Bio1001A+2023+midterm+code+444+FINAL

.pdf

School

Western University *

*We aren’t endorsed by this school

Course

1001A

Subject

Biology

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by JudgeYak4063

October 2023 Biology 1001A Midterm Page 1 of 18 Code 444 Multiple Choice: (There are 42 questions ) Circle the best single letter choice for each of the following questions before transferring your answers to your scantron sheet. Note, for “multiple-multiple” style questions, more than one option may be correct (e.g. 1, 2 & 3 only). Part marks may be available for choosing some of the correct answers but choosing any incorrect answer earns a grade of “0” for that question. 1. Insects in the genus Cimex feed on the blood of mammals and birds. Bedbugs live in human homes and feed on the blood of humans; batbugs live in bat roosts and feed on the blood of bats. Bedbugs are considered a pest, so they are often sprayed with chemical insecticides; but batbugs are not. A recent study reported that bedbugs have become resistant to insecticides, but batbugs have not. How would modern evolutionary theory explain this difference in resistance? A. Mutations that improve resistance to insecticides have probably occurred in both bedbugs and batbugs. Individual bedbugs with mutations conferring resistance have a survival advantage compared to susceptible individuals, but individual batbugs with mutations conferring resistance do not have a survival advantage compared to susceptible individuals. Thus, resistance to insecticides has increased in bedbugs but not in batbugs. B. Mutations that improve resistance to insecticides have occurred in bedbugs, because bedbugs need to be resistant to survive. Batbugs do not need to be resistant, so they did not experience these mutations. C. When insects are exposed to insecticides, each individual adjusts its physiology to improve its resistance. Because bedbugs have been exposed to insecticides and batbugs have not, resistance to insecticides has improved in bedbugs but not in batbugs. 2. In humans, colour blindness is a recessive trait carried on the X chromosome. Imagine that a couple, John (father) and Rebecca (mother), have a colour-blind child with Klinefelter syndrome (XXY). Although John and Rebecca both have normal colour vision and are karyotypically normal, Rebecca’s father was colour-blind. Assuming the child’s karyotype is due to a single error in chromosome partitioning, in which of the following stages of meiosis might the error have occurred? 1. Non-disjunction in John during Meiosis I 2. Non-disjunction in Rebecca during Meiosis I 3. Non-disjunction in John during Meiosis II 4. Non-disjunction in Rebecca during Meiosis II A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct.

October 2023 Biology 1001A Midterm Page 2 of 18 Code 444 3. The first comprehensive theory of evolution was put forward by Jean-Baptiste de Lamarck, who is now best known for his transformational view of evolution. How would Lamarck have explained the pattern through which human immunodeficiency virus (HIV) becomes resistant to AZT? Assume that you have already explained HIV's lifecycle to Lamarck and told him how AZT works. A. Exposure to AZT triggers each and every virion to be more selective in which nucleosides their reverse transcriptase attaches to the growing strand of DNA. This trait is passed on to the next generation of virions. B. A small proportion of virions already have a mutation that makes their reverse transcriptase more selective in which nucleosides it attaches to the growing strand of DNA. This allows them to replicate in the presence of AZT, and these virions pass this mutation on to the next generation. C. Exposure to AZT triggers each and every virion to be more selective in which nucleosides their reverse transcriptase attaches to the growing strand of DNA. This trait is not passed on to the next generation of virions, but if AZT is still present they too will be triggered to become selective. 4. Imagine two populations of pine trees. Population A is on the west side of a mountain range and experiences a rainy environment with high water availability. Population B is on the east side of the mountain range, where little rain falls; this population experiences frequent droughts (periods of low water availability). DNA proofreading enzymes are equally effective, and population size is similar, in each population. However, generation time is much faster in Population A (which experiences high rainfall) than in Population B (which experiences low rainfall). That is, pines in Population A produce offspring in about 50 years, and pines in Population B produce offspring in about 100 years. In which population is a mutation that improves drought-resistance more likely to occur? Why? A. The mutation is equally likely to occur in either population, because if the environment changes in the future, either population may need to resist droughts. B. The mutation is equally likely to occur in either population, because mutations are random with respect to the needs of the population. C. The mutation is more likely to occur in population A , because its faster generation time provides more opportunities for mutations to occur. D. The mutation is more likely to occur in population B , because it occupies a dry environment where drought-resistant alleles will improve survival. 5. Which of the following is NOT part of the scientific method? A. Making predictions B. Trying to falsify your predictions C. Trying to prove that your theory is true D. Revising theories as new evidence comes to light

October 2023 Biology 1001A Midterm Page 3 of 18 Code 444 6. Why do many biologists not consider viruses to be alive? 1. Viruses do not use enzymes to reproduce. 2. In viruses, mutations are random. 3. Viruses do not contain genes. 4. Viruses are not made of cells. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct. 7. This image below shows a replication bubble in a stem cell where the left side is the end of the chromosome. Which of the following could occur at a given time during the replication process? 1. DNA polymerase I can add bases at point B. 2. Telomerase can add bases at point A. 3. DNA polymerase III can add bases at point C. 4. Ligase can seal a nick at point D. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3, and 4 are correct.

October 2023 Biology 1001A Midterm Page 4 of 18 Code 444 8. Which of the following statements correctly describe (an) aspect(s) of evolutionary theory? 1. All life is related through descent from a common ancestor. 2. Adaptation occurs when individuals respond to changes in their environment. 3. Populations can only evolve if they contain heritable variation. 4. Animals are more highly evolved than plants, and plants are more highly evolved than single- celled organisms. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct. 9. A common misconception about evolution is that it is always random and unpredictable. In what way(s) is evolution predictable ? 1. We can confidently predict that populations will become better and better suited to their environment, because all evolution reflects the actions of natural selection. 2. We can confidently predict that a mutation that provides an advantage in a particular environment is more likely to occur than a harmful mutation. 3. We can confidently predict that when the environment changes, new mutations will appear to improve the population's fit to its new environment. 4. We can confidently predict that if a mutation occurs that provides an advantage in a particular environment, it is likely to spread throughout the population. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct. 10. If the C-value for a diploid (2n) organism is 60 pg, how much DNA would be present in its karyotype? A. 30 pg B. 60 pg C. 120 pg D. 240 pg

October 2023 Biology 1001A Midterm Page 5 of 18 Code 444 11. Shikha and her partner, Rowan, both submitted their DNA to 23andMe and were delighted to discover that they are both karyotypically normal and heterozygous for a SNP (A673T) in the APP gene that is associated with a lower risk of Alzheimer’s disease. The APP gene is on chromosome 21. Shikha and Rowan’s first child, Shiro, was born with trisomy 21. However, only one of his copies of chromosome 21 carries the A673T SNP. Assuming Shiro’s trisomy 21 karyotype resulted from a single error in chromosome partitioning, in which of the following stages of meiosis might the error have occurred? 1. Meiosis I in the father. 2. Meiosis II in the mother. 3. Meiosis I in the mother. 4. Meiosis II in the father. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct. 12. A simplified and small portion of the meiosis process is shown in the image. Which of the following corresponds to the correct ploidy and genome copy number during the different stages? NOTE: Each answer option has the format of cell A – cell B – cell C. A. 2n/4C – 2n/4C – 1n/2C B. 2n/4C – 2n/4C – 1n/1C C. 2n/2C – 2n/2C – 1n/2C D. 2n/2C – 2n/2C – 1n/1C

October 2023 Biology 1001A Midterm Page 6 of 18 Code 444 13. Imagine you are a scientist studying epistasis in the Siberian tiger ( Panthera tigris altaica) . You are told that under normal conditions, the tiger has a gene called Trw2 that is epistatic to a gene called Rps3. Another gene, Tpvr4, is unrelated to both the other genes. You isolate one Siberian tiger and take a closer look at a portion of its genome and notice the pattern shown below. The genes are labeled in white boxes and transposable elements are represented as solid boxes numbered 1 through 7. Which of the following statements about the DNA of this tiger is correct? 1 It is possible that transposable element 7 is interfering with the expression of Tpvr4 gene. 2. Over evolutionary time, the accumulation of transposable elements in the Siberian tiger could affect its C-value. 3. It is likely that most of the transposable elements have been inactivated by the Siberian tiger host. 4. It is possible that Trw2 is no longer epistatic to Rps3. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct. 14. The image shows the products of a cell division. Which of the following statements is correct? [Hint: Think of all organisms (plant, fungi, algae), not just animals] 1. This could be a result of mitosis in a 2n=8 organism. 2. This could be a result of mitosis in a n=4 organism. 3. This could be a result of meiosis II in a 2n=8 organism. 4. This could be a result of meiosis I in a 2n=4 organism. A. 1, 2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1, 2, 3 and 4 are correct.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version