FTM 11 Genomics and Human Variation Problem Set Questions PDF

Practice Questions FTM 11 Genomics and Human Variation Questions Created by all faculty from Grenada and NU 1. A 39-year-old woman comes to the physician as she has just learned that she is pregnant. She is offered standard G-banded karyotype analysis as a prenatal test. Which of the following gene...

Practice Questions FTM 11 Genomics and Human Variation Questions Created by all faculty from Grenada and NU 1. A 39-year-old woman comes to the physician as she has just learned that she is pregnant. She is offered standard G-banded karyotype analysis as a prenatal test. Which of the following genetic abnormalities would this best be helpful to establish a diagnosis? a) a 500,000 bp deletion in a chromosomal b) trisomy c) a substitution mutation d) a frameshift mutation e) gene duplications 2. Which of the following describes a human acrocentric chromosome? a) a chromosome that lacks a centromere b) a chromosome that possesses repetitive rRNA sequences on one end c) centromere located in the central portion of the chromosome d) a chromosome containing two centromeres e) the long arm and short arms of the chromosome are different lengths 3. A teenage female patient is affected with β-thalassemia. Molecular diagnosis of her β-globin genes identifies a mutation that removes the 100 nucleotides immediately preceding the start site for transcription. What β-globin regulatory elements is most likely removed by this deletion? a) Polyadenylation signal b) CAAT & TATA boxes c) 3' UTR (untranslated region) d) 3' intron/exon boundary for intron 2 e) Termination codon 4. A G-band karyotype confirmed Down syndrome in a 2 year-old female patient. Of the choices below, what is most descriptive of the karyotype of this patient? a) 46,XY  b) 46,XX c) 47,XX d) 45,X e) 45,XY 5. Cells are being studied by light microscopy, and mitiotic spindle fibres attached to the chromosomes are observed. What is the name of this attachment site? a) telomere b) pseudoautosomal region c) euchromatin d) centromere e) intron 6. A scientist is studying a nuclear gene (POLG) which encodes DNA polymerase-g. This gene functions to replicate the mitochondrial genome in a human cell. Which part of this gene is not transcribed? a) The promoter b) The 5' UTR (region between the transcription start site and the translation start site) c) Exons d) Introns e) The 3’ UTR 7. A 7-year-old boy with a rare form renal distal tubule acidosis is brought to the physicians. Cells are biopsied and cultured from the patient to facilitate molecular studies. Genetic testing shows a predicted pathogenic mutation thought to affect splicing. The mRNA for the gene is analyzed and it is found that it is larger than the normal form of that transcript. Which of the following would most likely cause aberration? a) Mutation of splice donor site of intron 1 b) A non-sense mutation within exon 1 c) Insertion of a repetitive sequence within intron 1 d) Mutation of the splice acceptor site of exon 2 e) RNA editing of the mature mRNA 8. Jessica (all names are made fictional) is an unmarried woman that just gave birth to a healthy baby boy named Billy. Jessica’s family owns the Big Megatron Company and they are very rich. As a result, a number of males have expressed paternity interest in Jessica’s boy. The figure below represents a DNA analysis (DNA fingerprint of SSR loci) and shows the genotypes of a several marker loci. Based on these data, which individual has the highest probability of being the father? a) Tom could be the father b) Jerry could be the father c) Patrick could be the father d) James could be the father e) Neither Tom, Jerry, Patrick or James could be the father 9. Linkage analysis was used to place a polymorphic 14 bp tandem repeat in close proximity to a gene thought to be involved with a rare genetic disorder. What best describes this tandem repetitive DNA sequence? a) SNP b) CNV c) LINE d) SINE e) VNTR f) SSR 10. The human X-chromosome has more than a thousand protein coding genes on it, and the Y- chromosome has only a handful. To maintain balanced gene expression between the sexes a) One X-chromosome is deactivated b) Gene expression from the Y-chromosome is increased c) Gene expression from each X chromosome in a normal female expresses half d) The X-chromosome has a lot of genes, but they are not expressed Practice Questions FTM 11 Genomics and Human Variation Questions Created by all faculty from Grenada and NU 1. A 39-year-old woman comes to the physician as she has just learned that she is pregnant. She is offered standard G-banded karyotype analysis as a prenatal test. Which of the following genetic abnormalities would this best be helpful to establish a diagnosis? 1. a 500,000 bp deletion in a chromosomal 2. trisomy 3. a substitution mutation 4. a frameshift mutation 5. gene duplications Answer: B. Commentary – To be visible on a G-band karyotype, a chromosomal abnormality has to be large enough to affect the banding pattern; for deletions or insertions this is a change that is typically in the range of 5 – 10 Mb (mega, meaning millions of base-pairs), so a 500,000 bp deletion is invisible by G- band karyotype. Later in the course, we describe technologies that reliably detect deletions in this range. The presence of an extra chromosome (trisomy, not to be confused with triploidy), is directly seen by G-band karyotype, as there is the complete presence of an extra chromosome – so this is the best answer. A substitution mutation means that one base is substituted by another which only affects a single base pair. A frameshift mutation typically means that a few bases, are either inserted or removed from a gene (not a multiple of 3) – this changes the reading frame and usually means that a stop codon is created by chance. These are not detectable by G-band. The term gene duplication implies that a single gene is duplicated, in some cases this might be visible on a karyotype, in some cases it would not be. 2. Which of the following describes a human acrocentric chromosome? 1. a chromosome that lacks a centromere 2. a chromosome that possesses repetitive rRNA sequences on one end 3. centromere located in the central portion of the chromosome 4. a chromosome containing two centromeres 5. the long arm and short arms of the chromosome are different lengths Answer: B Commentary – The acrocentric chromosomes have the centromere very close to one side. On one side of the acrocentric chromosome is the q-arm, and on the other is what is often called a “satellite structure”. These small bits of chromosome capping the centromere of acrocentric chromosomes consist of hundreds of copies of the genes encoding rRNA. The cell needs a lot of these genes (repetitive) because the cell needs a lot of ribosomes to make all the proteins that are needed for cell function. A chromosome without a centromere is “acentric”, and is not stabile. If the centromere is located in the middle of a chromosome we call it metacentric. A chromosome with two centromeres is called dicentric. If the long (q) and short (p) arms are of different lengths, it is a submetacentric chromosome. Note: do not confuse the “satellite structures” of acrocentric chromosomes with “satellite DNA” which will be discussed in the genetics of cancer section of this course. 3. A teenage female patient is affected with β-thalassemia. Molecular diagnosis of her β-globin genes identifies a mutation that removes the 100 nucleotides immediately preceding the start site for transcription. What β-globin regulatory elements is most likely removed by this deletion? a) Polyadenylation signal b) CAAT & TATA boxes c) 3' UTR (untranslated region) d) 3' intron/exon boundary for intron 2 e) Termination codon Answer: B Commentary: the CAAT box is ~80 bp & the TATA is ~25 bp 5’ to the transcription start site, these sequences on the DNA would be removed based upon the information in the vignette. The polyadenylation signal, the 3’UTR, and the termination codon are all found near the 3’ region of the gene. Spice junctions are found internal in the gene. 4. A G-band karyotype confirmed Down syndrome in a 2 year-old female patient. Of the choices below, what is most descriptive of the karyotype of this patient? a) 46,XY b) 46,XX c) 47,XX d) 45,X e) 45,XY Answer C. – Commentary: The best answer here is C; but read below to see why it really isn’t correct. Note that Down syndrome is the presence of an entire extra copy of chromosome 21, so to be more precise a geneticist would describe the condition in this female patient as 47,XX +21. In this question, we are trying to emphasize that normally, humans have 46 chromosomes. It is interesting that there are only three autosomes that are tolerated in extra copies (13, 18, and 21). This is because out of the 22 autosomes, chromosomes 13, 18, and 21 have the fewest genes. An important genetic principle is illustrated here: in most cases, gain of a copy of DNA is usually less deleterious than loss of a copy. Note that there are no examples of a monosomy of an autosome that is compatible with human life. The sex chromosomes behave differently, and will be discussed more as we proceed with this course. For the other choices: In brief, 46,XY is a normal male karyotype, 46,XX is a normal female karyotype, 45,X is monosomy X, which is Turner syndrome, and is the only monosomy known to be compatible with life (we discuss this more later). Finally karyotype 45,XX, or 45,XY would indicate that the (typically normal appearing) patient had a particular chromosomal rearrangement called a Robertsonian translocation. We will discuss this balanced Robertsonian translocation later in the course (we come back to it). 5. Cells are being studied by light microscopy, and mitiotic spindle fibres attached to the chromosomes are observed. What is the name of this attachment site? a) telomere b) pseudoautosomal region c) euchromatin d) centromere e) intron Answer D. – Commentary: The telomere is a repetitive DNA structure found on the end of chromosomes. Telomeres are important for maintaining chromosomal integrity, and to ensure that the chromosomes don’t shrink during replication. The pseudoautosomal region is found on the X- and Y- chromosomes and is used for alignment during meiosis and mitosis to ensure appropriate chromosome segregation during division. Many of the genes on the pseudoatosomal region of the X-chromosome escape X-inactivation, and these genes are responsible for normal female development. Euchromatin describes chromosomal regions that have high transcriptional activity. The centromere is a constriction of the chromosome; during metaphase, a kinetochore (proteins) forms on the centromere, and microtubule spindle fibers hold this region in place to ensure appropriate segregation (disjunction) of the chromosomes during telophase. Introns are found on mRNA as intervening sequences between exons. Introns are post-transcriptionally processed out as exons are spliced together during maturation of the pre-mRNA to the mature mRNA. 6. A scientist is studying a nuclear gene (POLG) which encodes DNA polymerase-g. This gene functions to replicate the mitochondrial genome in a human cell. Which part of this gene is not transcribed?  a) The promoter b) The 5' UTR (region between the transcription start site and the translation start site) c) Exons d) Introns e) The 3’ UTR Answer A. Commentary: Do not get confused by the mitochondrial mention in the stem of this question. Recall that the vast majority of the proteins that have a mitochondrial function are actually encoded by nuclear genes. These genes are encoded in the nuclear genome (on nuclear chromosomes), mRNA is expressed in the nucleus, the protein is synthesized in the cytoplasm, and the protein is then imported into the mitochondria. Since the gene is expressed in the nucleus; the RNA is processed just like any other nuclear encoded gene would be and must be exported and translated in the cytoplasm, Proteins which are destined for the mitochondria have a specific sequence of amino acids “zip code” which helps to direct that protein to the mitochondria so that it may function in the correct place in the cell. This is called protein trafficking, and may also be referred to as “targeting” the protein to the mitochondria. There are just a handful of protein coding genes on the mitochondrial chromosome, these encode a few of the proteins used in the electron transport chain. 7. A 7-year-old boy with a rare form renal distal tubule acidosis is brought to the physicians. Cells are biopsied and cultured from the patient to facilitate molecular studies. Genetic testing shows a predicted pathogenic mutation thought to affect splicing. The mRNA for the gene is analyzed and it is found that it is larger than the normal form of that transcript. Which of the following would most likely cause aberration? a) Mutation of splice donor site of intron 1 b) A non-sense mutation within exon 1 c) Insertion of a repetitive sequence within intron 1 d) Mutation of the splice acceptor site of exon 2 e) RNA editing of the mature mRNA Answer A - Commentary: If the splice donor site of intron 1 is not used, then the intron, or a portion of the intron, will be retained in the processed mRNA – this will produce a longer transcript. Choices B, C, and E would have no effect on mRNA size. Choice D would likely cause a smaller mRNA to be created. Note: It is common convention to refer to a splice site as belonging to the intron. That is, splice acceptor (or donor) of (or for) intron 3; this makes it easier to count the exons and introns, and keep track of everything. In some cases, books or practice MCQs will describe spice sites with respect to the exon; this makes for a very difficult or impossible interpretation. We understand that this stuff is difficult enough, and we (in the Department of Biochemistry) have agreed to always describe (and name) splicing with respect to the intron (hopefully). 8. Jessica (all names are made fictional) is an unmarried woman that just gave birth to a healthy baby boy named Billy. Jessica’s family owns the Big Megatron Company and they are very rich. As a result, a number of males have expressed paternity interest in Jessica’s boy. The figure below represents a DNA analysis (DNA fingerprint of SSR loci) and shows the genotypes of a several marker loci. Based on these data, which individual has the highest probability of being the father? a) Tom could be the father b) Jerry could be the father c) Patrick could be the father d) James could be the father e) Neither Tom, Jerry, Patrick or James could be the father Answer: B. Commentary – Please excuse the ridiculous scenario described in the question. SSR polymorphisms are inherited and can be tracked by co-dominance. The gel is showing 4 markers, that means that there are 2 polymorphic loci being tested (each loci should give 2 bands; one from each chromosome unless two loci have the same size). The mother is in lane A, she gives one allele from each loci to her offspring. Only Jerry (lane D) has markers that are shared with the baby. Notice the words used in this question. With only 2 loci being tested, it would not be appropriate to claim that Jerry IS the father; at this point, we can only say that Jerry MIGHT be the father, and that the others cannot be the father. This now becomes an issue of probability, and statisticians can calculate how many loci need to be tested to eliminate a match by chance (these calculations are not discussed further in this course). 9. Linkage analysis was used to place a polymorphic 14 bp tandem repeat in close proximity to a gene thought to be involved with a rare genetic disorder. What best describes this tandem repetitive DNA sequence? a) SNP b) CNV c) LINE d) SINE e) VNTR f) SSR Answer E. Commentary: Variation in the human genome can take many forms: SNP - (Single Nucleotide Polymorphism) is a relatively common single base pair change which is found in approximately one in a thousand base pairs. The term polymorphism is reserved for those DNA changes that are relatively frequent in populations (most experts agree that this is ~1% of the population). CNV - is Copy Number Variant. This describes changes in copy number but does not apply to polymorphic sites such as the SSR, and the VNTR. We continue to discuss the CNV as we progress with this course. LINE and the SINE are Interspersed Nuclear Elements (Long and Short, respectively). These are highly repetitive DNA sequences that are found interspersed throughout the genome. The long ones can create their own reverse transcriptase so that they can make copies of themselves, and propagate in the genome. The short ones have the ability to hijack reverse transcriptase from some other source, and they are also able to propagate in the genome. SSR - is the Simple Sequence Repeat. The SSR is a highly polymorphic sequence that consists of di, tri or tetra nucleotides repeated tandemly in the genome. VNTR – is the Variable Number Tandem Repeat. These are tandem repeats that are bigger than the SSR, but less than 100s or 1,000s of base pairs. These are polymorphic, but not as variable as the SSR. The vignette in the question above describes the VNTR. LCR – the Low Copy Repeat. These can be quite large, (1,000 of bp, and bigger), and are found in only a few places in the genome, and they are usually very similar (or the same) between individuals. As a general rule, the shorter the repeat sequence, the more variability it is likely to have. 10. The human X-chromosome has more than a thousand protein coding genes on it, and the Y- chromosome has only a handful. To maintain balanced gene expression between the sexes a) One X-chromosome is deactivated b) Gene expression from the Y-chromosome is increased c) Gene expression from each X chromosome in a normal female expresses half d) The X-chromosome has a lot of genes, but they are not expressed Answer A. Commentary: Since males have half the gene dosage than females for all genes on the X- chromosome, one X-chromosome must be turned off in females so that gene dosage is the same among the sexes. Some genes on the X-chromosome escape inactivation, this is what provides the genetic information to allow a female to develop normally.