BIO325_F23_Practice Problems_Unit III

BIO 325 Practice Problems for Unit III – Multiple Choice section 1-2. Shown below is part of the last protein-coding exon of the CFTR gene. 5' GGCTAAGATCTGAATTTTCCGAG .... TTGGGCAATAATGTAGCGCCTT 3' 3' CCGATTCTAGACTTAAAAGGCTC .... AACCCGTTATTACATCGCGGAA 5' 1. Which of the following sets of primers could you use to amplify the entire target DNA sequence shown above? A. 5' GGCTAAGATC 3' and 3' CCGATTCTAG 5' B. 5' GTAGCGCCTT 3' and 3' CCGATTCTAG 5' C. 5' GGCTAAGATC 3' and 3' CATCGCGGAA 5' D. 5' GTAGCGCCTT 3' and 3' CATCGCGGAA 5' 2. What is the length of your PCR product if each primer is 10 nucleotides in length, and the last protein-coding exon of the CFTR gene that lies between the two primers is 100 bp (base pairs)? A. 80 bp B. 90 bp C. 100 bp D. 110 bp E. 120 bp 3. One of the following sequences was obtained from a cloned piece of a genome that includes parts of two exons of a gene. The other sequence was obtained from the corresponding part of a cDNA clone. What is the sequence of an intron separating two exons? Sequence 1: 5’ TAGGTGAAAGAGTAGCCTAGAATCAGTTA 3’ Sequence 2: 5’ TAGGTGAAAGAAATCAGTTA 3’ A. 5’ AGCCT 3’ B. 5’ GTAGCCTAG 3’ C. 5’ TAGGTGAAAGA 3’ D. 5’ AATCAGTTA 3’ E. 5’ TAGGTGAAAGAAATCAGTTA 3’ 4. What information can be determined or predicted from the DNA sequence of a cDNA clone? A. Exon sequences B. Sequence of the promoter C. Intron sequences D. Amino acid sequence of the encoded polypeptide E. Both A and D F. All of the above

5-8. The human RefSeq of the entire first exon of a gene involved in Brugada syndrome (a cardiac disorder characterized by an abnormal electrocardiogram and an increased risk of sudden heart failure) is shown. The first exon includes the start codon. The genomic DNA of four people (1 – 4) was subjected to sequencing. The following sequences represent all those obtained from each person. Nucleotides different from the RefSeq are underlined. RefSeq: 5′ CA ACG CTT AGG ATG TGC GGA GCC T 3′ Individual 1: 5′ CA ACG CTT AGG ATG TGC GGA GCC T 3′ 5′ CA ACG CTT AGG ATG TGC GGA G A C T 3′ Individual 2: 5′ CA ACG CTT AGG ATG TG A GGA GCC T 3′ Individual 3: 5′ CA ACG CTT AGG ATG TGC GGA GCC T 3′ 5′ CA ACG CTT AGG ATG GCG GAG CCT 3′ Individual 4: 5′ CA ACG CTT AGG ATG TGC GGA GCC T 3′ 5′ CA ACG CTT AGG ATG TG T GGA GCC T 3′ 5. Which individual(s) show SNP compared to RefSeq? A. 1, 2 B. 3 C. 1, 2, 3 D. 1, 2, 4 E. 1, 2, 3, 4 6. Which individual is homozygous for SNP? A. 1 B. 2 C. 3 D. 4 7. Which individual is a compound heterozygote (AKA trans- heterozygote with two different mutant alleles)? A. 1 B. 2 C. 3 D. 4 E. None of them 8. If three out of four individuals have the disorder, what is the inheritance of Brugada syndrome? A. Dominant B. Recessive

12-13. A genetic counselor is examining a family in which both parents are known to be carriers for a recessive CFTR mutation. Their first child was born with the disease, and the parents want to assess the genotype of the fetus inside the mother’s womb. DNA samples from each family member and the fetus are tested by PCR and gel electrophoresis for an SSR marker within one of the CFTR gene’s introns. The following results are obtained: 12. How many individuals are heterozygous for the given SSR locus? A. 0 B. 1 C. 2 D. 3 E. 4 13. How many individuals are heterozygous carriers of the CFTR mutation? A. 0 B. 1 C. 2 D. 3 E. 4 14. You compare the brain cDNA library with the heart cDNA library. The brain cDNA library has clones that contain Gene A, which none of the heart cDNA library clones have. The heart cDNA library has clones that contain Gene C, which none of the brain cDNA library clones have. Both libraries have clones that contain Gene B, but the brain library has 10 times more of such clones than the heart library does. Which of the following is false ? A. Gene A is primarily transcribed in the brain. B. Gene C is primarily transcribed in the heart. C. Gene B is transcribed in both tissue types. D. Gene B is expressed 10 times more in the brain than it is in the heart. E. The difference in the cDNA clones represented is due to some genes present in one tissue type but absent in the other.

15-17. Wild oysters are diploids with 20 chromosomes and are fertile. A triploid oyster is produced by fertilization of a diploid egg with a haploid sperm. 15 . What is the basic chromosome number ‘ x’ for oysters? A. 1 B. 2 C. 10 D. 20 16. What is the number of chromosomes of balanced gametes produced from triploid oysters? A. 10 B. 15 C. 20 D. Both A and C E. A, B, and C 17. At what frequency are balanced gametes produced by triploid oysters? A. 1 ÷ 2 0 B. 1 ÷ 2 1 C. 1 ÷ 2 9 D. 1 ÷ 2 19 18. Which of these events can produce individuals with full-fledged Down syndrome? A. Nondisjunction of chromosome 21 during first meiotic division B. Nondisjunction of chromosome 21 during second meiotic division C. Mitotic nondisjunction of chromosome 21 D. Mitotic loss of chromosome 21 E. Both A and B F. Both C and D 19. Why is X chromosome aneuploidy more common than autosomal aneuploidy? A. Dosage compensation allows silencing of all but one X chromosome to maintain balance of X- linked gene expression. B. Dosage compensation allows equal amount of X-linked gene expression by either upregulation or downregulation of transcription depending on the number of X chromosomes in a cell. C. X chromosome is more likely to undergo nondisjunction than autosomes are. D. X chromosome is devoid of essential genes other than those that determine sex. E. It is okay to have increased or decreased expression of X-linked genes due to aneuploidy, but it is not okay for autosomal genes.

25. The strand shown is the RNA-like strand. How many possible open reading frames (ORFs: frames without intervening stop codons) that extend through the following sequence exist? 5’……… CTTACAGTTTATTGATACGGAGAAGG ………3’ A. 0 B. 1 C. 2 D. 3 26. One way by which a gene can be identified is sequence comparison between different species. Functionally important sequences evolve more slowly than nonfunctional sequences. Which of the following displays the highest degree of sequence conservation across species? A. Introns B. Protein-coding exons C. Noncoding exons D. Intergenic region between genes 27. What method is NOT useful to identify a human gene within a cloned genomic DNA fragment? A. Search for long ORFs. B. Align the sequence with the nucleotide sequences from the online database. C. Compare the genomic sequence with various cDNA clones. D. Look for repetitive sequences such as SSR (simple sequence repeat). E. Search for known protein domains from the predicted amino acid sequence. 28. How does SNP help us understand ancestry? A. Species or populations that are more closely related would share smaller number of SNP alleles than those that are more distantly related. B. Species or populations that share greater number of SNP alleles must have separated from each other more recently than those whose DNA sequences are more divergent. C. If some humans share a T allele of a SNP locus while other humans and all chimps have a G allele at the same locus, T is the ancestral allele and G is the derived allele. D. Different species or populations can be distinguished based on the number of SNP loci in the genome. E. Different SNP alleles arise through random insertion or deletion mutation. 29. Why is it essential to sequence many clones from multiple cDNA libraries to annotate a genome? A. Different genes are transcribed in different tissues. B. A gene may be spliced differently in different tissues. C. Some genes are transcribed rarely in certain tissues. D. Every cDNA library will represent an identical set of transcribed genes and multiple sequencing will support correctness of sequence data. E. A, B, and C F. All of the above

30. What is a true statement about PCR (polymerase chain reaction)? A. Two primers bind to the opposite ends of the same template strand. B. The target sequence is defined by specificity of the DNA polymerase. C. It requires a large quantity of intact DNA as the template. D. Both dNTPs (deoxy) and ddNTPs (dideoxy) are used. E. The target region is amplified exponentially with repeating cycles. 31. Why do most DNA polymorphisms not influence phenotype? A. Less than 2% of the human genome consists of codons within genes. B. Many mutations in codons don’t change the amino acid ( i.e. silent mutation). C. Deleterious mutations will disappear from population due to negative selection. D. DNA polymorphisms do not alter the genome. E. A, B, and C F. All of the above 32. Allele difference for which polymorphism cannot be distinguished by PCR amplification followed by gel electrophoresis? A. SNP (single nucleotide polymorphism) B. SSR (simple sequence repeat) C. InDel (insertion/deletion) D. CNV (copy number variant) 33. Which of the following descriptions of a large chromosomal deletion is true? A. Heterozygotes may show various defects if the deleted region contains genes that are haploinsufficient. B. Homozygous cells with large deletion will not survive. C. It may be caused by double-stranded breaks or aberrant crossing over. D. It can be detected by FISH (fluorescent in-situ hybridization). E. All of the choices are true. 34-35. The human IGF2 gene is autosomal and maternally imprinted. Two alleles of this gene encode two different forms of the IGF2 protein. One allele encodes a larger 60K (Kilodalton) blood protein; the other allele encodes a smaller 50K blood protein. In an analysis of blood proteins from a couple named Bill Sr. and Joan, you find only the 60K protein in Joan’s blood and only the 50K protein in Bill Sr.’s blood. They have two children, Jill and Bill Jr. Jill produces only the 50K protein, and Bill Jr. produces only the 60K protein. 34. If Bill Jr. is heterozygous, what must be the genotype of his mother Joan? A. She is homozygous. B. She is heterozygous for the two alleles. C. It is inconclusive with these data alone. 35. When Bill Jr. (heterozygous) makes sperm, which allele will be silenced? A. 50K allele will be silenced. B. 60K allele will be silenced. C. Neither allele will be silenced. D. Both alleles will be silenced.

41-43. The following figure shows the pedigree of a family in which a completely penetrant, autosomal dominant disease is transmitted through three generations. The alleles for a SNP locus are shown (A or C). The disease allele is rare and has a single origin in the pedigree. 41. Which of the four mating is informative (can provide information about linkage) given that the male involved in mating 1 is heterozygous for the disease gene? A. Mating 1, 4 B. Mating 2, 3 C. All four mating are non-informative. D. All four mating are informative. 42. Assuming linkage between the SNP and the disease locus, what is the genetic distance between the two loci? A. 6.25 m.u. B. 7.1 m.u. C. 10 m.u. D. 14.3 m.u. 43. What is the likelihood ratio for linkage between the SNP and the disease locus? A. ( 1 16 ) 1 × ( 15 16 ) 15 ÷ ( 1 2 ) 16 B. ( 1 14 ) 1 × ( 13 14 ) 13 ÷ ( 1 2 ) 14 C. ( 1 10 ) 1 × ( 9 10 ) 9 ÷ ( 1 2 ) 10 D. ( 2 14 ) 2 × ( 12 14 ) 12 ÷ ( 1 2 ) 14 Mating 1 Mating 3 Mating 4 Mating 2

44. Which of these events cannot produce either XXX individuals (full-fledged) or mosaics? A. Nondisjunction of X chromosome during first meiotic division B. Nondisjunction of X chromosome during second meiotic division C. Mitotic nondisjunction of X chromosome during embryonic development D. Mitotic loss of X chromosome during embryonic development 45. Why is the incidence of Down syndrome very high among the offspring of a parent with Down syndrome? A. One chromosome 21 cannot attach to spindle and will end up in roughly 50% of gametes. B. One chromosome 21 cannot attach to spindle and will end up in nearly all of the gametes. C. Two chromosome 21 cannot attach to spindle and will end up in roughly 50% of gametes. D. Two chromosome 21 cannot attach to spindle and will end up in nearly all of the gametes. 46. Packaging of DNA with histones to form nucleosomes A. is the first level of compaction for DNA to fit into a nucleus. B. occurs in eukaryotes but not in prokaryotes. C. increases basal transcription rates. D. is permanent and chromatin cannot be unpacked. E. Both A and B 47. Human X chromosomes contain a 450 kb region of DNA called the X inactivation center (XIC) that mediates dosage compensation. XIC contains the Xist gene, which is only transcribed from the future inactive X that will become a Barr body. Which of these scenarios will cause the affected chromosome to become a Barr body? A. X chromosome has a deletion in XIC. B. X chromosome has a mutant Xist promoter so it cannot be transcribed. C. A part of X chromosome containing XIC is fused to a part of autosome through reciprocal translocation. (Translocation does not interfere with the XIC function of this X chromosome.) D. This is the only X chromosome in a somatic cell. E. None of the choices 48-49. 48. How many of these are DNA sequences that regulate transcription? A. 1 B. 2 C. 3 D. 4 E. 5 49. How many of these are transcription factors that bind directly to their target DNA? A. 1 B. 2 C. 3 D. 4 E. 5 promoter, enhancer, insulator, splice donor/acceptor ( count as one ), RNA pol, basal factor, activator, repressor, coactivator, corepressor

55. The part of the last protein-coding exon of the CFTR gene contains a disease-causing SNP. What is the most efficient method to identify the mutation? A. PCR amplify the last exon and run the PCR product on an agarose gel to compare the size with that of the wild-type exon. B. PCR amplify the last exon and sequence the PCR product to compare the sequence with that of the wild-type exon. C. Genome sequencing and comparison to the RefSeq D. Exome sequencing and comparison to the RefSeq 56-57. The human IGF2 gene is autosomal and maternally imprinted. Two alleles of this gene encode two different forms of the IGF2 protein. One allele encodes a larger 60K (Kilodalton) blood protein; the other allele encodes a smaller 50K blood protein. In an analysis of blood proteins from a couple named Bill Sr. and Joan, you find only the 60K protein in Joan’s blood and only the 50K protein in Bill’s blood. They have two children, Jill and Bill Jr. Jill produces only the 50K protein, and Bill Jr. produces only the 60K protein. 56. What is the genotype of Bill Sr.? A. He is homozygous. B. He is heterozygous for the two alleles. C. It is inconclusive with these data alone. 57. If Bill Jr. is heterozygous, what must be the genotype of his sister Jill? A. She is homozygous. B. She is heterozygous for the two alleles. C. It is inconclusive with these data alone. 58. Which of the following facts supports discovery of a rare disease-causing mutation? A. A SNP changes a critical amino acid into a non-conservative (chemically dissimilar) amino acid. B. The SNP locus lies between genes and does not affect any known regulatory sequences. C. A small deletion results in dramatic reduction in mRNA expression, which leads to reduced amount of the protein. D. An SSR locus located in an intron of a gene has several alleles that are common in population. E. Both A and C 59. Which of the following descriptions of chromosomal rearrangements is false ? A. Inversion or translocation may place a gene in a new regulatory environment and alter gene expression. B. Cells homozygous for a large deletion are not likely to survive. C. Duplication usually does not alter phenotype unless the duplicated gene is dosage-sensitive. D. They are caused by double-stranded breaks or aberrant crossing over. E. They are relatively small changes in genome compared to point mutations.

For 60-61: See the following pedigree for a disease allele that is recessive and rare. (There is only one origin for the disease allele.) 60. This gene is paternally imprinted. In the somatic cells of individuals showing the trait (filled- in symbols), which allele is imprinted (silenced)? A. Wild-type allele from dad B. Mutant allele from mom 61. In what cells of the descendants in generation II and III will the disease allele from the female in generation I be expressed? A. II-2 male germ cells B. II-2 male somatic cells C. III-1 female germ cells D. III-1 female somatic cells E. Both A and B F. Both B and C G. A, B, and C 62-64. Seedless watermelons that you find in the supermarket are triploids with 33 chromosomes. 62 . What is the basic chromosome number ‘ x’ for watermelon? A. 3 B. 11 C. 22 D. 33 63. What is the number of chromosomes of balanced gametes produced from triploid watermelons? A. 11 B. 22 C. 16.5 D. Both A and B E. A, B, and C 64. At what frequency are balanced gametes produced by triploid watermelons? A. 1 ÷ 2 2 B. 1 ÷ 2 10 C. 1 ÷ 2 21 D. 1 ÷ 2 32 I II III IV 1 2 1 2

BIO 325 Practice Problems for Unit III – Free Response section 71. Two chromosome pairs, one solid homologous pair and one dotted homologous pair, are prepared from a metaphase cell for FISH. The solid and dotted chromosomes are nonhomologous. Each chromosome is divided into six regions. The gene moana is located in region A. A. You design a FISH probe specific for the gene moana . How many FISH signals will you see in this wild-type cell? B. Draw the metaphase chromosomes of a translocation heterozygote involving AB and PQ reciprocal translocation. Show all four chromosomes with letters labeled clearly. (Assume that sister chromatids have the same translocation.) C. How many FISH signals will you see in this translocation heterozygote ? D. Write the genotypes of two balanced gamete types produced from the translocation heterozygote in part b. Indicate clearly which regions are linked together on each chromosome found in each balanced gamete. E. How many FISH signals will you see in a translocation homozygote involving the same regions as in part b? F. Write the genotype of gametes produced from the translocation homozygote in part e. Indicate clearly which regions are linked together on each chromosome found in these gametes. What is the percentage of balanced gametes made by the translocation homozygote ? A B C D E F P Q R S T U

Different drugs have been developed to treat cystic fibrosis, and each drug functions distinctively to restore the function of the CFTR protein, which is a chloride ion channel. For example, ivacaftor has been approved for patients who have a G551D allele (a missense mutation), which produces a protein inefficient in transporting chloride ions across the membrane. Ivacaftor enhances the efficiency of chloride ion transport. E. Do you think ivacaftor will be effective for patients who are homozygous for a ΔF508 allele (a single amino acid deletion), which encodes a protein that cannot fold up properly and therefore is not inserted into the cell membrane? How about patients who have both mutant alleles (G551D /ΔF508)? F. What polymorphism is responsible for each mutation: G551D and ΔF508? G. ΔF508 allele is present in the frequency of 1/500, while a less prevalent G551D allele is found in the frequency of 1/10,000. What is the frequency of a compound heterozygote (AKA trans - heterozygote) that has both of these alleles? Show your calculation.

73. The following figure is part of a screenshot of the UCSC Genome Browser for the human LCT gene that codes for the enzyme lactase-phlorizin hydrolase (LPH), commonly known as lactase. The LCT gene contains 17 exons; the first exon is to the right of the figure and not shown, and the last exon is indicated by an arrow. The coding exons are depicted in taller blue rectangles and are distinguishable from a non-coding exon, such as a part of exon 17, depicted in a shorter blue rectangle. The horizontal lines between exons are introns. A. If the LCT gene is on the right arm of chromosome 2 (the LCT locus is labeled with a red vertical line at 21.3 on the idiogram found at the top), what is the direction of transcription: toward or away from the centromere ? And where is the promoter located in the RefSeq shown above (with blue lines and rectangles): to the left or to the right of the diagram ? B. In the lower part of the figure is the DNA sequence comparison for the LCT gene locus between the human genome and the genomes of eight different species. Regions in black are homologous between the human genome and the genome of the indicated species. Which vertebrate shares the highest degree of sequence similarity to the human LCT gene? And which region, the coding exons, non-coding exon, or introns, has the highest degree of sequence conservation? C. Based on the figure, which vertebrates would make an LPH enzyme that is smaller than the human enzyme? There is more than one possible answer, but name only two. D. The bar graph in the box illustrates the RNA-Seq expression data for 53 tissues averaged from 570 donors. What is the starting material for generating this data? What kind of information can be obtained from RNA-Seq experiment? E. Two vertical bars in the box are for small intestine ( taller bar ) and colon ( shorter bar ) tissues. The other 51 tissues generate no detectable LCT sequence reads. What can you infer from this RNA- Seq data with regard to the LCT expression? Your answer should compare the three tissue types: small intestine, colon, and the rest .