Genetics Exam #3
Define the term "core promoter." List the components needed in a test tube in order to transcribe non-chromatin eukaryotic DNA.
"Core promoter" refers to the minimal set of DNA sequences required to recruit RNAPol II and initiate transcription of a gene when in a non-chromatin structure. The components needed are the "core promoter," general transcription factors, and RNAPol II.
Describe one method a cell can use to repair a massive number of mutations in DNA.
- "SOS" repair which isn't always correct, but when a cell knows that there's too much damage, it does this so that it can quickly repair the mutations enough that it won't kill the cell. - Non-homologous end joining - Synthesis-dependent strand annealing - Homologous recombination
If nondisjunction occurs in meiosis I of a mother, what proportion of her gametes could have been fertilized by a normal sperm to produce a child with Turner syndrome?
A) 1/2
Given the DNA template sequence 3' - TCC ATA TAC TAG GGT - 5', which of these sequences represents a missense mutation? A) 3' - TCG ATA TAC TAG GGA - 5' B) 3' - TCC ATA TAC TAA GGC - 5' C) 3' - TCT ATG TAC TAT GGT - 5' D) 3' - CCA TAT ACT AGG GTT - 5' E) 3' - TCT ATT TAC TAT GGC - 5'
A) 3' - TCG ATA TAC TAG GGA - 5'
Given the DNA coding sequence 5' - TAC AAA ATA CAG CGG - 3' A. 5' - TAG AAA ATA CAG CGG - 3' B. 5' - TAC AAA TAC AGC GGG - 3' C. 5' - TAC AAG ATA CAG CGG - 3' D. 5' - TAC AAA ATA CAC CGG - 3' E. 5' - TAC AAA ATA CAG AGG - 3'
A. 5' TAG AAA ATA CAG CGG - 3'
Which of the following events could result in a frameshift mutation? A. Base deletion B. Tautomeric shift C. Incorporation of base analogs D. Point mutation
A. Base deletion A base deletion would shorten the DNA sequence and change the reading frame of the mRNA.
Reverse genetics
Changing a gene sequence, then analyzing the resulting phenotype
ChIP-Seq
Chromatin Immunopreciptiation in combination with Next Generation sequencing to identify proteins associated with specific DNA sequences.
Which of the following is FALSE of Down syndrome?
Down syndrome is the most common triploidy (3n) in which individuals have three sets of chromosomes.
3. List the targets of activation domains of DNA-binding transcription activator proteins.
Histone code "writers" (HATs and certain HMTs), TFIID, Mediator, Chromatin Remodeling Complexes
Which type of vector (Plamid, lambda phage, cosmid, BAC, YAC) would be ideal for cloning up to 15 kilobases DNA fragments?
Lambda phage
If you want to prevent chromatin loops from being anchored to the chromosome scaffold, which regions of the DNA would you target?
MARS (matrix attachment regions)
During mismatch repair, how is the parent strand of DNA recognized by recognition enzymes?
Methylation of the adenines of the parent strand.
Explain why "Dolly" the sheep - the first mammal ever cloned is not a transgenic animal.
She did not carry a foreign gene, instead had the same DNA as the original sheep
T or F. The end result of meiosis is four haploid daughter cells.
T
List at least three different ways in which the gut micro biome benefits its host(refer to the assigned reading to learn more).
The microbes in our gut benefit us in many ways, including: • Helping digest and obtain calories and nutrients from the food we eat of food particles that enter intestine) • Influencing gut cell development • Regulating immune system homeostasis or maturation • Synthesis of some key nutrients (vitamins K and B12) • Protecting against pathogens (pathogen exclusion)
Assume that an organism exists in which crossing over does not occur, but that all other processes associated with meiosis occur normally. Consider how the absence of crossing over would affect the outcome of meiosis. If crossing over did not occur, which of the following statements about meiosis would be true? Select all that apply -The four daughter cells produced in meiosis II would all be different. -Independent assortment of chromosomes would not occur. -The daughter cells of meiosis I would be diploid, but the daughter cells of meiosis II would be haploid. -The two sister chromatids of each replicated chromosome would no longer be identical. -The two daughter cells produced in meiosis I would be identical. -There would be less genetic variation among gametes.
There would be less genetic variation among gametes.
What is the "sticky ends" created by restriction enzyme cut.
When the strand is cut there are overlapping ends which are complementary to each other
Which of the following lists the stages of Prophase I of Meiosis correctly? a. Leptotene, zygotene, pachytene, diplotene, diakinesis b. Leptotene, zygotene, diplotene, pachytene, diakensis c. Zygotene, Leptotene, diplotene, pachytene, diakinesis d. Zygotene, diplotene, leptotene, diakinesis, pachytene e. Diakinesis, diplotene, pachytene, leptotene, zygotene
a. Leptotene, zygotene, pachytene, diplotene, diakinesis
For each of the following genes, characterize them as either a tumor suppressor gene (TS) or an oncogene (O).
a. Rb(TS) b. Ras(O) c. p21 (TS) d. Cdkinhibitor(TS) e. p53(TS)
[Matching] Mutation that results in a complete loss of protein function a. amorphic mutations b. dominant negative mutations c. Neomorphic mutations d. Hypermorphic mutations e. Hypomorphic mutations
a. amorphic mutations
DNA polymerase is very accurate and rarely makes a mistake in DNA replication. Occasionally, however, an error in replication, known as a ___, is introduced. There are two general categories of point mutations—frameshift mutations (also called ___ or ____) and base ____ mutations
point mutation frameshifts= base-pair insertions or base-pair deletions substitution
What kind of DNA lesion does UV energy cause?
pyrimidine dimerisation
Explainwhythe16SrRNAgeneisagoodchoicewhenusingametagenomic approach to characterize a microbiome. Feel free to illustrate your answer with a diagram.
• The 16S rRNA gene is present and well conserved in all bacteria, and so is a good choice as a general diagnostic gene for monitoring which taxa are present in a sample. • Regions of the 16S rRNA gene are highly conserved (e.g. the 3' and 5' ends), so that universal primers can be used to amplify the gene from essentially all bacteria (see slide from 4/26/17 lecture). • The 16S rRNA gene also contains variable regions, that are evolving at a relatively quick rate that can be used to distinguish different taxa from each other. • Additionally, since the 16S DNA sequence is known for many well-characterized (e.g. reference) organisms, comparing new sequences to known sequences can tell us a lot about the taxonomy of an organism, how closely related it may be to known, characterized bacteria.
Transgenic fusions between a gene and a reporter are used to determine whether a gene is expressed. LacZ, luciferase, and GFP are reporter genes. Describe how a reporter gene can be used to help researchers to study the transcription activity or cellular localization of the gene of interest in living specimens.
- A reporter gene can be used to replace a GOI or inserted between the last coding sequence and the stop codon by homologous recombination. This will allow control of the reporter gene expression by the regulatory DNA sequence of the GOI, or create a fusion protein between the GOI and reporter. The reporters can give out specific signals, ex. * LacZ - produces blue color if incubated with X-Gal * Luciferase - gives out luminescence in the presence of luciferin * GFP - emits green fluorescence under the UV - The signals allow us to identify the expression activity of GOI in a living cell or tissue/organ, as well as the cellular localization of the GOI if expressed as a fusion protein
Transgenic fusions between a gene and a reporter are used to determine whether a gene is expressed. LacZ, luciferase, and GFP are reporter genes. Describe how these three reporter genes allow the researchers to study of interest in living specimens?
- A reporter gene encodes a protein what is easily detectable or produces a detectable product - The coding sequence of your gene of interest can be replaced by a reporter gene (transcriotional fusion) or fused to the reporter gene (translational fusion), so that the reporter gene is controlled by the regulatory sequence of your gene of interest - The LacZ gene encodes b-glactosidase, which cut the X-gal to a blue product; the luciferase catalyzes the interaction of luciferin and ATP to emit light; whereas the GFP protein emit green fluorescence under UV or a lower energy light
How do epigenetic modifications at DNA and histone levels assist in predicting whether a gene is active or inactive in a cell?
- DNA methylation - suppresses gene transcription. DNA demethylation produces the opposite effect - Histone methylation - usually inactivates gene expression, e.g., H3K27 and H3K9 methylation, but K3K4 methylation activates gene expression - Histone acetylation - activates gene expression. Histone deacetylation produces the opposite effect - Chromatin remodelers reposition nucleosomes to either activate or repress gene transcription
In forward genetics, you perform backcrossing after mutagenesis. At which generation can the mutant phenotypes be identified from a dominant mutation? At which generation can the mutant phenotypes be identified from a recessive mutation?
- Dominant - F1 - Recessive - F3
During which stage of an organism's life is genetic imprinting reset? Can paternal and maternal imprinting be kept in both the somatic cells and germ cells of the offspring, why?
- During meiosis/gametogenesis - Imprinting is kept in the somatic cells, however, in germ cells, imprinting is rewritten so that the genes are imprinted either paternally or maternally, depending on the sex of the offspring, regardless of the parental origin of the imprinted genes
A trans-acting factor is one that binds DNA and affects transcription. What are some examples of this?
- Enhancers - Repressor proteins - Silencers - Transcription factors
Describe how homogenous recombination is used to specifically knockout a gene in a reverse genetics study.
- Homogenous recombination uses a donor DNA construct with two pieces of sequences that are identical to the target gene and flanking a positive selection marker. A negative selection marker is usually attached to the end of the homogenous sequence. Once introduced into the target cells the DNA recombination events happen between the homogenous sequences of the donor DNA and the cell chromosome, thus the target gene is knocked-out, and the positive selection marker gene is transferred into the target chromosome - The positively targeted cells can be identified through antibiotic selection, whereas the random non-specific recombination event will be eliminated by the activity of the negative selection marker.
Describe 2 technologies that a researcher can use in reverse genetics to knockout or knockdown a gene expression.
- Knockdown a gene expression - Homologous recombination - RNAi
Which cloning vectors use E.coli as a host?
- Lambda phage - Cosmid - BAC
During which stage of an organism's life is genetic imprinting reset? Can paternal and maternal imprinting be kept in both the somatic cells and germ cells of the offspring?
- Meiosis/Gametogenesis - In somatic yes, but in germ cells it depends on the gender and doesn't matter which parent it came from
Compare the processes, advantages, and disadvantages of microinjection and nuclear transfer for the creation of transgenic organisms.
- Microinjection is the injection of foreign DNA segment into the pronuclei of a fertilized embryo, while nuclear transfer is to replace the nucleus of an oocyte with a diploid cell nucleus, in this case a transgenic cell nucleus - Microinjection is simpler process and produces a low success rate for transgenic offspring, < 1-5%, with heterogeneous genotypes due to random foreign DNA insertion. On the other hand, generation of transgenic organisms using nuclear transfer is a more complicated process involves oocyte enucleation and generation of transgenic cells in vitro. However, it produces 100% of transgenic offspring with the same genotype
Compare and contrast the processes of microinjection and nuclear transfer in the creation of transgenic organisms.
- Microinjection is the injection of foreign DNA segment into the pronuclei of the embryo, while nuclear transfer is to replace the nucleus of an oocyte with a diploid cell nucleus, in this case a transgenic cell nucleus - Microinjection produces a low success rate for transgenic offspring, <1-5%, whereas nuclear transfer produces 100% of transgenic offspring
Describe what is meant by a "misfolding disease" and give an example. Why do you think these diseases exhibit late age of onset?
- Proteins not folded correctly, are not active, functional - Alzheimers - Misfolded proteins accumulate in brain and cause signs and symptoms of disease
What are the 2 chemical structures found in RNA but not DNA?
- Uracil instead of thymine - Ribose instead of deoxyribose
You have performed a mutagenesis screening in a population of organism to study a gene function. Describe how you would identify phenotypes from dominant mutation. Describe how you would identify phenotypes from recessive mutation.
- You would backcross with a wild type to identify both dominant and recessive mutations - Dominant mutation can be studied at F1 - Recessive mutation can be studied at F3
What is the role of cis-acting and trans-acting factors in transcription? Explain how these factors play a roe in the lac operon, and discuss how interactions between genes on chromosomes can change the phenotype of the bacterial cell. (Structural genes vs regulatory regions).
- cis acting factors control/regulate transcriotion as part of the DNA. In the lac operon examples would be CAP-cAMP binding site, promoter region and the operator site. These are regulatory genes, not transcriber - trans acting factors control/regulate transcription by binding regulatory regions or cis acting factors.
OutlinethethreemainstepsofeukaryoticmRNAprocessing.
1) 5'mRNAcapping(7meGppp) 2) poly A tail addition and cleavage (3' end processing) 3) Splicing of introns
Prokaryotes and eukaryotes show differences in posttranscriptional modifications of mRNA. What are these differences, and why do eukaryotes show more modifications than prokaryotes?
1. 5' cap - modified guanine added to 5' end of pre-mRNA 2. 3' poly-A-trail - string of adenines added to 3' end of pre-mRNA 3. Introns removed, exons spliced together - Eukaryotes have more modifications in mRNA because mRNA must last longer in the cell, 5' cap + poly-A-tail help to ensure mature mRNA leaves the nucleus and is oriented correctly on the ribosome and alternative exon splicing increases the number of possible proteins made from a limited number genes.
List 2 of the 4 types of mutagens.
1. Chemical 2. Biological 3. Radiation
Several steps in eukaryotic transcription are driven by the energy of ribonucleotide tri- phosphate hydrolysis. Describe two of these steps.
1. Chromatin remodeling by ATPase activity of chromatin remodeling complexes 2. Opening of transcription bubble by TFIIH helicase 3. Phosphorylation of CTD tail of PolII large subunit to promote initiation of transcription and to facilitate downstream chromatin manipulations and RNA processing events 4. Phosphodiester bond formation during RNA synthesis drives polymerization/elongation process.
What are the common types of epigenetic modifications at histone and DNA levels? Describe how these epigenetic modifications can assist in predicting whether a gene is active or inactive in a cell.
1. DNA methylation - suppresses gene transcription, DNA demethylation produces the opposite effect 2. Histone methylation, usually inactivates gene expression, exceptions like H3K4 methylation 3. Histone acetylation, activates gene expression, deacetylation produces the opposite effect 4. Chromatin remodelers reposition nucleosomes to activate or repress gene transcription
What are the 3 steps of PCR?
1. Denature 2. Anneal 3. Extend
Describe the major steps involved in positional cloning of a gene.
1. Genetic Mapping 2. Establishing a contig 3. Identify the gene of interest - One needs to map the gene of interest in between two genetic markers, using linkage analysis. He will then use these two markers to screen the genetic library to obtain clones with these marker sequences, and use the end sequences of these clones to reprobe the library to expand the sequence coverage. He then repeat this process to establish a series of clones with overlapping sequences flanking the two markers. he will then need to screen the sequence of each candidate gene found within the contig, to identify mutations that associate with specific disease phenotype, in order to identify and finally clone the gene of interest
What is one example of a mutation that occurs spontaneously and randomly?
1. Missense 2. Tautomers
Which 2 factors determine the DNA migration rate in a gel electrophoresis?
1. Size of DNA fragments 2. Strength of electric field
Which two factors determine the DNA migration rate in gel electrophoresis?
1. Strength of the electric field 2. Size of the DNA fragments
Eighteen of the amino acids have two or more synonymous codons. Which 2 amino acids are the exceptions?
1. Trp 2. Met
During gametogenesis, what percentage of gametes would be aneuploid if the nondisjunction event occurs during meiosis I?
100%
During gametogenesis, what percentage of gametes would be trisomic if the nondisjunction event occurs during meiosis II?
25%
Donkeys have a somatic chromosome number of 62. Hybrids produced by crossing donkeys and zebras are called zonkeys, are sterile, and have many characteristics intermediate between the two parental species. If a zonkey has 52 chromosomes, what is the somatic chromosome number of zebras?
42
A diploid organism with a genome size of n= 23 experienced a Robertsonian translocation. How many chromosomes would you expect to see in the karyotype of a somatic cell in an affected individual?
44
Nucleosomes are comprised of how many molecules of histones?
8
Given a chromosome with the gene order, A B C D • E F G H, and an inverted chromosome with this gene order, A F E • D C B G H, which of the following recombinant chromosomes would result from a crossover between C and D?
A B C D • E F A and H G F E • D C B G H
A chromosome contains the following gene order: A B C D • E F G H Which of the following rearrangements represents a paracentric inversion?
A C B D • E F G H
A chromosome contains the following gene order: A B C D • E F G H Which of the following rearrangements represents a pericentric inversion?
A F E • D C B G H
Why is only a fraction of an organism's genes represented in any cDNA library?
A cDNA library only has all the information of an mRNA at an exact time/location/tissue
Microarrays
A grid of usually DNA fragments, also called a gene chip, that can be used to identify mixtures of either DNA or RNA. Can be used to identify mRNA present in a cell under certain conditions.
WhatisthedifferencebetweenamicroRNAandansiRNA?p53isa53kdprotein,whichis activated in cells exposed to DNA damage. That activation causes cells to stop their progression through the cell cycle so that the damage can be repaired. Inactivating mutations in p53 lead to cancer. You would like to investigate how this happens. How could you use an siRNA to inhibit the production of the p53 protein in cells? What precautions would you take to ensure that any results you get are the result of an siRNA targeting the p53 mRNAa?
A microRNA is a small RNA that is derived from a small double stranded RNA that is processed from a cellular polymerase II transcribed nuclear RNA via 2 endonuclease cleavage events. After being loaded onto a RISC complex, the miRNA is selected as a single strand from the small dsRNA and presented for partial Watson-Crick base pairing to target sites, usually within the 3'UTRs of mRNAs. Typically their association results in translational repression of the mRNA. Small interfering RNAs, as we discussed them, are generated by experimenters, either directly by chemical synthesis, or indirectly in cells by the introduction using vectors of short hairpin RNAs that are processed in siRNAs by the same kind of processing system that generates miRNAs. They are also loaded as double stranded RNAs onto a RISC complex that selects one of the strands for extensive Watson-Crick base pairing to target mRNAs. The goal here, accomplished by designing RNAs that base pair perfectly with the target RNA, is to promote the target RNA's degradation.
What is a Contig in gene cloning?
A set of DNA clones with overlapping sequences between 2 DNA markers
Explain the difference between a tumor suppressor gene and an oncogene.
A tumor supressor gene acts as a break to cell division and must be mutated to inactivate to promote cancer. An oncogene results from an activating mutation of a proto-oncogene. An oncogene is an overactive form of a gene that promotes cell division.
Feather color in parakeets is produced by the blending of pigments produced from two biosynthetic pathaways shown below. Four independently assorting genes (A, B, C, and D) produces enzymes that catalyze separate steps of the pathways. For the questions below and a lowercase letter to indicate a recessive allele producing no functional enzyme. Feather colors produced by mixing pigments are green (yellow + blue) and purple (red + blue). Red, yellow, and blue feathers result from production of one colored pigment, and white results from absence of pigment production. Pathway I: Compound I (colorless) → via Enzyme A → Compound II (red) → via Enzyme B → Compound III (yellow) Pathway II: Compound X (colorless) → via Enzyme C → Compound Y (colorless) → via Enzyme D → Compound Z (blue) A. What is the genotype of a pure-breeding green parakeet strain? B. What is the genotype of a pure-breeding yellow strain of parakeet? C. If a pure-breeding blue strain of parakeet aaBBCCDD is crossed to one that is pure-breeding purple, predict the possible genotype(s) of the F1. D. What color are the F1 birds whose genotype you identified in part C above? E. EXTRA CREDIT (only attempt if time allows) If F1 birds (genotype identified in parts C) are mated, what phenotypic classes and in what ration do you expect in the F2 generation?
A. AABBCCDD B. AABBccdd C. AaBbCCDD d. Green e. 9:3:1 ratio 9/16 Green 3/16 Purple 1/16 Blue
In the multiple allelic series that determines coat color in rabbits: C = Full Color (Agouti) Cch = Chinchilla Ch = Himalayan cc = Albino Dominance is in the order = C > Cch > Ch > c A. What is the coat color of the following rabbits: Cch/C = ? Ch/c = ? B. If a cross of Cch/c x Ch/c individuals was made, what genotypes and phenotypes and in what proportions will we expect? Clearly note if genotype and phenotype ratios are different.
A. Cch/C = Full Color (Agouti) Ch/c = Himalayan B. Genotypes: 1/4 Cch Ch 1/4 Cch c 1/4 Ch c 1/4 cc Phenotypes: 1/2 Chinchilla 1/4 Himalayan 1/4 albino Genotypic ratio: 1:1:1:1 Phenotypic ratio: 2:1:1
Which type of protein listed below encases a recently made peptide to allow proper folding into a mature protein? A. Chaperones B. Exonucleases C. Nuclear D. Phosphorylases E. Polymerases
A. Chaperones
Which of the following regulatory sequences may be located in the 3' un-transcribed region from the genes they regulate? A. Enhancers and silencers B. Core promoters C. Proximal elements D. TATA boxes E. ALl of the above
A. Enhancers and silencers
You have conducted an Ames test on a given compound. Which of the following would be classified as a positive result on the Ames test? A. His- strain (mutation in the his gene) grows on an his- plate (no his in the media) B. His- strain grows on an his+ plate C. His+ strain grows on an his- plate D. His+ strain grows on an his+ plate E. His+ strain grows on either an his- or an his+ plate
A. His- strain (mutation in the his gene) grows on an his- plate (no his in the media)
In humans, Occipital horn syndrome OHS) is a rare X-linked recessive disorder that is caused by a deficiency in the transport of the essentail mineral copper and associated with mutations in the ATP7A gene. The dominant and recessive alleles for OHS are represented by H and h. Acatalasia is an autosomal recessive condition that results from mutations of the gene producing catalase, and presents with recurrent periodontal infections. A and a allele symbols represent the Acatalasia alleles. A healthy man named Simon is married to a healthy woman named Sally. Simon's parents are healthy, but his brother Sean has OHS, and his sister Syndey has Acatalasia. Sally's brother Scott has Acatalasia and OHS. Sally's father has OHS, but her mother is normal. The pedigree shown below is incomplete. For each individual, let the left side of the circle/square represent their phenotype for OHS, and the right side represent their phenotype for Acatalasia. Darkened areas mean they are affected by the disorder. A. Complete the pedigree to reflect ALL of the data and individuals described above including filling in or leaving symbols as indicated. B. Using typical numbering system designations state the genotypes of the following individuals as completely as possible. Be sure to include information for both genes. Using H and h for OHS; A and a for Acatalasia; include X and Y symbols in addition of the genes are on sex chromosomes. I-1 = ? I-3 = ? I-4 = ? II-1 = ? C. Determine the probabilities that SImon and Sally's first child will be a Boy with OHS and Acatalasia.
A. Since I don't have full version of Quizlet and can't upload a picture of the pedigree, here's a description of what the pedigree should look like I-1 (Sally's father): Shade left side I-2 (Sally's mother): Empty I-3 (Simon's father): Empty I-4 Simon's mother): Draw a circle; leave empty II-1 (Sally's brother, Scott): Shaded entire square II-2 (Sally): Empty II-3 (Simon): Empty II-4 (Simon's sister, Syndey): Draw a circle; shade right side II-5 (Simon's brother, Sean): Shade left side B. I-1 = XhY/Aa I-3 = XHY/Aa I-4 = XhXh/Aa II-1 = XhY/aa C. 1/2 (for a boy) x 2/3 (Mom carries "a" allele) x 2/3 (Dad carries "a" allele) x 1/2 (chance for OHS) x 1/4 (child is aa) = 1/36
Snuppy, a male Afghan hound, was created by transferring the nucleus from an ear cell of the male Afghan, Tai, into an enucleated egg from a donor. The embryo was transplanted into a surrogate mother to develop to term. Which of the following animals would be a genetic match for Snuppy if nuclear DNA were tested? A. Tai B. The egg donor C. The surrogate D. Both Tai and the egg donor E. Both Tai and the surrogate
A. Tai
What type of mutation results in a single amino acid substitution? What type of mutation results in a single amino acid substitution? A. Missense. B. A single nucleotide insertion. C. Silent. D. Nonsense.
A. missense
In the lac operon, what acts as the inducer?
Allolactose
Changing conformation at the active site of an enzyme as a result of binding a substance at a different site is known as ________?
Allostery
What is a characteristic of protein secondary structure?
Alpha-helix
You have developed a new drug and have been asked to test the drug's mutagenicity before putting it on the market. Which test could you use to determine your compound's mutagenic potential?
Ame's test
CRISPR/Cas9
An easily engineered sequence-specific nuclease for use in eukaryotic cells. This system can add double strand breaks at specific points in the genome. sgRNA (guide RNA) can be designed to anneal to the target point in the genome. A nuclease then cleaves double-stranded within 20 bp of binding. This system is being exploited in many ways in research for targeted genome editing.
Chaperone proteins A) Aid in mismatch repair B) Allow correct folding of proteins C) Are imbedded in the membrane of cells and allow transport of products out of the cell D) Repair mutations in DNA caused by X-rays E) Reside in the golgi apparatus and glycosylate proteins
B) Allow correct folding of proteins
In the lac operon, the CAP-cAMP complex evaluates the concentration of which of the following compounds? A) Galactose B) Glucose C) Lactose D) Mannose E) Table sugar
B) Glucose
Given the DNA coding sequence 5' - TAC AAA ATA CAG CGG - 3', which of these sequences represents a frameshift mutation? A. 5' - TAG AAA ATA CAG CGG - 3' B. 5' - TAC AAA TAC AGC GGG - 3' C. 5' - TAC AAG ATA CAG CGG - 3' D. 5' - TAC AAA ATA CAC CGG - 3' E. 5' - TAC AAA ATA CAG AGG - 3'
B. 5' - TAC AAA TAC AGC GGG - 3'
You want to design a repressor protein mutant. Which protein domain is the best target for preventing binding of the inducer? A. Activator binding site B. Allosteric domain C. DNA-binding domain D. Helix-turn-helix domain E. Promoter domain
B. Allosteric domain
The compound 5-bromodeoxyuridine (BrdU) is a derivative of uracil (structure similar to thymine), and if BrdU becomes incorporated during DNA replication, it pairs with adenine. This compound is best classified as which type of mutagen? A. Alkylating agent B. Base analog C. Deaminating agent D. Intercalating agent E. Oxidative agent
B. Base analog
Which of the following is primarily responsible for genomic imprinting? A. Acetylation of histone B. Methylation of DNA C. Methylation of histone D. Acetylation of DNA E. None of the above
B. Methylation of DNA
You want to design a drug that prevents transcription of mRNAs but does not affect transcription of other RNAs. What enzyme would you target? A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III D. Ribozyme E. Methyl transferase
B. RNA polymerase II
In using Agrobacterium tumefaciens to transfer genes into plants, what is transferred from bacterium to plant DNA? A. The entire Ti plasmid containing the gene of interest B. T-DNA containing the gene of interest C. Only the antibiotic resistance gene D. The origin of replication E. The entire bacterium genome
B. T-DNA containing the gene of interest
Which of the following would be an implication of the universality of the genetic code? A) A DNA mutation may be silent B) Aflatoxin B1 insertion into the DNA double helix C) Bt Corn: a soil bacterium gene inserted into the corn genome in order to provide resistance to corn borer (an insect) D) Makes organ donation between humans possible E) All of the above
C) Bt Corn: a soil bacterium gene inserted into the corn genome in order to provide resistance to corn borer (an insect)
Which single-base mutation would insert a premature stop codon into the following protein sequence? N-Met-Gln-Leu-Arg-Cys-C A. 5' - AUG CAG AUA GCG UGC UAG - 3' B. 5' - AUG AAG UUA GCG UGC UAG - 3' C. 5' - AUG CAG UAA GCG UGC UAG - 3' D. 5' - AUG CAG UUA UUG UGC UAG - 3' E. 5' - AUG CAG UUA GCG UGC AAG - 3'
C. 5' - AUG CAG UAA GCG UGC UAG - 3'
Given the DNA coding sequence 5' - TAC AAA ATA CAG CGG - 3', which of these sequences represents a silent mutation? A. 5' - TAG AAA ATA CAG CGG - 3' B. 5' - TAC AAA TAC AGC GGG - 3' C. 5' - TAC AAG ATA CAG CGG - 3' D. 5' - TAC AAA ATA CAC CGG - 3' E. 5' - TAC AAA ATA CAG AGG - 3'
C. 5' - TAC AAG ATA CAG CGG - 3'
Given the DNA sequence 5-TAC AAA ATA CAG CGG-3, which of these sequences represents a missense mutation? A. 5-TAC AAG ATA CAG CGG-3 B. 5-TAC AAA TAC AGC GGG-3 C. 5-TAC AAA ATA CAC CGG-3 D. 5-TAG AAA ATA CAG CGG-3 E. 5-TAC AAA ATA CAG AGG-3
C. 5-TAC AAA ATA CAC CGG-3 because.....CAG->CAC = Gln->His *amino acid change in protein*
During DNA cloning, you will first linearize the vector using a restriction enzyme that has restriction site localized in: A. A plasmid vector at the origin of replication site B. A plasmid vector at the antibiotic resistance gene site C. A plasmid vector at the multiple cloning sites D. All of the above E. None of the above
C. A plasmid vector at the multiple cloning sites
Molecular biologists can determine whether a region of chromatin is heterochromatin or euchromatin by assessing the hypersensitivity of this region to ______. A. Nucleosomes B. Histone deacetylase C. DNase I D. RNA polymerase II
C. DNase I
In forward genetics, what is the correct order in which the researcher proceeds? A. Cloning a gene through observing the wild type phenotype B. Introduce a mutation to a known gene of interest, followed by screening for newly acquired traits of the organism C. Establish a defect phenotype, then search for the genes responsible for this defect D. All of the above E. None of the above
C. Establish a defect phenotype, then search for the genes responsible for this defec
Proofreading during translation can reverse mutations in which of the following ways? A. Enzymatic cleavage removes incorrect bases B. Misfolded proteins are degraded by the cell C. Slow reaction rate of incorrect tRNA at the A site allows replacement by correct tRNA D. Thymine dimers are cut from DNA strand and replaced by unattached thymine bases E. None of the above
C. Slow reaction rate of incorrect tRNA at the A site allows replacement by correct tRNA
The enhancer region of a eukaryotic gene is bound by which of the following? A. RNA polymerase II B. Basal transcription factors C. Transcription factors D. DNA polymerase II E. All of the above
C. Transcription factors
Which of the following cloning vector does NOT use bacterial (E.coli) as a host? A. Plasmid B. Lambda phage C. YAC D. Cosmid E. BAC
C. YAC
Which of the following reagent is required in the reaction tubes used in Sanger DNA sequencing but not in PCR? A. dNTPs B. Template DNA C. ddNTPs D. Primer E. DNA polymerase
C. ddNTPs
The Shine-Dalgarno sequence: A. is a consensus sequence found in the 16S rRNA subunit B. is a consensus sequence found in the 3' UTR of the mRNA C. is a consensus sequence found in the 5' UTR of the mRNA D. is a consensus sequence involved in the termination of translation E. is a region of the tRNA molecule involved in formation of charged tRNAs
C. is a consensus sequence found in the 5' UTR of the mRNA
What does the guide strand of a miRNA target for destruction? A. Heterochromatic regions of DNA B. Methylated DNA C. mRNAs with complementary nucleotide sequence D. Methylated histones E. Ribosomes
C. mRNAs with complementary nucleotide sequence
Binding of which complex increases the ability of RNA polymerase to transcribe the lac operon?
CAP-cAMP
In an allopolyploid organism, what is true regarding the fertility of interspecies hybrids?
Chromosome doubling by nondisjunction in gametocytes can lead to homologous chromosome pairing, disjunction, and fertile hybrids.
What kind of analysis is needed to determine whether a phenotypic defect is caused by mutation in the same gene or in different genes?
Complementation analysis
What kind of analysis is needed to determine whether a phenotypic defect is caused by mutation to the same gene or to two different genes?
Complementation analysis
In prokaryotes, the TATA box is a DNA sequence that is conserved across many species and used for the same purpose. What is another name for this stretch of DNA?
Consensus sequence
What is meant by the statement: "Bacterial translation and transcription are tightly coupled"?
Coupling means that translation of an ORF in a bacterial mRNA can be translated as soon as the RNA polymerase has synthesized it and made available its Shine-Delgarno sequence and initiator AUG codon. These two processes are not separated in space nor time in bacterial cells.
DNA methylation most frequently happens on what kind of dinucleotide sequence?
CpG
DNA methylation most frequently happens to what nucleotide sequence?
CpG
Which of the following is FALSE regarding individuals who are heterozygous for a large inversion?
Crossovers between homologous chromosomes that occur outside the inverted region produce nonviable gametes.
What is a "sticky end" created by restriction enzymes?
Cut DNA with overhanging strands that are complementary
Given the DNA template sequence 3' - TCC ATA TAC TAG GGT - 5', which of these sequences represents a frameshift mutation? A) 3' - TCG ATA TAC TAG GGA - 5' B) 3' - TCC ATA TAC TAA GGC - 5' C) 3' - TCT ATG TAC TAT GGT - 5' D) 3' - CCA TAT ACT AGG GTT - 5' E) 3' - TCT ATT TAC TAT GGC - 5'
D) 3' - CCA TAT ACT AGG GTT - 5'
What is true about the DNA agarose gel electrophoresis? A. DNA will migrate towards the positive electoral pole B. DNA will migrate towards the negative electoral pole C. DNA in the gel can be visualized under UV by adding ethidium bromide D. Both A and C are correct E. Both B and C are correct
D. Both A and C are correct
Which of the following are characteristics of the euchromatin? A. DNA demethylation B. Condensed chromatin/nucleosome structure C. Histone acetylation D. Both A. and C. are correct E. None of the above
D. Both A. and C. are correct
The enzyme B-galactosidase catalyzes which reaction? A. allolactose --> glucose + lactose B. Galactose --> glucose + lactose C. Glucose --> galactose + lactose D. Lactose --> galactose + glucose E. Lactose --> glucose + fructose
D. Lactose --> galactose + glucose
Which of the following does NOT belong to non-coding RNAs? A. MicroRNA (miRNA) B. Small interfering RNA (siRNA) C. Long non-coding RNA (lncRNA) D. Message RNA (mRNA) E. None of the above
D. Message RNA (mRNA)
Which of the following is NOT an essential characteristic of a balancer chromosome? A. Presence of inverted chromosome segments to prevent crossing over B. Presence of a dominant mutant allele to identify heterozygous balancer chromosome C. Presence of a recessive mutant allele which is lethal when homozygous D. Presence of a recessive mutant allele to identify heterozygous balancer chromosome E. None of the above
D. Presence of a recessive mutant allele to identify heterozygous balancer chromosome
In selecting bacteria transformed with recombinant plasmid, cells are chosen that are resistant to a specific antibiotic. How are the bacteria made resistant? A. They are pre-selected for the experiment on this bases B. Resistance is activated by the recombination event C. Resistance is activated when the cells are provided with the antibiotics D. The antibiotic resistance gene is expressed by the plasmid vector E. None of the above
D. The antibiotic resistance gene is expressed by the plasmid vector
During embryo development, the global DNA methylation status will be reset. Which of the following statement is NOT correct: A. In pre-implantation stage, embryo DNA will undergo global DNA demethylation B. In post-implantation stage, embryo DNA will undergo global DNA remethylation C. The methylation status of imprinted genes will not change in different somatic cell types D. The methylation status of imprinted genes will not change in germ cells E. All of the above
D. The methylation status of imprinted genes will not change in germ cells
A researcher wants to study gene function by mutagenesis. He is particularly concerned with being able to easily find and amplify the mutated gene with a tag of known sequence. Which of the following mutagens would therefore be more useful? A. Ionizing radiation B. UV radiation C. Chemical mutagen D. Transposon DNA E. All of the above
D. Transposon DNA
Separation of sister chromatids occurs _______. A. at anaphase II in mitosis and anaphase in meiosis B. in meiosis, but not in mitosis C. in mitosis, but not in meiosis D. at anaphase in mitosis and anaphase II in meiosis
D. at anaphase in mitosis and anaphase II in meiosis
A bacterium is unable to transport lactose into the cell to be broken down. Which gene is likely mutated in this bacterium? A. lacl B. lacO C. lacP D. lacY E. lacZ
D. lacY
Given the DNA template sequence 3' - TCC ATA TAC TAG GGT - 5', which of these sequences represents a nonsense mutation? A) 3' - TCG ATA TAC TAG GGA - 5' B) 3' - TCC ATA TAC TAA GGC - 5' C) 3' - TCT ATG TAC TAT GGT - 5' D) 3' - CCA TAT ACT AGG GTT - 5' E) 3' - TCT ATT TAC TAT GGC - 5'
E) 3' - TCT ATT TAC TAT GGC - 5'
What is the function of Ames test? A) Detect transition mutation B) Measure the affect of UV radiation C) Test level of hydroxylation of DNA bases D) Test potential mutagens E) All of the above
E) All of the above
Individuals with Xeroderma Pigmentosum possess mutations that do not allow repair of which of the following mutations? A) Double stranded DNA breaks B) Hydroxylation of adenine nucleotides C) Insertion of intercalating agents D) Methylation of guanine bases E) Thymine dimers
E) Thymine dimers
Given the following mRNA sequence, which of the following mutations would affect the protein sequence? 5' - AUG CAG UUA CGC UGC UAG - 3' A. 5' - AUG CAA UUA GCG UGC UAG - 3' B. 5' - AUG CAG UUA GCA UGC UAG - 3' C. 5' - AUG CAG UUG GCG UGC UAG - 3' D. 5' - AUG CAA UUA GCG UGU UAG - 3' E. 5' - AUG CAC UUA GCA UGC UAG - 3'
E. 5' - AUG CAC UUA GCA UGC UAG - 3'
The genetic conflict hypothesis attempts to describe imprinting of DNA. IGF2 is only paternally expressed and is associated with large fetuses, whereas IGF2 receptor is only maternally expressed and is associated with small fetuses. How is this correlated with the genetic conflict hypothesis? A. Female silences genes that allocate resources to fetus at the expense of the mom B. Female activates genes that allocate resources to fetus at the expense of the mom C. Male silences genes that conserve maternal resources at the expense of the other male's fetuses D. Male activates genes that conserve maternal resources at the expense of the other male's fetuses E. Both A. and C. are correct
E. Both A. and C are correct
What are the features of a heterochromatin? A. DNA methylation B. Histone acetylation C. Histone deacetylation D. Unmethylated DNA E. Both A. and C. are correct
E. Both A. and C. are correct
Cloning genes by cDNA library complementation is most feasible in organisms with which of the following characteristics A. Organisms not amenable to transformation B. Presence of a recessive mutant gene in that organism C. Presence of a dominant mutant gene in that organism D. Organisms amenable to transformation E. Both B. and D. are correct
E. Both B. and D. are correct
During gametogenesis, the methylation status of the imprinted genes are reset: A. Alleles from both parents will be imprinted maternally in oocytes and sperms B. Alleles from both parents will become imprinted maternally in oocytes C. Alleles from both parents will be imprinted paternally in oocytes and sperms D. Alleles from both parents will become imprinted paternally in sperms E. Both B. and D. are correct
E. Both B. and D. are correct
A mutagen has introduced a frame-shift mutation by adding two nucleotide bases. Which of the following would be classified as a reversion mutation for this particular mutant? A. Adding 2 bases B. Deleting 1 base or adding 1 base C. Deleting 1 base or adding 2 bases D. Deleting 1 base or adding 3 bases E. Deleting 2 bases
E. Deleting 2 bases
During translation initiation in prokaryotes, the amino acid on the initiator tRNA is: A. acetylated B. added using ATP as the energy source C. IF-1 D. methionine (Met) E. N-formylmethionine (fMet)
E. N-formylmethionine (fMet)
What type of mutation is seen here? Wild type: 5' - TAC AAA ATA CAG CGG - 3' Mutation: 5' TAC AAC ATA CAG CGG - 3' A. deletion B. insertion C. nonsense D. transition E. transversion
E. transversion
Which of the following is FALSE regarding organization in an interphase nucleus?
Each chromosome occupies exactly the same region in all nuclei within an organism.
What is used in DNA agarose gel electrophoresis to help visualize under UV?
Ethidium Bromide
T or F. In order to create the possibility of generating a trisomy, nondisjunction must occur during meiosis II.
F Nondisjunction during either meiosis I or meiosis II creates gametes that will generate trisomies if fertilized.
BONUS QUESTION: A cell in the G1 has the option or entering which two phases?
G0 S phase
What features are shared by genes identified as "hotspots" for mutations? Genes are often large. Genes often lack introns. Genes are typically highly conserved. Genes are often located near the centromeres.
Genes are often large.
Which type of DNA library (genomic or cDNA) would include introns, promoters, and other non-coding sequences?
Genomic
A cell can form 10 nm chromatin fibers, but not 30 nm fibers. Which molecule has likely been removed or mutated in this cell?
H1
Which histone protein is not part of the core nucleosome structure?
H1
After DNA replication, during reassembly of nucleosomes, the ________ tetramers are bound by H2A-H2B dimers to form complete nucleosomes.
H3-H4
Explainhowheterochromatinspreads,histonemark,andthereaderinthe process. What two proteins help maintain the boundary of heterochromatin?
Heterochromatin spreads by the writer HMT (histone methyltransferase) methylating H3K9. The mark (MeH3K9) is read by HP1 protein which binds to MeH3K9. Then HMT binds to HP1. HP1 then methylates H3K9 of the neighboring nucleosome and spreads the heterochromatin signal. Boundaries are maintained by HATs and kinases.
Describe how Histone acetylation changes chromatin status through altering the interaction between DNA and histone.
Histone acetylation neutralizes the lysine positive charge of the amino group in its side chain, therefore weakens the electrostatic interaction between histone and the negatively charged DNA backbone. This helps the formation of euchromatin and gene expression
Many antibiotics cause mispairing between codons and anticodons. What fatal effect would this mispairing have on the cell?
Improper amino acid sequence in protein
Explain how genetic imprinting causes IGF2 gene to be activated in paternal chromosome 15, and silenced in maternal chromosome 15.
In paternal chromosome, methylation of the ICR region prevents the insulator protein from binding, therefore it no longer blocks the activation of the IGF2 gene by the transcription activator. In maternal chromosome, the ICR region is unmethylated, therefore the insulator protein can bind and block the activation of IGF2 gene expression by the activator protein.
Explain how genetic imprinting causes H19 genes to be imprinted, and Igf2 gene to be activated in paternal chromosome 15?
In paternal chromosome, the H19 gene is methylated, so its expression is suppressed. They methylation of ICR region prevents the insulator protein from binding, therefore it no longer blocks the activation of IGF2 gene by the transcription activator
Total cholesterol in blood is reported as the number of milligrams (mg) of cholesterol per 100 milliliters (mL) of blood. The normal range is 180-220 mg/100 mL. A gene mutation altering the function of cell-surface cholesterol receptors restricts the ability of cells to collect cholesterol from blood and draw it into cells. This defect results in elevated blood cholesterol levels. Individuals who are heterozygous for a mutant allele and a wild-type allele have levels of 300-600 mg/100 mL, and those who are homozygous for the mutation have levels of 800-1000mg/100 mL. Identify the genetic term that best describes the inheritance of this form of elevated cholesterol level.
Incomplete dominance
Strand slippage causes which type of mutation?
Increase in trinucleotide repeats
What is the function of the transcription factor in an enhanceosome for gene expression?
It binds to the enhancer or silencer region which is far away from the transcription initiation site, and helps to form a DNA loop to interact with RNA polymerase II that is associated with the core promoters, to either stimulate or inhibit a particular gene expression
One example of polycistronic mRNA in prokaryotes is:
Lac Operon
Missplicing can lead to many genetic disease. What is the cause of missplicing?
Mutation at exon/intron border region
Given the following mRNA sequence, what is the amino acid sequence for the corresponding polypeptide? 3' - AAC GGG GAG CAC UCC - 5'
N - Pro His Glu GLy Gln - C
In the presence of glucose or low lactose, where would the lac repressor be bound?
Operator
What is a Contig?
Overlapping clones from genomic library between two markers flanking the gene of interest
How does the example of Xeroderma Pigmentosum demonstrate the frequency of DNA damage within skin cells?
People with this disease cannot repair UV damage from the sun due to inherited mutations in the genes which produce DNA repair proteins. These individuals, which have many more cases of skin cancer than others, demonstrate the increased incidence of UV-caused mutations in normal individuals which are repaired.
BONUS QUESTION: What is the name of the autosomal dominant condition in which affected individual have more than 5 fingers and/or toes?
Polydactyly
In DNA agarose gel electrophoresis, does the DNA migrate towards the negative or positive electoral pole?
Positive
Following the insertion of an incorrect amino acid into a growing polypeptide chain, proofreading at the P site would cause:
Premature termination
Which portion of DNA could be mutated to cause a change in transcription regulation?
Promotor region
Which molecule binds to the promoter region of a gene?
RNA polymerase
What are the four pathways that need to be affected in vitro to transform a healthy cell into a cancer cell?
Ras, Rb, p53, and telomere maintenance
What would you expect to find bound to the stop codon at the A site during translation of mRNA?
Releasing factor
Binding of what protein initiates translation-termination events that result in polypeptide release and dissociation of ribosomal subunits?
Releasing factors
What kind of enzyme is used in generating the first strand of cDNA?
Reverse Transcriptase
Describe how "Golden Rice" is generated using T-DNA technology.
Rice lacks certain genes for beta-carotene synthesis. Researchers cloned the daffodil PSY gene and bacteria CRTI gene into the T-DNA of transformation vector (derived from Ti-plasmid), and used the disarmed Ti-plasmid (which expresses virulence proteins) and the transformation vector to deliver the T-DNA region containing these two genes into the rice cells, which is then integrated into the rice genome. The transgenic rice can then produce beta-carotene from GGPP in its endosperm.
Which repair mechanism is an error-prone system that uses bypass polymerases to replicate short segments of daughter strands whose normal replicaiton is blocked by cuts in template strand DNA?
SOS repair
Massively parallel sequencing
Sequencing thousands to millions of DNA sequences at a time. A technique used by Next Generation sequencing methods.
Explain why "Dolly" the sheep - the first mammal ever cloned is not a transgenic animal.
She did not carry a foreign gene from another species
Which types of mutation are possible thanks to the redundant nature of the genetic code?
Silent
You discover a new compound that yeast requrie for survival. To find genes involved in the synthesis of this compound (Compound X) you have screened for mutants that cannot grown on minimal media, but can grow on minimal media + compound X. You find 10 mutatnts and perform a complementation tests. The results are shown below (+ = growth, - = no growth) A. How many genes are represented in the data above? B. Group the mutants complementation groups.
Sorry that the figure is not actually shown. But here's the answer. A. Four B. 1, 4 2, 3, 7 5, 6, 9 8, 10
Which type of mutation may result in expression of intronic sequence or loss of part of an exon? Triplet-repeat expansion mutation Promoter mutation Polyadenylation mutation Splice site mutation
Splice site mutation
What is the function of an enhanceosome?
Stimulates gene transcription by allowing transcription factors to interact and regulate RNA polymerase activity
What process, which is responsible for many trinucleotide repeat disorders, alters the number of DNA repeats?
Strand slippage
Functional genomics
Studying the function of all the genes in a cell or organism (through their protein or RNA products)
____________ are structures that have the same composition and general arrangement but a slight difference in bonding and placement of a hydrogen.
Tautomers
BONUS QUESTION: Imagine you are a male painted turtle.... (did you actually try to imagine it?).... What can be said about the conditions in your part of the nest that were required to make you you?
Temperature of the nest was cooler than the critical point
What is meant by the "beads on a string" model of chromatin?
The beads are the nucleosomes, and the string is the linker DNA.
What is the Synthetic Lethality?
The combination of two viable mutant results in inviability
Genome
The complete set of genetic material in a cell or organism (haploid)
Which of the following best describes the histones associated with eukaryotic DNA?
The five types of histone proteins are small, basic proteins with a positive charge that allows them to bind to DNA.
Which of the following is true of the gametes of a human female who has nondisjunction of her X chromosomes in meiosis I?
The gametes contain 22 or 24 chromosomes.
Compare the size of the human genome with the size of the E.coli genome in base pairs and in number of genes. Does this size difference make sense when thinking about organism complexity? Provide an explanation for this disparity.
The human genome is 3 x 109 bp compared with the E.coli genome of 4.6 x 106 bp. The human genome contains approximately 20,000 genes, while E.coli contains ~4,000 genes. It seems like humans should have more than just five times the number of genes in E.coli based on the differences in complexity. In lecture, one of the explanations for this disparity is the alternative splicing used in humans that results in a far greater number of different protein products than is estimated by just the gene number.
SuVar mutations in Drosophila are those which lead to suppression of variegation of expression of the white gene when it is repositioned next to heterochromatin by a chromosome inversion. Less variegation means less white eye color, that is, less repression of red pigment gene expression by spread of heterochromation to the adjacent white gene sequences. eVar mutations are those that lead to enhanced variegation;i.e. more white eye color. What genetic alterations might lead to an eVar phenotype?
The most obvious way to generate an eVar phenotype would be to overproduce the major structural heterochromatin components, in this case Hp1 and the H3K9 lysine methyl transferase. In our class discussion, it was mentioned that one thing that limits the ability of flanking heterochromatin to silence a neighboring white gene is limiting quantities of the heterochromatin components, making more of these components would promote the silencing.
What is the Position Effect Variegation (PEV)?
The spread of heterochromatin characters to the euchromatin region placed to its vicinity due to chromosome translocation
Which of the following is true regarding heterozygous carriers of large chromosome inversions or translocations?
They typically have a reduction in the number of viable gametes.
Describe how the first generation of "Golden Rice" is generated using recombinant DNA technology.
They use the PSY gene from daffodils and CRTI gene from bacteria into T-DNA, and using disarmed Ti-plasmids and transformation vectors, the genes are delivered into the rice's endosperm so that it can produce beta-carotene.
In using Agrobacterium tumefaciens to transfer genes into plants, what is transferred from bacterium to plant?
Ti plasmid containing the gene of interest
What is the purpose of a modifier screen?
To see if the mutation of a second gene will modify the phenotype of the first gene mutant
What proteins aid in the recognition of the promoter sequence and binding of RNA polymerase II in eukaryotes?
Transcription factors
What type of mutation is seen here: Wild type: 5' - TAC AAA ATA CAG CGG - 3' Mutation: 5' - TAC AAA ATA CAG AGG - 3'
Transversion
What is the Synthetic Lethality when you study forward genetics?
Two viable gene mutations cause lethality when combined together
Thymine dimers are most commonly caused by which of the following?
U.V. irradiation.
Marfan syndrome is an autosomal dominant disorder in humans. It results from mutation of the gene on chromosome 15, which produces the connective tissue protein fibrillin. In its wild-type form, fibrillin gives connective tissues, such as cartilage, elasticity. When mutated, however, fibrillin is rigid and produces a range of phenotypic complications, including excessive growth of the long bones of the leg and arm, sunken chest, dislocation of the lens of the eyes, and susceptibility to aortic aneurysm, which can lead to sudden death in some cases. Different sets of symptoms are seen among various family members, as shown in the pedigree. Each quadrant of the circles and squares represents a different symptom, as the key indicates. Sicne all cases of Marfan syndrome are caused by mutation of the fibrillin gene, and all family members with Marfan syndrome carry the same mutant allele, what genetic term best describes the differences shown in the pedigree?
Variable expressivity
Explainthewholegenomeshotgun(WGS)approach,itsroleinsequencingthehuman genome, and what is meant by "10x coverage"?
WGS was used as a competitive method to map-based sequencing in sequencing the human genome. It is quicker than the traditional approach because the mapping step can be skipped. In WGS, the genome is fragmented in multiple different ways, then sequenced. Once the sequences are obtained, overlapping fragments are aligned in order to create a complete sequence. The process is repeated until each area of the sequence has at least 10x coverage, meaning that it has 10 data sets of sequence for each base position.
If a researcher is cloning a gene of interest (GOI) into a plasmid vector capable of blue/white selection, and uses this plasmid to transform bacteria, which are then plated on X-gal medium, would the recombinant plasmids containing the GOI be forming blue or white clones?
White
If a researcher is cloning a sequence into a plasmid vector capable of blue/with selection, and uses this plasmid to transform bacteria, which are then plated on X-gal medium, would the recombinant plasmids be found in blue or white colonies or both?
White
What dosage compensation mechanism is employed by female placental mammals? Y-inactivation synteny X nondisjunction X-inactivation X chromosome crossing over
X-inactivation
Which wavelengths of energy are most damaging internally?
X-rays
The formation of Bar Body is regulated by which type of RNA?
Xist or lncRNA
Which cloning vector uses yeast (S. cerevisiae) as a host?
YAC
Which type of vector would be ideal for cloning up to 2,000 kilobases of DNA fragments?
YAC
Which of the following is the BEST description of a proto-oncogene? a. A gene that normally stimulates progression through the cell cycle but when mutated can lead to overstimulation of division b. A gene that blocks progression through the cell cycle when it is inactivated c. I forgot to study these!!! d. A gene that stimulates progression through the cell division cycle only when expressed at high levels e. A gene that normally slows progression through the cell division cycle
a. A gene that normally stimulates progression through the cell cycle but when mutated can lead to overstimulation of division
During which cell cycle phase(s) are chromosomes composed of two DNA molecules? a. G2 b. G1 c. Anaphase d. G1 and G2 e. G1, G2, and Anaphase
a. G2
You discover a new species of snail, Biologica terificia, and find that its haploid number is 4 (n=4). The somatic cells of Biologica terificia are diploid. How many chromosomes are in a somatic cell of the snail in G1 phase? a. 4 b. 8 c. 16 d. 10 e. 23
b. 8
Which structure is responsible for chromosome movement during cell division? a. Synaptonemal complex b. Kinetochore microtubules c. Metaphase plate d. Polar microtubules e. Cohesin
b. Kinetochore microtubules
King George III of England and other members of the royal family were afflicted with a series of strange, seemingly unrelated symptoms including abdominal pain, rapid pulse, convulsions, and insanity. It has been determined that he likely suffered from porphyria, caused by a mutation in a single allele. What is the genetic term describing the alteration of multiple, distinct traits of an organism by a mutation in a single gene? a. Epistasis b. Pleiotropy c. Environmental influence d. Neomorphism e. Incomplete dominance
b. Pleiotropy
[Matching] Also known as "spoiler" mutations due to their negative effect on protein complexes as a whole a. amorphic mutations b. dominant negative mutations c. Neomorphic mutations d. Hypermorphic mutations e. Hypomorphic mutations
b. dominant negative mutations
[Matching] Mutations that acquire new gene activities not found in wild a. amorphic mutations b. dominant negative mutations c. Neomorphic mutations d. Hypermorphic mutations e. Hypomorphic mutations
c. Neomorphic mutations
What dosage compensation mechanism is employed by female placental mammals? a. Synteny b. Y-inactivation c. X-inactivation d. X non-disjunction e. X Chromosome crossing over
c. X-inactivation
If a trait is X-linked recessive, who would express the trait? a. females homozygous for the dominant allele and males hemizygous for the recessive allele b. heterozygous females and males hemizygous for the dominant allele c. female homozygous for the recessive allele and males hemizygous for the recessive allele d. females homozygous for the recessive allele and males hemizygous for the dominant allele e. the same proportions of females and males
c. female homozygous for the recessive allele and males hemizygous for the recessive allele
Which type of DNA library (genomic or cDNA) would only contain information for message RNA sequences?
cDNA
What is the term for the phenomenon that describes the significant reduction, or absence, of recombinant chromosomes in the gametes produced by inversion heterozygotes, when a crossover is within the inverted region?
crossover suppression
During which stage of mitosis and/or meiosis do homologs separate? a. Anaphase of Mitosis only b. Anaphase I of Meiosis and Anaphase of Mitosis c. Both anaphase I and II of Meiosis d. Anaphase I of Meiosis only e. Anaphase II of Meiosis only
d. Anaphase I of Meiosis only
[Matching] Mutations that produce more than normal levels of gene activity a. amorphic mutations b. dominant negative mutations c. Neomorphic mutations d. Hypermorphic mutations e. Hypomorphic mutations
d. Hypermorphic mutations
What is required in the reaction tubes used in Sanger DNA sequencing but not in PCR?
ddNTPs
How will the centromeres be distributed in the two recombinant chromosomes following a crossover in the inversion loop of a paracentric inversion heterozygote?
dicentric and acentric fragment
Bateson and Punnet crossed two white-flowered lines and saw all purple flowers in the F1 generation. What ration of phenotypes did they observe in the F2 generation that led them to conclude that this was an example of complementary gene interactions? a. 8 purple to 8 white b. 16 purple to 0 white c. 0 white to 16 purple d. 7 purple to 9 white e. 9 purple to 7 white
e. 9 purple to 7 white
[Matching] Mutation that results in a partial reduction of protein function a. amorphic mutations b. dominant negative mutations c. Neomorphic mutations d. Hypermorphic mutations e. Hypomorphic mutations
e. Hypomorphic mutations
What technique would you use to detect a target sequence in an intact chromosome using a labeled molecular probe?
fluorescent in situ hybridization (FISH)
In individuals heterozygous for a reciprocal balanced translocation, adjacent-2 segregation of tetravalent structures is very rare since it requires which type of centromeres to move to the same pole of the cell at anaphase I?
homologous
The phenomenon of ________ in allopolyploids consists of more rapid growth, increased fruit and flower production, and improved disease resistance.
hybrid vigor
Which type of chromosome deletion is caused by two concurrent chromosome breaks (rather than a single break)?
interstitial deletion
What structure, seen during synapsis, is indicative of a chromosome inversion?
inversion loop
Which type of RNA is translated within a cell?
mRNA
You are attempting to isolate cells for your karyotype, and you find one cell in each of the phases (prophase, metaphase, anaphase, and telophase). Which one is best suited for a karyotype?
metaphase
A karyotype shows that a child has Klinefelter syndrome (47,XXY). If the child is also color-blind (due to a recessive X-linked allele), despite his parents having normal color vision, in which parent and stage of meiosis did nondisjunction occur?
mother in meiosis II
Chromosomal translocation involves chromosome breakage and reattachment of the broken segment to a ________ chromosome.
non-homologous
Which of the following correctly lists the levels of compaction in eukaryotes from naked DNA to the most compact?
nucleosome, solenoid, looped chromatin (300-nm fiber), metaphase chromosome
"Seedless" fruits and vegetables have how many chromosomes and what type of chromosomal distribution?
odd number polyploidy
A region of a chromosome spanning the centromere is broken and reattached in the reverse direction. This is an example of which type of chromosomal defect?
pericentric inversion
Identify two DNA repair mechanisms that remove UV-induced DNA lesions. Check all that apply. photoreactivation nonhomologous end joining synthesis-dependent strand annealing nucleotide excision
photoreactivation and nucleotide excision
________ is used to map genes in deleted chromosome regions by a method known as deletion mapping. It is a genetic phenomenon that occurs when a normally recessive allele is "unmasked" and expressed in the phenotype because the dominant allele on the homologous chromosome has been deleted.
psuedodominance
Which proteins would you target if you wanted to disrupt the chromosome superstructure for chromatin condensation that ultimately produces the characteristic shape of the metaphase chromosome?
scaffold proteins
Karyotypes are a method for displaying chromosomes by grouping them into homologous pairs based on what two distinguishing factors?
size+banding pattern
What is the name of the structure formed when the 10 nm fibers of chromatin form a cylindrical filament of coiled nucleosomes?
solenoid structure
Which type of chromosome has no p arms?
telocentric
Cri-du-chat syndrome is a human disorder caused by which type of chromosomal defect?
terminal deletion
What type of mutation is seen here? Wild type: 5-TAC AAA ATA CAG CGG-3 Mutation: 5-TAC AAG ATA CAG CGG-3
transition
A chromosome has broken, and a piece of one chromosome is translocated to a non-homologous chromosome. This is an example of what type of chromosomal alteration?
unbalanced translocation
________ is rare and occurs most commonly when repetitive regions of homologous chromosomes misalign, resulting in partial deletions and partial duplications.
unequal crossover
Prader-Willi and Angelman syndromes are caused by which type of chromosomal mutations, both in connection with chromosome 15?
uniparental disomy
What structure will form during synapsis in Prophase I if there is a partial chromosome deletion or duplication?
unpaired loop
Active transcription occurs where with respect to matrix attachment regions (MARs)?
within euchromatin regions a large distance from MARs
Listanddefinethefunctionandbindingorderofthefourgeneraltranscriptionfactors discussed in lecture.
"DeBoRaH" TFIID binds to the promoter, TFIIB binds to TFIID, then RNA Pol II binds. Finally, TFIIH binds and "opens" the DNA for transcription. This forms the "pre-initiation complex." TFIIH is both a helicase and a kinase. The kinase phosphorylates amino acids in the C-terminal domain (CTD) of RNA Pol II subunit 1. Phosphorylation of the CTD untethers Pol II. Helicase activity unwinds the duplex DNA.
Name two differences between RNA and DNA.
1. RNA is single stranded, DNA is double stranded. 2. They have the same base compounds except RNA has U and DNA has T.
If nondisjunction occurs in meiosis II, how many of the four gametes will be aneuploid? If nondisjunction occurs in meiosis II, how many of the four gametes will be aneuploid? 1, 2, 3, or 4
2 Nondisjunction in meiosis II produces one nullosomic gamete and one disomic gamete. The two products produced by the other meiosis I daughter cell will produce two normal haploid gametes.
Given the sequence of a DNA coding strand, 5'-TAC AAA ATA CAG CGG-3', which of these sequences represents a nonsense mutation? Given the sequence of a DNA coding strand, 5'-TAC AAA ATA CAG CGG-3', which of these sequences represents a nonsense mutation? A. 5'-TAG AAA ATA CAG CGG-3' B. 5'-TAC AAG ATA CAG CGG-3' C. 5'-TAC AAA ATA CAC CGG-3' D. 5'-TAC AAA ATA CAG AGG-3' E. 5'-TAC AAA TAC AGC GGG-3'
A. 5'-TAG AAA ATA CAG CGG-3'
Which of the following is a common consequence of a viable trisomy? A. Reduced fertility B. Increased rate of cell division C. Polyploidy D. Increased cell size
A. Reduced fertility Trivalent synaptic structures form in meiosis I. Segregation of homologs results in the production of aneuploid gametes.
WhatroledoesthecarboxylterminaldomainofthelargesubunitofRNApolymerase2 play in the addition of a Cap structure to the 5' end of animal cell mRNAs?
As described in lecture, the CTD coordinates Cap addition by providing binding sites for the enzymes that add the Cap.
8. Explain the statement: Aminoacyl-tRNA synthetases are the only components of gene expression that interpret the genetic code.
As discussed in class, tRNA synthetases are the enzymes in the translation process that are responsible for connecting the nucleotide code to the amino acid code. Translation is really accomplished by matching the anti-codon sequence with the correct amino acid added to the 3' end of the tRNA, and tRNA synthetases are solely responsible for that task.
Which nucleotide will base‑pair with the enol form of 5‑bromouracil? A. Thymine B. Cytosine C. Guanine D. Adenine
C. Guanine The enol form of 5‑bromouracil forms a base pair with guanine.
How is the effective dosage of X-linked gene products balanced in placental mammals between females that carry two X chromosomes and males that carry only a single X chromosome? A. Genes on the single male X chromosome are expressed at twice the rate of genes on the two female X chromosomes. B. The paternally derived X chromosome is inactivated in all female somatic cells, so only the maternally derived X chromosome is expressed. C. One X chromosome is randomly inactivated in female somatic cells. D. Only half of the genes on each X chromosome are expressed in female somatic cells.
C. One X chromosome is randomly inactivated in female somatic cells. One X chromosome is randomly inactivated early in embryogenesis. This leaves female somatic cells with one copy of the X chromosome that is expressed, which is equivalent to what is found in male cells, which carry a single X chromosome.
DNA methylation of eukaryotic DNA is associated with gene repression. Where are methyl groups added to DNA and what is one mechanism by which this modification results in gene repression?
Methylation of DNA occurs at the 5 position of the cytidine ring in cytosines located within CpG dinucleotide sequences in eukaryotic DNA. One way in which this modification promotes gene repression is by binding Methyl Binding Domain (MBD) containing proteins that recruit Histone deacetlyases to the chromatin harboring the methylated DNA.
How many Barr bodies are found in a normal male nucleus? One Barr body. Two Barr bodies. Three Barr bodies. No Barr bodies.
No bar bodies
Shownbelowisthealternativesplicingpatternforagene.Livercellsincludethecross- hatched exon in 50% of the gene's mRNAs. In all other cells the exon is included in 90% of the gene's mRNAs. What are two possible explanations for this result?
One explanation would be that liver cells contain lower concentrations of a splicing activator that binds to an exon splicing enhancer (ESE) in the cross- hatched exon. Alternatively, liver cells might contain a higher concentration of a splicing inhibitor protein that binds to sites required for defining the exon. A possible target for the inhibitor protein would be the sequences that bind the protein complex that helps to define the 3' splice site and recruit the U2 snRNP to the branch point.
Enhancersequenceswouldprobablybefoundintheintronanduniquenon-coding fractions of the human genome. Given the existence of the complete human genome sequence and those of other mammalian and vertebrate species, can you think of a way to identify candidate enhancer sequences simply by using the sequences?
Regulatory sequences like enhancers that promote expression of conserved in genes in conserved patterns should be conserved in evolution. Therefore you might predict they would be found in enhancer sized sequence blocks (100-200 bp) that are more conserved, i.e. more similar in sequence between species, than flanking non-regulatory sequences. Doing an alignment of the sequences found in the introns and flanking sequences of genes that are conserved between species should reveal candidate enhancer elements. Shown below is such an alignment for a region of the cystic fibrosis gene that shows a conserved exon sequence and several conserved intron sequence blocks.
Tor F. Tautomers of nucleotide bases are isomers that differ from each other in the location of one hydrogen atom in the molecule.
T
T or F. Many chemicals are more mutagenic after being processed in the liver.
T Many potential mutagens are poorly mutagenic until passing through the liver.
What are 3 different mechanism used by eukaryotic cells to traffic their proteins to appropriate cellular compartments? What role is played by small GTP binding proteins in these processes?
The 3 different mechanisms are 1) Gated transport through a membrane pore, used for transport from the cytoplasm to the nucleus. 2) Transport across membranes using "channels" that allow cotranslational passage across the membrane as an unfolded linear polypeptide chain; used for transport of proteins into the endoplasmic reticulum for eventual delivery to the plasma membrane or secretion from the cell. 3) Vesicular transport involving budding of small membrane vesicles from membrane compartments for delivery to other membrane complartments or for delivery to the cell exterior. The first and third of these use small GTP binding proteins to give direction to these processes. The binding and hydrolysis of GTP to these proteins ensure that the transport processes are uni- directional.
Type I self-splicing introns require a guanosine cofactor. What is the function of guanosine in splicing and what is the equivalent "cofactor" in mRNA splicing?
The 3' hydroxyl group of the guanosine initiates the splicing reaction by attacking the phosphodiester bond between the upstream exon and the intron. The consequence of this reaction is the joining of the guanosine to the 5'end of the intron and the exposure of a 3' hydroxyl group at the end of the upstream exon which will attack the downstream intron/exon junction phosphodiester bond to "ligate" the exons together and release the intron. The equivalent "cofactor" in mRNA splicing is the branch point A through its 2' OH group.
Proteome
The complete set of proteins in a cell or organism
Assumethatyouhavemappedamutationinahumangenethatisassociatedwitha disease phenotype. The mutation lies within the third exon of a five exon gene but is in the third position of an amino acid codon and doesn't change the predicted protein coding sequence. Experiments show the mRNA for this gene in mutant cells is shorter than the mRNA in wild type cells. Based on our discussion of splicing regulation, what is one possible explanation for this result?
The most likely explanation would be that the mutation disrupts an exon splicing enhancer (ESE) that provides a binding site for a splicing activator protein which is necessary to define the third exon. That is, it is required to help recruit the U1 snRNP to the downstream 5' splice site and the protein complex and U2 snRNP to the upstream 3'splice site.
5. Nonsense suppressors tRNAs are coded for by mutant tRNA genes whose anti- codon sequences are changed in such a way that they can recognize stop codons and bring their their cognate amino acids to the ribosome those sites. This allows them to "suppress" the mutant phenotype caused by mutations in genes that convert a sense codon (codes for an amino acid) to a nonsense codon (codes for a stop). For example, a UUA codon is converted to a UAA codon. How many different tRNA types can be converted by a single nucleotide change in their anticodon sequence to one that will suppress a UAA nonsense codon?
To suppress a UAA nonsense codon, tRNAs recognizing CAA, GAA, AAA, UCA, UUA, UAC, and UAU can change by a single base to recognize a UAA nonsense codon to suppress a nonsense mutation. tRNAs recognizing UAG and UGA codons won't suppress a nonsense mutation because UAG and UGA are themselves nonsense codons. A single tRNA should recognize UAC and UAU because C and U are wobble bases; UAC and UAU should be decoded by a 3' AUG 5' anticodon. The answer should therefore be six.
Advantages of cDNA library / EST sequencing include:
• Direct access to protein coding genes, which are of primary interest to many scientists • RT using processed mRNA removes introns and generates ORFs that are useful for genetic engineering applications
Advantages of genomic library sequencing include:
• Represents the true genome, including all genic and non-genic sequences. Allows access to important gene sequences that fall outside of the coding sequence. • All sequences are present in proportion to their copy number in the genome, avoiding the imbalance in representation due to expression level differences.
Many antibiotics target steps in the translation process. Streptomycin is an antibiotic that interferes with the binding of fMet-tRNA, while tetracycline inhibits the aminoacylated-tRNA from binding. What area of the ribosome do these antibiotics bind? Propose an explanation for why these compounds are effective antibiotics targeting bacteria infections in humans without being toxic to the patient.
• Streptomycin binds to the P site • Tetracycline binds to the A site These drugs specifically inhibit prokaryotic translation, and do not interfere with eukaryotic protein synthesis. Why? Eukaryotic ribosomes, although closely related to the bacterial ribosome, differ enough in size and structure that they don't bind to these same antibiotics.
Explain how RNA-Seq analysis of mRNA populations gives information about patterns of alternative RNA processing?
1) Isolation of polyA containing mRNA. 2) Fragmentation of the RNA into ~200 base fragments. 3) Generation of double strand cDNA from the RNA fragments. 4) Sequencing of the cDNAs using massively parallel next generation sequencing (Illumina). 5) Computational analysis to map and quantitate the sequences to specific genes. ii) Because RNA-Seq uses sequencing to quantitate mRNA levels, the presence of and structures of alternatively processed mRNAs are directly revealed in the sequences. Exon/intron boundaries are detected as discontinuities between genomic and mRNA sequences.
What are two types of post-transcriptional modifications made in eukaryotic cells?
1. Splicing 2. Poly-A-tails 3. 5' Cap
The purpose of the Ames Test is to _______. A. test the mutagenic effects of chemicals B. determine whether Salmonella typhimurium his- mutants can revert to his+ C. determine whether histidine has mutagenic effects in S. typhimurium D. study how the liver affects potential mutagens
A. test the mutagenic effects of chemicals
How many Barr bodies are found in a normal human female nucleus? One Barr body. Two Barr bodies. Three Barr bodies. No Barr bodies.
One Barr body
Read the NPR story referenced below, and then list at least two ways in which eating dark chocolate may be beneficial to human health. In your answer include a brief explanation as to how gut microbes might be involved.
The NPR story cites reports suggesting that consuming dark chocolate may be beneficial to human health. The benefits include reducing blood pressure, reducing body weight and promoting a healthy heart. Chocolate is made from cocoa powder (after adding sugar and milk). Cocoa powder contains polyphenols, which have antioxidant properties. However, it is difficult for our gut to absorb polyphenols, as they are large molecules. Some of the microbe in our gut can break down these large molecules into smaller ones that still have beneficial properties, but can be more easily taken up by our gut cells, and thus can more readily enter the blood stream.
What role does the Shine-Delgarno sequence play in bacterial translation? How does the use of the Shine-Delgarno sequence allow the existence of poly-cistronic mRNAs in bacteria?
The Shine-Delgarno (SD) sequence is a short consensus sequence that lies 5- 13 bases upstream of the first AUG of the open reading frame (ORF) of bacterial genes. It recruits the small (30S) ribosomal subunit to the first AUG by base-pairing with a complementary sequence near the 3' end of the 16S rRNA. Ribosome binding to the SD sequence only requires that it be accessible, not that it be near the 5' end of the message. Downstream ORFs in polycistronic mRNAs need only have good SD sequences downstream of the stop codon from the ORF before it and upstream of their AUG initiation codons to be efficiently translated (See diagram).
How long would a sequence need to be unique in the human genome? Hint: 45 is equal to about 103 and the human genome is 3 billion base pairs long, 50% of which is repeat sequence DNA.
The probability of a particular 15 base sequence is 1/415 ~ 1/109 so a 16 base sequence would occur about 1 time in 3 billion base pairs. Therefore a sequence of around 20 bases would have a very good probability of being unique in the 1.5 billion base pairs of single-copy (non-repeat) human DNA. For the purpose of chromatin immunoprecipitation or gene expression experiments that use direct sequencing of ChIPped fragments or cDNA copies of mRNAs, 20bp of sequence is sufficient to map the sequence onto the genome. Initially the most popular next generation sequencing technology (callled Solexa or Illumina) sequence reads were only about 25bp in length, short but long enough to map to the genome. Current Solexa technology generates 100bp reads.
Why is it important to be able to characterize microbial communities using culture-independent techniques? Give an example of such a technique.
We are not been able to culture many microbes in the lab, so if we were to rely simply on culture-dependent methods, we would miss most of the microbes present in various environments and communities. Metagenomics is one of the most successful approaches for identifying or characterizing microbial communities without needing to culture individual microbial strains or species. This involves preparing DNA from the entire community, and then sequencing the DNA from the entire set of combined genomes. Targeted sequencing of specific genes, such as the 16S rRNA gene, provides information regarding what taxonomic groups (e.g. phyla, class, genus etc) of bacteria are present. Alternatively, shot gun sequencing of the entire genomes provides information about the combined biochemical functions of the community.
Disadvantages of genomic library sequencing include:
• For large complex genomes, only a small fraction of genomic clones will contain genic sequences. • In organisms with large introns, it is difficult or impossible to clone a gene on a single insert.
Disadvantages of cDNA library / EST sequencing include:
• Genes that are expressed at low levels will be extremely rare in a cDNA library, while highly expressed genes will be over-represented. • Important sequence elements of the gene are not present in the insert, including promoters, enhancers, and introns. • Most of the genome will be absent from the cDNA libraries... all intergenic regions.
Translationisaveryaccurateprocess.Whichstepsintranslationincludea process that helps assure the accuracy of the translation process? Explain your answer.
1) Establishing the 30S initiation complex at the first AUG of an open reading frame (ORF). The spacing between the Shine-Delgarno sequence (aka Ribsome Binding Site or RBS) and the complementary sequence on the 16S rRNA in the 30S ribosomone ensures that the correct start codon is placed in the P site, and thus the correct ORF is translated. 2) Linking of the correct amino acid to the appropriate tRNA, to generated a "charged tRNA". This is carried out by aminoacyl-tRNA synthestases, which have a mechanism that helps ensure that the correct amino acid is linked to a tRNA with the corresponding anticodon sequence. The "mischarging" rate is low: 1/105- 1/104 . 3) Insertion of charged tRNA into A site. Correct interactions between the tRNA anticodon loop and the codon sequences must occur before the hydrolysis of GTP on EF-Tu and the release of EF-Tu from the charged tRNA.
You have conducted an Ames test on a given compound. Which of the following would be classified as a positive result on the Ames test? A. His- strain grows on an his- plate. B. His+ strain grows on an his+ plate. C. His- strain grows on an his+ plate. D. His+ strain grows on either an his- or an his+ plate. E. His+ strain grows on an his- plate.
A. His- strain grows on an his- plate.
Which of the following accurately describes a possible meiotic nondisjunction event? Which of the following accurately describes a possible meiotic nondisjunction event? A. Homologs fail to separate during meiosis I. B. Fusion of gametes results in trisomy. C. Sister chromatids fail to separate during meiosis I. D. Meiosis fails to proceed to completion.
A. Homologs fail to separate during meiosis I.
The 5' splice site consensus sequence is AGGUAAGU. Where in this sequence is the exon/intron junction? A change of this sequence to AGGUACUU generates an inactive 5' splice site. What component of the splicing machinery would you have to change to reactivate this splice site? How would you change it? What consequence would that change have on the splicing of other mRNAs?
AG/GUAAGU Since the U1snRNP uses the 5' end of the U1snRNA to bind to this sequence you would have to change the U1 RNA sequence to restore Watson-Crick base-pairing. The changed sequence of the U1snRNA would have to be 5'-AAGUAC instead of 5'-ACUUAC to restore pairing. Obviously, although this change might restore splicing of the mutant intron, it would interfere with splicing of other mRNAs. You would also have to take into account the effect the mutation would have on the interaction of the U6 snRNA with the 5' splice site.
What are the functional consequences of acetylating and methylating lysine for transcription of eukaryotic genes?
Acetylation of lysine neutralizes its positive charge and provides a binding site for proteins. Neutralization of the positive charge results in a less compact chromatin structure by disrupting histone/DNA and internuceosome protein/protein interactions. Methylation of lysine provides binding sites for histone code readers, which may activate or repress gene transcription.
Which domain on the lac operon repressor protein affects the DNA binding domain shape?
Allosteric Domain
How does an epigenetically silenced gene differ from a mutant gene (a null, i.e. non- functioning, version of the same gene)?
An epigenetically silenced gene is non-functional in the sense that none of its gene product is produced. The lack of production is a consequence of the gene being "packaged" into chromatin that is unable to be transcribed, either because its transcriptional activators can't access the DNA or because they are unable to make the gene promoter accessible to the general transcription machinery. The gene sequence is the same as that of a gene copy that is normally expressed. A mutant gene has a change in the DNA sequence of the gene, either in its regulatory elements or in its coding region, that prevents the normal expression of the protein or the activity of the protein if it is expression is normal.
ThinkingaboutthedifferentrolesthatmRNAandtRNAplay,ifagenomic mutation led to an incorrect tRNA anticodon sequence, would this mutation likely have smaller or larger repercussions than a genomic mutation leading to a mutant mRNA (a mutation in a protein-coding gene)?
An incorrect anticodon sequence could lead to the incorrect amino acid being added to many different proteins in the cell. This would likely have large repercussions, and could even be lethal. In contrast, a mutation in an mRNA sequence would only affect one protein type in the cell.
During which stage of mitosis and/or meiosis do homologs segregate? A. Both anaphase I and II of meiosis B. Anaphase I of meiosis C. Anaphase of mitosis D. Anaphase II of meiosis
B. Anaphase I of meiosis
Like lac gene expression in bacteria, Gal gene expression in yeast is repressed by glucose (Described in Griffiths Chap12). Experimenters have shown that the binding of protein called Mig1 to sequences that lie between the Gal1/Gal10 UASG and the Gal1 TATA element is required for glucose repression of the Gal1 and Gal10 promoters. Deletion of those sequences prevents rapid glucose induced repression of Gal1 and Gal10 transcription; however, over a longer period of time, their transcription is repressed by glucose. Mig1 binds to a co-repressor called Tup1. What activity might be associated with Tup1? What might account for the long term glucose repression in the Mig1 binding site mutants? Contrast this system with the lac system in bacteria.
Based on our discussions in class, histone deacetylase activity might be associated with Tup1. Other possible activities would be chromatin remodeling activity whose function would be to make DNA sequences required for promoter activity less accessible to the transcription machinery or histone methyl transferase activity that put methyl groups on histone lysines that provided binding sites for proteins that made the promoter less accessible. One way to for glucose to repress the promoter in the absence of Mig1 binding would be for transcription of the Gal4 activator protein or the Gal3 inducer protein to themselves be glucose repressed, presumably by having Mig1 binding sites in their promoters. In fact, Gal4 transcription is glucose regulated by Mig1 binding to its promoter In the bacterial lactose system, lactose induction results from inactivation of a repressor protein while glucose repression results from failure to activate an activator protein. In the yeast galatose system, galactose induction results in activation of a activator (albeit by inactivation of an inhibitor of the activator), while glucose repression results from glucose activation of a repressor protein.
The parent cell that enters meiosis is diploid, whereas the four daughter cells that result are haploid. Which statement correctly describes how cellular DNA content and ploidy levels change during meiosis I and meiosis II? (Note: Ignore any effects of crossing over.) A. DNA content is halved only in meiosis I. Ploidy level changes from diploid to haploid in meiosis I, and remains haploid in meiosis II. B. DNA content is halved in both meiosis I and meiosis II. Ploidy level changes from diploid to haploid only in meiosis II. C. DNA content is halved in both meiosis I and meiosis II. Ploidy level changes from diploid to haploid in meiosis I, and remains haploid in meiosis II. D. DNA content is halved only in meiosis II. Ploidy level changes from diploid to haploid in meiosis I, and remains haploid in meiosis II. E. DNA content is halved only in meiosis I. Ploidy level changes from diploid to haploid only in meiosis II.
C. DNA content is halved in both meiosis I and meiosis II. Ploidy level changes from diploid to haploid in meiosis I, and remains haploid in meiosis II. During anaphase of both meiosis I and meiosis II, the DNA content (number of copies of chromosomes) in a cell is halved. However, the ploidy level changes only when the number of unique chromosome sets in the cell changes. This occurs only in meiosis I (where separation of homologous chromosomes decreases the ploidy level from 2n to n and produces daughter cells with a single chromosome set).
Why is aneuploidy in animals generally detrimental? A. It results in chromosome segregation errors in mitosis due to aberrant chromosome numbers. B. Synthesis of extra chromosomes creates a metabolic burden that decreases cell viability. C. It results in an imbalance of gene products from affected chromosomes, which alters normal development. D. It results in DNA replication errors due to the aberrant number of chromosomes.
C. It results in an imbalance of gene products from affected chromosomes, which alters normal development.
Generally speaking, which of the following mutations would most severely affect the protein coded for by a gene? Generally speaking, which of the following mutations would most severely affect the protein coded for by a gene? A. a base substitution at the beginning of the gene B. a base substitution at the end of the gene C. a frameshift deletion at the beginning of the gene D. a frameshift deletion at the end of the gene
C. a frameshift deletion at the beginning of the gene A frameshift mutation at the beginning of a gene would affect every codon after the point where the mutation occurred. During protein synthesis, incorrect amino acids would be inserted from the point where the frameshift mutation occurred on; the resulting protein would most probably be nonfunctional. For this reason, a frameshift mutation at the beginning of a gene is generally the most severe type of mutation.
Novel combinations of genes can arise from _______. A. reciprocal exchange of DNA between homologs during prophase II B. the alignment of chromosomes at Metaphase II C. reciprocal exchange of DNA between homologs during prophase I D. reciprocal exchange of DNA between sister chromatids during prophase
C. reciprocal exchange of DNA between homologs during prophase I
In the Ames Test, the appearance of his+ revertants in the presence of a non-mutagenic control compound indicates that _______. A. the growth medium contains factors that are mildly mutagenic B. liver extract increases the potency of some mutagens C. some of the reversion mutations are not caused by the mutagen being tested D. there is some low-level contamination in most experiments
C. some of the reversion mutations are not caused by the mutagen being tested His+ revertants on the control plate are the result of spontaneous mutation.
All of the following events occur during normal meiosis except _______. A. homologous chromosomes separate from one another during meiosis I B. one diploid cell produces four haploid cells C. sister chromatids separate from one another during meiosis II D. two haploid gametes fuse to form a diploid cell
D. two haploid gametes fuse to form a diploid cell
Propose a hypothesis for why a person's diet might influence the composition of their gut microbiome. Might this explain, at least in part, why the gut microbiota vary a lot from individual to individual?
Different diets consist of different relative amounts of carbohydrates, protein, fat and fiber. Further, in the case of vegetarians, individuals consume a lot more plant-based or dairy-based protein versus meat- based protein. It is not difficult to imagine that different consortia of bacterial develop within a person's gut in response to the type of food ingested by the person. Fiber seems to be very important- there are a lot of new articles about the (e.g. fermentation importance of fiber in maintaining a healthy gut microbiome. For example, see the Scientific America article below. NOTE: This is for your own information, only. We will not test you on this article.
A research scientist is performing a cloning experiment in which a fragment of the yeast genome is ligated into a vector. The vector is then transformed into E.coli. Explain the role of E.coli in this experiment.
E.coli is serving as the host organism and is being used as a factory to produce more of the recombinant vector (vector + yeast gene insert). By growing large amounts of E.coli (which is easy and cheap), the research scientist is able to obtain large amounts of the vector with the cloned yeast gene insert. It is also relatively easy to break open E.coli and purify the recombinant vector.
Based on our description of galactose gene transcription in yeast, inactivating mutations in which of these "Gal" genes would result in yeast mutants which would be unable to grow on galactose. Explain your reasoning. GAL1, GAL4, GAL80, GAL3.
GAL1-unable to grow. Gal1 codes for an enzyme involved in metabolizing galactose; lack of the enzyme would prevent yeast from growing on galactose. GAL4-unable to grow. Gal4 codes for a transcriptional activator protein that binds to sites upstream of genes required for growth on galactose and activates their transcription in the presence of galactose. GAL80-able to grow. Gal80 binds to and inhibits the activation domain of Gal4 in the absence of galactose. Its absence would result in constitutive (unregulated) synthesis of enzymes used to metabolize galactose; wasteful but not growth inhibitory in the presence of galactose. GAL3-unable to grow. Gal3 binds to galactose to inhibit Gal80 inhibition of Gal4. In the absence of Gal3, galactose cannot "induce" Gal4 activity and galactose cannot be metabolized.
For each of the following terms, indicate whether it is associated with gene activation, gene repression, or both. Give a one sentence defense of each answer.
Mediator-activation, It is a complex that binds to both activator proteins and RNA polymerase. DNA methylation-repression, methylation at C5 position of cytosine, provides a docking site for Histone deacetylase which promotes repressed chromatin state. Nucleosome remodeler-both, depending on nature of remodeling, in principle it could either activate or repress transcription. Histone acetylation-Activation-neutralizes lysine charge leading to less condensed chromatin. Also provides binding sites for Readers that are part of the transcriptional activation machinery. Heterochromatin-Repression-Condensed chromatin state resulting from specific histone methylation marks (H3K9 mostly), which are read by factors that set up the repressed state.
Cis-regulatorysequencesareshortsegmentsofDNAtowhichregulatory proteins bind to activate or repress the transcription of genes. In this case, "cis" refers to the fact that they regulate expression of genes that are on the same chromosome on which they reside. Eukaryotic cis-regulatory sequences controlling transcription are often referred to as "modular". Explain what is meant by this characterization. Use the Sonic Hedgehog (Shh) example from class to illustrate your explanation.
Modular = composed of smaller parts that function together as a single unit. Transcriptional control regions are composed of multiple binding sites for different trans-acting factors. The combination of the different protein binding sites allows the assembly of modular transcriptional control regions (enhancers) that specify different patterns of gene expression. In development a module can be responsible for expression of a gene in a particular set of cells (location) at a particular time, as was seen for the Sonic Hedgehog gene. The ability of different enhancers to separately interact (through looping) to the same gene promoter allows a single gene to be multifunctional during development.
Listseverallocationswheremicrobialcommunitiesarefound:a)onourplanetand b) in or on the human body. If possible, explain how the community might impact or modify the surrounding environment.
a) Microbial communities that play important roles on our planet are located in: • Soil- where they are important for nutrient cycling, such as the nitrogen cycle • In ponds, • In the ocean, where they can play a role in helping decompose oil from oil spills (See question 8) • In the gut of cockroaches or ruminant animals, where they important for breaking down cellulose into digestible sugars b) There are many microbial communities in or on the human body, including: • Skin, Mouth, GI tract • They serve many important roles including "excluding" pathogens, and especially in our gut- helping digest our food, producing compounds such as vitamins, regulating the development of gut cells or our immune system.
Elongation Factors
• EF-Tu is also a G-protein. EF-Tu in the GTP bound form binds to charged tRNAs, stabilizes the labile linkage between the amino acid and the 3'OH on the tRNA. EF-TU brings the charged tRNA to the A site of the ribosome. If the codon/anticodon sequences are complementary, the GTP on the EF-Tu is hydrolyzed, and the EF-Tu- GDP bound form leaves, with the correct tRNA loaded into the A site. • EF-G is another G- protein. It is also called the "translocase". It binds to the A site, displacing the tRNA in the A site, thus shifts the tRNA that was in the A site into the P site (Note: this tRNA now has the growing peptide attached to it). Hydrolysis of the GTP on EF-G provides the energy required for EF-G to insert itself into the A site, forcing the mRNA and its bound tRNAs to move over one codon in the ribosome
Initiation Factors:
• IF3 prevents the premature assembly of the ribosome. IF3 binds to the 30S su and prevents the 50S su from binding until the 30 S Initiation complex has been formed. • IF2 binds to the initiator tRNA charged with formyl methionine (fMet), and brings it to the P site on the 30S subunit. This forms the 30S initiation complex, and establishes the correct ORF for the protein. IF2 is a G-protein. It is in the GTP-bound form when it brings tRNA-fMet to the P site. Upon binding of the 50S ribosome to the complex, GTP is hydrolyzed, and the IF2-GDP is released.
Release Factors:
• Release factors (RF) are involved in terminating translation. • We discussed RF1 and RF2 in class, although we did not mention the specific numbers. This class of RFscan bind to stop codons position in the A site, and in doing so promote the hydrolysis of the acyl bond between the final amino acid on the nascent peptide the 3-OH of the tRNA in the P site. RF1 recognizes UAA and UAG and RF2 recognizes UAA or UGA.
