BIOL 2153 Principles of Genetics Final Exam | Sections 7–12 | Complete Questions & Answers | 100% Correct | Fall 2026.

BIOL 2153 Principles of Genetics Final Exam Final Exam (Grade 1) for Principles of Genetics, Fall Term 2025, Section 7-12. Bubble in your name and student identification number on SIDE TWO of the scantron. Also, you need to mark in the grade section whether your exam is Grade 1 or 2. The correct answer for each question will be awarded 3 points (3 x 33 = 99), and 6 points from two bonus questions (3 x 2 = 6) will be added to make 105 points. Keep this exam booklet for your record. 1. The following components are usually required for DNA replication. Which of the following lists all the components for a successful reaction for PCR? a. dNTPs b. ddNTPs c. NTPs d. heat-stable DNA polymerase e. genomic DNA f. a set of single-stranded DNA primers g. a set of single-stranded RNA primers A. a+d+e+f B. b+d+e+f C. c+d+e+f D. a+d+e+g E. b+d+e+g 2. In PCR, we need to increase temperature up to 94oC to convert double-stranded genomic DNA into single-stranded DNA by breaking ____________, which is equivalent to the action by ____________. A. Hydrogen bond Polymerase B. Hydrogen bond Helicase C. Hydrogen bond Ligase D. Covalent bond Polymerase E. Covalent bond Helicase (Q3-7) Human ABO gene is responsible for blood types, and this gene is located human chromosome 9. SSLP#1 is located within one of the introns of human ABO gene. Among human population, the ABO gene has three individual alleles, IA, IB, and i. Similarly, PCR of SSLP#1 also produces 100, 120, 140 base pair(bp) products among human population. We also have two additional SSLPs, SSLP#2 and SSLP#3. SSLP#2 is from human chromosome 21, and PCR of SSLP#2 produces 200, 220, 240 bp products among human population. SSLP#3 is from human X chromosome, and PCR of SSLP#3 produces 300, 320, 340 bp products. We genotyped one family with these SSLPs, and the results are shown below. Husband: SSLP#1 (100, 140), SSLP#2 (220, 240), SSLP#3 (300) Wife: SSLP#1 (100, 120), SSLP#2 (200, 220), SSLP#3 (320, 340) Son1: SSLP#1 (100), SSLP#2 (200, 220), SSLP#3 (340) Son2: SSLP#1 (120, 140), SSLP#2 (200, 240), SSLP#3 (?) Daughter1: SSLP#1 (100, 120), SSLP#2 (220, 240), SSLP#3 (?) Daughter2: SSLP#1 (100, 140), SSLP#2 (200, 220, 240), SSLP#3 (300, 320) 3. Husband and Wife are A and B blood type. Then, which child is AB and O type, respectivly? A. Son1 Son2 B. Son2 Son1 C. Daughter1 Daughter2 D. Daughter2 Daughter1 E. none of the above 14. Which child is affected by the Down syndrome? A. Son1 B. Son2 C. Daughter1 D. Daughter2 E. none of the above 5. While we were doing genotyping, we didn't label the PCR tubes for Son2 and Daughter 1 for SSLP#3. Let’s assume Son2 and Daughter1 are normal without any major health effects. Then, what would be possible SSLP#3 for Son2? A. (300, 340) B. (300, 320) C. (300) D. (320, 340) E. (320) 6. Also, what would be possible SSLP#3 for Daughter1? A. (300, 320) B. (300) C. (320, 340) D. (340) E. (320) 7. If Son2 is a patient of Klinefelter syndrome, what would be possible the genotype for SSLP#3? A. (300) B. (320, 340) C. (340) D. (320) E. no detectible band 2(Q8-10) The diagram below illustrates the genotyping results of one family with the Cystic Fibrosis disease. Both parents and the first child are normal, but the second child is affected with this disease. It has also been shown that Allele 2 usually co-segregates with this disease. While performing this genotyping, unfortunately, we made two major mistakes: 1) we did not label the PCR reaction tubes, and also 2) we accidently included two additional samples that are not related to this family. Thus, we do not know which genotype represents the allele combination of each family member. Answer the following questions. 8. Which genotypes represent two parents? A. 1/1 and 1/3 B. 1/3 and 2/3 C. 1/2 and 2/3 D. 1/1 and 3/3 E. 2/2 and 3/3 9. Which genotype would represent the first child without the cystic fibrosis disease? A. 1/2 B. 1/3 C. 2/3 D. 1/1 E. 3/3 10. Which genotypes are derived from those unrelated to this family? A. 1/1 and 1/3 B. 1/3 and 2/3 C. 1/2 and 1/3 D. 1/1 and 2/2 E. 1/1 and 3/3 3(Q11-14) The diagram shows you a pedigree of the family affect with Neurofibromatosis type 1 (NF1). One SNP is linked to this disease, and thus its genotype is also shown along with the pedigree (G or T). 11. Based on the pedigree, the inheritance pattern of NF1 is considered to be __________. A. autosomal dominant B. autosomal recessive C. X-linked dominant D. X-linked recessive E. none of the above 12. Which allele combination does the individual II-1 have? A. NF-G/ NF+-T B. NF-G/ NF+-G C. NF-G/ NF-T D. NF-G/ NF-G E. none of the above 13. Which individual has the crossing-over product from the individual II-1 and genetic distance? A. III-3 12.5% B. III-5 12.5% C. III-8 12.5% D. III-5 25.0% E. III-8 25.0% 14. Which individuals have the following linkage, NF+-G? A. III-1 B. III-3 C. III-5 D. III-8 E. all of the above 415. Genetic markers are two or more variations (alleles) for a given locus in chromosome. Genetic markers usually have detectable phenotypes at the molecular level, meaning that we can differentiate two DNA sequence variations through molecular biology techniques. Which of the following requirements need(s) to be satisfied to be used as genetic markers? A. need to be variable within populations, thus becoming different alleles B. need to be easily detectable with molecular biology techniques C. need to have clear information regarding chromosomal position D. all of the above E. none of the above 16. As we have a defense system against incoming organisms, bacteria also have a defense system. In the bacterial system, the incoming organisms are usually bacteriophages, which is parasitic virus. To protect against this bacteriophage, bacteria produce enzymes, which can digest the DNA of the incoming virus. This DNA-cutting enzyme has been named Restriction enzyme, since it restricts the proliferation of virus. At the same time, bacteria can protect their own genomes against these DNA-cutting enzyme through modification on their own DNA, which is usually adding ________ on the _______ of DNA. A. methylation ribose B. methylation deoxyribose C. methylation base D. phosphorylation deoxyribose E. phosphorylation base 17. For the class, Dr. Kim brought a set of embryos that have been modified at the mouse locus called Peg3 through a mouse _________ experiment. This particular experiment has inserted a ____________ gene into the Peg3 locus so that the embryos show the blue color as a reporter gene in the tissues where Peg3 is normally expressed. This technique has been successful for demonstrating the predicted function of Sry and Huntingtin genes. A. knockout experiment bacteria B. knockout experiment human C. transgenic experiment bacteria D. transgenic experiment jellyfish E. transgenic experiment human 18. The Sanger sequencing method is mimicking DNA replication, thus requires all the components for DNA replication. Furthermore, this method utilizes one innovative idea involving substrate analogs, which is called dideoxyNTPs, or ddNTPs. These analogs are exactly same as dNTPs except these analogs are missing another ____________ group at Carbon ___________ of the ribose. Normal dNTPs use this group to connect to an adjacent nucleotide through phosphodiester bond. Since ddNTPs do not have this group, DNA polymerase cannot extend anymore and ends up stopping, thus named the chain termination method. A. hydroxyl No 2 B. hydroxyl No 3 C. hydroxyl No 5 D. phosphoryl No 3 E. Phosphoryl No 5 519. The following three technologies can be used for curing genetic diseases in the future. First, Personal genomics allows us to sequence the entire genome of individuals at a very low cost within a couple of days. By applying this technique, we can have all the genotypes of the entire gene set. Second, with the ________ technology, we can have our own stem cell population that can develop into any organ or tissue for our body. If our body runs out of muscle cells, which is a common cause for many devastating diseases, we can derive the needed muscle stem cells in petri dish and put back into our body. Third, _______ -based genome editing can allow us to edit or correct mutations in our genome so that we may be able to avoid genetic disorders or diseases. A. SCNT CRISPR B. SCNT zinc finger nuclease C. iPS CRISPR D. iPS zinc finger nuclease E. none of the above (Q20-23) We performed a trihybrid cross experiment as shown below. The first gene (A) follows the inheritance pattern of complete dominance/recessive, but the second and third genes (B, C) follows the incomplete dominance pattern. P AABBCC x aabbcc F1 AaBbCc x AaBbCc F2 20. How many different phenotypes and genotypes were found at the F2 progeny? A. 12 phenotypes and 18 genotypes B. 12 phenotypes and 27 genotypes C. 18 phenotypes and 12 genotypes D. 18 phenotypes and 18 genotypes E. 18 phenotypes and 27 genotypes 21. What fraction of F2 progeny will have the AABbCC genotype? A. 1/4 B. 1/6 C. 1/8 D. 1/16 E. 1/32 22. Let’s suppose the first gene follows the inheritance mode of lethal allele, but not complete dominance, meaning that AA is not viable. Also, let’s suppose that A is dominant to a. Then, how many different phenotypes and genotypes were found among all the viable F2 progeny? A. 12 phenotypes and 18 genotypes B. 12 phenotypes and 27 genotypes C. 18 phenotypes and 12 genotypes D. 18 phenotypes and 18 genotypes E. 18 phenotypes and 27 genotypes 623. Also, if the first gene follows the inheritance mode of lethal allele, what fraction of F2 progeny will have the AaBbCc genotype? A. 1/4 B. 1/6 C. 1/8 D. 1/16 E. 1/32 (Q24-26) 24. A woman has the following genotype (A B/a b, C/c, D/d), and the two genes, A and B, are located on a same chromosome. The genetic distance between these two genes is close to 50%. Then, how many genetically different eggs can be formed from this woman? A. one B. two C. four D. eight E. sixteen 25. If the genetic distance between A and B is close 0%, then how many genetically different eggs can be formed from this woman? A. one B. two C. four D. eight E. sixteen 26. Mendel used two traits of garden pea, color and shape, to derive the Independent Assortment Law. Let’s assume that the two genes responsible for Color (Yellow and Green) and Shape (Round and Wrinkled) are located on a same chromosome with their genetic distance being 10%, and that Yellow is dominant to Green, and Round is dominant to Wrinkled. Let’s also assume that you have repeated the dihybrid experiment. The pure-bred Yellow Round was crossed with the pure-bred Green Wrinkled, and later the dihybrid F1 progeny with Yellow Round was selfcrossed. Also, you have counted each phenotype category. Then, which categories will show up as recombinant categories with crossing-over among F2 progeny? A. Yellow Round Yellow Wrinkled B. Yellow Wrinkled Green Round C. Green Round Green Wrinkled D. Yellow Round Green Wrinkled E. none of the above 7(Q27-30) Take a look at the pedigree shown below and answer the following questions. 27. Which type of the inheritance pattern does the above disease follow? A. autosomal and dominant B. autosomal and recessive C. X chromosomal and dominant D. X chromosomal and recessive E. none of the above 28. Which individual(s) are heterozygous for the disease allele with 100% confidence? A. 2 B. 3 C. 15 D. 16 E. none of the above 29. What would be the probability of having an affected child from the potential marriage between individuals 12 and 14? A. 1/16 B. 1/12 C. 1/6 D. 1/4 E. none of the above 30. If this disease occurs in one out of 100 newborn babies, what is the allele frequency of the disease allele? A. 1/10 B. 1/100 C. 1/1,000 D. 1/10,000 E. none of the above 8(Q31-33) The following double-stranded DNA sequence code for a protein, which is 9-aminoacid long. To help you choose the right combination of START and STOP codons, these are marked with underlines and bold-type, respectively. 5’- AGCTCATGGCCGGTAGCTACGGTGACAGCTGGCTCCATGACTCCATGCC - 3’ 3’- TCGAGTACCGGCCATCGATGCCACTGTCGACCGAGGTACTGAGGTACGG - 5’ 31. Which strand codes for this protein? A. top B. bottom C. both D. all of the above E. none of the above 32. Which direction does this RNA polymerase move to make this mRNA? A. left to right B. right to left C. both directions D. all of the above E. none of the above 33. What is the right anticodon for the 4th amino acid? A. 5'-AGU-3' B. 3'-AGU-5' C. 5'-ACC-3' D. 3'-ACC-5' E. none of the above 34. For the class, Dr. Km brought one novel detailing several families’ struggles for their children, who happen to have deleterious genetic mutations. The title of this novel is The ___________ fix written by Ricki Lewis. This novel chronicled their trials of ________ to fix the mutations. A. final gene therapy B. forever gene therapy C. final CRISPR therapy D. forever CRISPR therapy E. none of the above 35. According to your instructor, Genetics is understanding 1) inheritance patterns with the behaviors of ____________ chromosomes and 2) dominance and recessive with the molecular features of _______________. A. meiotic DNA B. meiotic DNA and RNA C. meiotic DNA, RNA and protein D. mitotic DNA E. mitotic DNA and RNA 9