states IN GENERAL, so don't over-think it and get distracted with exceptions.
Which of the following merodiploids is inducible with the addition of lactose? A) I+ P+ O+ Z- Y+ A+/ I+ P- O+ Z+ Y+ A+ B) I+ P+ O+ Z- Y+ A+/ I- P+ O+ Z+ Y- A+ C) I- P+ OC Z- Y+ A+/ I- P+ O+ Z+ Y+ A+ D) I- P+ OC Z+ Y+ A+/ I+ P+ O+ Z+ Y+ A
B What does "inducible" mean?
Alternative splicing is accomplished through the removal of: A) Exons from the mRNA molecule B) Introns from the primary transcript C) Introns from the gene D) Introns from the mRNA molecule
B What molecule does this occur in?
In DNA replication, Okazaki fragments are formed by: A) Discontinuous synthesis on only one replicating strand B) Discontinuous synthesis on both replicating strands C) DNA polymerase D) RNA primer formation
B You must consider both replication forks of the replication bubble.
Promoters in genes are: A) The start of translation signal in genes B) Transcribed by RNA polymerase but not translated C) Recognized by RNA polymerase but not transcribed D) Transcribed and translated
C Assume the question is referring to the promoter of RNA pol itself and not the stuff inbetween. Are promoters present in mRNA?
A single crossover event occurs within gene C between the chromatids below. The resulting chromatids will be: A B C D and A C B D A) A C D and A B B D B) A B C D and A C B D C) A B C B D and A C D D) A B B D and A D
C Draw the two chromatids out with the two "C" genes lined up, one on top of the other.
In polycistronic mRNA, a single base deletion in the open reading frame of gene A occurs. Will this affect the downstream gene B? If so, why? A) The ribosome binding site for gene B will begin in the wrong place B) The promoter of B is now out-of-frame C) Gene B will not be affected D) The reading frame of B is changed and incorrect amino acids will be added
C Each gene has its own start and stop codon.
The following nitrogenous base is: A) A pyrimidine nucleoside B) A purine nucleoside C) A pyrimidine D) A purine
C Hint: Belongs to the nucleotide C
The process in which cancer cells cause disruption of local tissue and invasion of distant tissues is called: A) Angiogenesis B) Contact inhibition C) Metastasis D) Autocrine stimulation
C Movement
The mode of DNA replication was proven by: A) Watson and Crick B) Avery C) Meselson and Stahl D) Griffith
C Semiconsevative, conservative, or dispersive.
The cap structure of eukaryotic mRNA consists of: A) A ribose sugar, a guanine base, a phosphate group B) A ribose sugar, a methylated guanine base, a phosphate group C) A ribose sugar, a methylated guanine base, three phosphate groups D) A ribose sugar, a guanine base, a methyl group
C The cap is a modified purine.
Which statement best applies to the phenomenon of genetic drift? A) The effects of genetic drift are greatest in populations that are neither very small nor very large B) The smaller the population size, the lesser the effect of drift C) The smaller the population size, the greater the effects of drift D) The effects of genetic drift are greatest in either very small or very large populations
C The question is testing to see if you know what the definition of genetic drift is and how it can impact populations.
In DNA replication: A) Okazaki fragments are formed on both continuous and discontinuous strands B) RNA primers are synthesized for discontinuous strands only C) RNA primers are synthesized for both continuous and discontinuous strands D) Okazaki fragments are formed on only one of the two replicating strands of DNA
C Think carefully. EVERY initiation of a DNA strand must begin with an RNA primer. That is a RULE for ALL cellular DNA polymerases - they can't synthesize DNA from nothing.
The repair mechanism that humans can use to remove/correct thymine dimers is: A) Activation of p53 B) DNA polymerase proof-reading C) Nucleotide excision repair D) Mismatch repair
C You need to know what a thymine dimer is. Next, you need to be really clear on which of the 5 repair mechanisms discussed in lecture can fix these.
The lacZ gene is often included in plasmids used for cloning in order to: A) To restrict cell growth to clearly see the colonies B) Prevent the growth of cells not carrying vector C) Prevent the growth of cells not carrying insert D) Distinguish cells carrying insert from those without
D Not a selectable marker.
In prokaryotes, Rho-independent transcription termination depends on: A) A secondary structure formed in the DNA template B) A protein that destabilizes the RNA polymerase that is transcribing the gene C) A protein factor that binds to the RNA that is being transcribed D) A secondary structure formed in the RNA that is being transcribed
D Rho is a protein.
Translation in eukaryotes requires: A) 16S rRNA B) The promoter sequence on the mRNA C) The Shine-Dalgarno sequence D) 5' cap structure on the mRNA
D Some of these elements are in prokaryotes only.
A male fruit fly that is hemizygous for a recessive allele specifying double wings is mated with a heterozygous female fly with wild-type wings. What phenotypes would be expected in the resultant offspring? A) All females would have wild-type wings and males would have double wings B) All females would have double wings and males would have wild-type wings C) All flies would have wild-type wings D) There would be wild-type and double wing flies of both sexes
D The term hemizygous (look it up!) for the description of the male tells you the allele for double wings is on the X chromosome.
Individuals having one extra chromosome in addition to the normal diploid complement are best termed: A) Tetraploid B) Aneuploid C) Triploid D) Trisomic
D This question is testing to see if you can correctly apply the definitions for aneuploid versus euploid and triploid versus trisomic.
The reason prokaryotes can use polycistronic mRNA is because: A) They carry out simultaneous transcription and translation B) Their ribosomes bind at the 5' cap C) It allows better control of gene expression D) Their ribosomes can recognize internal ribosome binding sequences
D You need to know what polycistronic mRNA is. Next, you need to know how prokaryotes initiate translation.
What binds the Shine-Dalgarno consensus sequence? Be specific. A) The 18S rRNA of the ribosome B) The sigma factor C) The tRNA of the ribosome D) The 16S rRNA of the ribosome
D prokaryotes
Using a diagram, trace the general steps required for transcription in eukaryotes beginning at the level of chromatin. Also include how transcription can be enhanced or repressed
DNA wound around histones to form chromatin Addition of acetyl groups to histones dissociates the DNA from the histones (both DNA and acetyl groups are negatively charged) Gene is now exposed-basal factors (TBP, TAFs and RNA polymerase II) associate with TATA box of the core promoter. This will result in a low level of transcription-if activator proteins are present, they will dimerize in order to bind to the enhancer sequence through their DNA-binding domain The activation domain of the activator protein can now bind the basal complex to increase transcription Reducing this high level of transcription involves a repressor protein This protein can turn down transcription in two ways: Either by competition with the activator protein for the enhancer element OR by quenching. Quenching can occur by the repressor binding directly to the activator's DNA binding domain or by binding to its activation domain. When this happens, transcription returns to the basal level Gene silencing occurs when DNA of control regions become methylated at adjacent CpG dinucleotides (the C is methylated). This makes the DNA bind so tightly to the histones that it cannot be dissociated and the gene is nevertranscribed.
Describe the PCR procedure for amplification of a 500 base pair sequence of interest.
First you will need to design two short primers at each end of the 500 bp fragment. You must know the sequence of the region to which you are designing the primers. The primers must be designed to hybridize to opposite strands. Add the template DNA (e.g., chromosomal DNA), the primers, Taqpolymerase, and all four nucleotides. Begin the cycling procedure, which consists of denaturation at high temperature (to separate the template strands), annealing at a lower temp (to allow the primers to bind to the templates), then raising the temp for synthesis (to allow the Taqpolymerase to synthesize new strands using the primers and template). Repeat this cycle about 30 times and you will have about a billion copies of your 500 bp fragment. Taqpolymerase is used because it can withstand high temperature, unlike regular cellular DNA polymerase
The following diagram represents a PCR reaction after the synthesis step. Which of thefollowing correctly corresponds to numbers 1, 2, 3 and 4, respectively?
Forward primer, reverse primer, 3'OH, 5'P Hint: ALWAYS orient your strands 5' and 3' when you are given questions involving DNA. To do this, use your rule of only the 3' ends are extended by polymerase. Feedback: #3 is a 3' end and the line above #4 is a 3'end. Now use your anti-parallel rule. #1has to be the forward primer (meaning it is use to extend the DNA into the amplified sequence in the 5' to 3' direction). That leaves #2 as the reverse primer
What is the sequence transcribed from the DNA coding strand information shown below? 5' C T G G C G T T A A T T A A G C T A*A* G T CT G C T A A C G C T A..... 3' Bold = +1
H The question identifies the sequence as the coding strand. What does that mean?F This question is testing to see if you know what the term coding strand means. The sequence of the coding strand (shown above) looks just like the sequence on the mRNA, except T is U. The mRNA sequence begins at the +1 of transcription, identified in the question.
The proteome (the collection of cellular proteins) is usually much larger than the genome. Give reasons why
Hint: Splicing Feedback: Post-translational modifications of proteins can result in different proteins produced from the same gene. (Make sure you can identify some examples of post-translational modifications). Alternative splicingevents also allow for production of more than one protein from a single gene. (Make sure you know what alternative and trans-splicing is).
Using a diagram, explain how translation of the mRNA 5' .....AUGCUCUACGGUUAG...3' (only the coding sequences are shown) would be carried out by the ribosome.
In prokaryotes, the small subunit binds to the Shine-Dalgarno and AUG start codon area, and is directed by complementary base-pairing with the ribosome's 16S rRNA. In eukaryotes, the small subunit binds the region of the 5'cap and is directed by the 18S rRNA Next, in prokaryotes, an N-formyl methionine-charged tRNA binds to the AUG codon. When both the tRNA and small subunit are in place, then the large subunit binds such thatthe tRNA is placed in the peptidyl (P) site of the ribosome. Translation then begins In eukaryotes, the small subunit scans along the 5' untranslated region until an AUG codonis found with its bound tRNA carrying unmodified methionine. Then the large subunit binds with the tRNA in the P site, and translation begins. Binding between the tRNA and its codon is governed by recognition of the anticodon on the tRNA for the codon on the mRNA. (always pay attention to the 5' and 3' orientations) -From this point on, translation in prokaryotes and eukaryotes is similar -The next charged tRNA binds to the aminoacyl site by codon/anti-codon base pairing. Peptidyl transferase activity of the large subunit forms a peptide bond between COOH of the P-positioned amino acid and the NH2from the A-position amino acid- In doing so, it transfers the first amino acid onto the second amino acid (in the A site). The first tRNA in the P site now dissociates from the ribosome and the tRNA from the A site carrying the growing polypeptide chain, now moves into the P site This allows for the next charged tRNA to enter the A site. The cycle continues until a nonsense codon (stop) is reached at which point a release factor enters the A site and the entire complex dissociates, releasing the polypeptide. The amino acid sequence should be met leu tyr gly (given in the amino (N) terminal to carboxyl(C) terminal direction)
A frame-shift mutation is caused by:
Insertion or deletion of one or two basepairs, or multiples of these in a protein coding sequence (open readingframe) Hint: This question is confirming that you know what a frame-shift mutation is. Feedback:This question is confirming that you know a frame-shift mutation can only occur in an open reading frame (coding sequence for amino acids). That rules out the rest of the choices.Remember insertions or deletions of 3 directly consecutive basepairs will not shift the reading frame.
Describe the experiment that provided proof for semiconservative DNA replication. Who were the researches who did it?
Meselson and Stahl.
f the vector diagrammed below was used for cloning of a DNA fragment, what would the phenotype (in terms of colour and antibiotic resistance) of the resulting transformed bacterial cells carrying the recombinant plasmid be if the restriction endonuclease site PstI was used forcloning and the cells were plated onto media containing ampicillin and X-gal.
The cells would be ampicillin sensitive sowould not grow on this media
From the result of a sequencing reaction shown below, determine the sequence and the polarity of the template strand.
The newly synthesized strands are represented on the gel. Since the fragments at the top of the gel are longest, they would be closest to the 3' end of sequence. The fragments at the bottom of the gel represent the shortest and therefore are closer the beginning of the sequence (5' end). The sequence from top to bottom therefore reads 3' CACAGTCTGA 5'. Since this sequence represents the newly synthesized strand, then the template would be the opposite (complementary sequence). The template therefore reads 5' GTGTCAGACT 3'.
What does it mean when a population is said to be in Hardy-Weinberg equilibrium? What are the assumptions this law is based upon?
The population is not evolving (no change in alleles or their frequencies). Mathematically these populations can be described as p2+ 2pq+ q2= 1.It is based on 5 assumptions: the population is infinitely large; individualsmate at random (no influence of the gene in question on choice of mate); mutations do not occur; no immigration or emigration; individuals' ability to reproduce (survival) is not related to the gene in question
Describe the characteristics of a cancerous cell.
They exhibit uncontrolled growth because of autocrine stimulation and/or lack of apoptosis and/or lack of cell-to-cell communication and/or loss of contact inhibition.They carry multiple mutations in several genes (frequently p53, RB, telomerase, and those encoding cell growth factors). This is often evident by looking at the karyotype. Translocations, duplications, aneuploidy etc., are commonly found in cancerous cells.They exhibit metastasis, meaning they can dissociate from surrounding cells and be relocated somewhere else in the body.They can encourage angiogenesis that is necessary for growth of the mass.
Describe how you could use colony hybridization to locate a cloned gene of interest from a genetic library
You would use colony hybridization to screen a DNA library for a cloned fragment of interest(either a cDNA or genomic library).Colony hybridization involves replica plating to transferDNA from the colonies ona plate (usually from a library) onto nitrocellulose filters. The DNA is rendered single-stranded and permanently bound to the filter by NaOHtreatment. Next, a single-stranded andlabeled DNA probe is added to the filter and incubated with buffer. The probe will find its matching sequence and hybridization will occur. Next, the hybridized filter is washed and exposed to X-ray film. This will produce a black spot where the colony containing the gene lies on the master plate.This way, the precise colony containing the cloned gene of interestcan be retrieved
What experiments were done to show: a) DNA is the molecule of heredity in cells, not protein or carbohydrates or RNA b) DNA replication is semiconservative c) Adaptation is a result of selection of pre-existing mutations, and not the result of mutations occurring because of selective pressure
a) Avery'stransformation experimentsusing heat-killed S. pneumoniaeextracts treated with various enzymes; Hershey and Chase's Waring blender experiment with phage and E. coli b) Meselson-Stahl experiment using 14N and 15N labeled DNA and CsCl centrifugation gradients c) Luria-Delbruck fluctuation or 'jackpot' experiment using E. coliand phage; replica plating onto a selective media MAKE SURE YOU CAN EXPLAIN THESE EXPERIMENTS AND WHY THEY ARE PROOFS
List the possible genotypes of single-mutant E. coli cells demonstrating the following phenotypes for β-galactosidase production: a) constitutive b) repressive (non-inducible) c) inducible
a) I-P+O+Z+Y+A+or I+P+OcZ+Y+A+(or I+P+O-Z+Y+A+) b) IsP+O+Z+Y+A+or I+P-O+Z+Y+A+or I+P+O+Z-Y+A+or I+P+O+Z+Y-A+ c) must be wildtype for all genes/elements
The following is the nucleotide sequence of a non-coding strand (template) of E. coli DNA. 3' CGTCCTCCGCAGTCGTACGTCTCCAGCGGAGATCTTTTCCGGTCGCAACTGATTGATC5' The strand codes for a small peptide. a) Which end is the 3' end and which is the 5' end of the strand shown? b) What is the sequence of the coding DNA strand (make sure to orient it)? c) Identify the Shine-Dalgarno sequence, the start and stop codons. d) What is the amino acid sequence of the peptide?You'll need a genetic code chart for this (in text or lecture notes)
a)Hint: It is probably easiest to draw the complementary strand first so you can look for the presence of start sequences and the Shine-Dalgarno to determine orientation. Once you have this, then you can answer the other questions. b) See above sequence in red c) These are underlined above d) You'll need a genetic code chart for this (in text or lecture notes) Met Gln Arg Ser Pro Leu Glu Lys Ala Ser Val Asp
a) Describe the DNA damage that is done by radiation (X-rays and UV light). b) Explain how each can result in mutations.
a)X-rays destroy the phosphodiester bonds of DNA, producing strand breaks. UV light causes thymine dimers to form, but not strand breaks. The dimers produce deformations (kinks) in the conformation of the DNA. b) The cell will attempt to repair X-ray induced strand breaks by ligation. Ifchromosomes arebrokenin one place, the pieces can be ligated onto other chromosomes.This can result in a reciprocal or non-reciprocal translocation. The broken pieces can also be ligated back together in the same chromosome, but may be ligated in reverse orientation, producing an inversion mutation. If chromosomes are broken in two places, the piece can be ligated in the original orientation, or upside-down and backward, resulting in an inversion mutation. Deletion mutations can results if the broken piece(s)of DNA is not ligated.Thymine dimers on one strand are excised by an exinuclease and then DNA polymerase repairs the removed area, using the intact strand as template. This process is known as nucleotide excision repair. When the DNA polymerase copies the template properly, no mutations are produced. However, if excision repair is employed frequently because of UV exposure, then the chance of DNA polymerase making a mistake, increases and mutations will result. This is why excessive exposure to UV rays (e.g., sun tanning) can lead to skin cancer.In bacteria, photolyase can also repair the T-T dimer directly, by rearranging the bonds back to norma. This system is only active in the presence of UV light.
If the vector diagrammed below was used for cloning of a DNA fragment, what would the phenotype (in terms of colour and antibiotic resistance) of the resulting transformed bacterial cells carrying the recombinant plasmid be if the restriction endonuclease site EcoRI was usedfor cloning and the cells were plated onto media containing ampicillin and X-gal.
blue, ampicillin resistant
Name 2 tumor-supressor genes that are involved in cell cycle control. Describe the functionofeach and how homozygous mutations in these genes can lead to benign tumors and evencancer
p53andpRB.p53codes for the p53 protein. p53 is produced when there has been damage to the cell's DNA. In the presence of p53, DNA replication is prevented (progression through S phase is prevented). p53 is a transcription factor for p21. p21 inactivates cyclin-dependent kinase (CDK) and this prevents entry into S phase. If the damage in the DNA cannot be repaired within a short time because it is too great, then the increasein concentration of p53 will induce genes needed for apoptosis (cell death). This ensures that any cells that have damaged DNA cannot reproduce. Homozygous mutations in both p53alleles mean that p53 is not functional and so any cell that undergoes DNA damage isreplicated as S phase is not prevented and the cell cannot die. This provides an opportunity for further mutations (e.g., mutations in pRB) to accumulate and the cell may loose all ability to control cell cycle. When this happens, then uncontrolled cell replication occurs, and a tumor develops. If the right combination of further mutations occurs in one of the tumor cells, then it gives rise to cancer. 12pRBencodes a protein product called pRB. In the presence of pRB, entry into S phase is prevented. As long as pRB is present, the cell cycle is controlled, only replicating when it receives the correct signals including release of pRB from the genesneeded for DNA replication. If homozygous mutations occur in both copies ofpRB, then the cell cannotstop S phase and cell division occurs continually. This allows unregulated growth and a tumor results (e.g., retinoblastoma). As mutations are more likely to occur in a large population rapidly replicating cells, it is often just a matter of time before other genes involved in cancer (e.g., p53, telomerase) acquire random mutations, and cancer develops from the tumor.
PCR amplification combined with DNA probe hybridizationwas used to determine thegenotypes of 3 family members with regard to sickle-cell anemia. The results are shownbelow. Each pair of dots represents the result from one person. The upper row represents hybridization using the probecontaining the mutant sequence and the lower rowrepresentsthe results of hybridization using the probe containing the wild-type sequence.What are thephenotypesof the 3 individuals (1 to 3)? What are the genotypes?
sickle-cell anemia, homozygous mutant 2. Normal*, heterozygous 3. Normal, homozygous wildtype•heterozygous individuals can show some symptoms of sickle cell under conditions of stress (e.g., low oxygen levels)
f the vector diagrammed below was used for cloning of a DNA fragment, what would the phenotype (in terms of colour and antibiotic resistance) of the resulting transformed bacterial cells carrying the recombinant plasmid be if the restriction endonuclease site SalI was used forcloning and the cells were plated onto media containing ampicillin and X-gal
white, ampicillin resistant
