year 1: bio midterm understanding (cycle 1-5) flashcards

Three flasks each contain equal numbers of bacteria, reproducing at equal rates.

• Flask X → treated with the antibiotic ampicillin;

• Flask Y → treated with two antibiotic drugs (ampicillin and bacticillin);

• Flask Z are not treated with any antibiotic drugs.

1. Bacteria in which flask is most likely to develop antibiotic resistance through mutation?

2. After a long time, which flask will see bacteria resistant to ampicillin only dominate?

A. All flasks are equally likely for this.

B. Flask X.

C. Flask Y.

D. Flask Z.

E. Flasks X and Y.

Question 1 – the answer is option A.

• Mutations happen not in response to any environment changes, but they happen randomly

• Each flask is likely to undergo this random mutation, as nothing truly “forces” a mutation to happen

• Students associate “chance of mutation happening” with the “effect/end product of mutation” happening, explained in Question 2:

Question 2 – the answer is option B.

• Each flask is equally likely to see a mutation (resistance, or something else)

• But the question asks what we would “see,” and we’d see amp-resistant-only bacteria only in Flask X as non-resistant bacteria will die

• In Flask Z, all strains of bacteria are equally likely

• In Flask Y, we’ll only see bacteria resistant to both

• Just because we see amp-resistant bacteria in Flask X, doesn’t mean it is the only flask in which that resistance evolved!

Which of the following traits is shared by all life and viruses?

A. Genetic information is contained within a protein shell.

B. Genes are transcribed from a RNA code to a DNA message.

C. DNA or RNA genomes are copied without error or mutations.

D. Copies of DNA or RNA genomes are inherited by offspring.

The answer is D.

A — this is incorrect, as not all life forms have a protein shell around genetic information. Viruses do this (called a capsule) but bacteria (DNA contained in plasma membrane and cell wall, which are lipids and proteins) and eukaryotes (DNA contained in a nucleus, which is not 100% protein) are exceptions.

B — this is incorrect, as while some life forms rely on reverse transcription (retroviruses), most forms of life take DNA code to make RNA messages (transcription).

C — this is incorrect, as mutations happen everywhere for all organisms.

D — this is correct, as during cell division, daughter cells (and organisms!) receive genetic information from their parent.

The genome of the human immunodeficiency virus (HIV) is approximately 9500 nucleotides long and encodes genes for 15 proteins that are used by HIV in its life cycle. Currently, standard treatment for an HIV infection requires a patient to take several antiviral drugs together that interfere with proteins functioning at multiple points of the HIV life cycle. Under this drug cocktail regime, HIV will continue to evolve.

Where will mutations most likely appear in the genomes of HIV collected from a patient taking a drug cocktail?

A. In the reverse transcriptase gene.

B. In genes where a mutation will be beneficial.

C. In genes that are targeted by the antiviral drugs.

D. In any gene, but not in non-coding regions.

E. Anywhere in the genome.

E

Which statement is NOT consistent with the theory of “evolution?”

A. Nectar-sucking birds, on average, have longer beaks than 100 years ago.

B. All life on Earth originated from one single species billions of years ago. 

C. Species change is mostly driven by external factors, not necessarily because the organisms want to change.

D. Organisms can simply decide to change over their lifespan, and these purposeful changes drive species-level change.

The answer is D.

A — this is incorrect. While parts of this statement suggests some part of the Lamarckian theory, and does not have concrete aspects of the Darwinian theory, it is trumped by Option D which is a direct reference to Lamarck.

B — this is incorrect. Lamarck does not outline that all species are related, Darwin did.

C — this is incorrect. External factors changing species is the core aspect of Darwin’s ideas.

D — this is correct. Lamarck proposed an individual-driven evolution, and this is the most relevant statement.

Which statement describes possible evidence that could falsify the theory of evolution?

A. A new type of microorganism is discovered whose cells are identical to one another and never mutate.

B. There are species that undergo mutation, but their reproductive speeds are so slow that they do not appreciably change.

C. A new species of frogs is discovered in one specific place on Earth, nowhere else.

D. The theory of evolution can never shown to be false.

The answer is A.

A — this is correct. Evolution states that all life is related, and all life changes. If we find one form of life that never changes, that would falsify the theory’s claim that ALL life changes.

B — this is incorrect. The first part of the statement that mutation happens in some species, even if very, very slow, agrees with the theory of evolution that life changes.

C — this is incorrect. Even though these frogs are isolated in one habitat, they can still be related (and will be!) to other organisms.

D — this is incorrect, as the theory of evolution is a scientific theory, which intrinsically can be suggested false.

T OR F: In our modern world, evolution is occurring at a significantly lower rate than it did millions of years ago.

FALSE: Evolution is certainly ongoing! All conditions for evolution (variation, non-random survival and heritability) are still present in today’s organisms. Even humans are still evolving; this will be explored in greater detail in further lectures.

T OR F: Individuals with a greater fitness are almost always stronger, faster, and larger, seeing that these traits are conducive to reproductive success.

FALSE: Fitness is simply defined as reproductive success (how many offsprings an individual has). It has nothing to do with the word ‘fitness’ in its everyday use. In the evolutionary sense, an animal that has 5 offspring is more ‘fit’ than an animal that has 4 offspring, even if the first animal is smaller, uglier, less intelligent, etc.

T OR F: Chimpanzees evolved into humans

FALSE: Chimpanzees did not evolve into humans; rather, chimpanzees and humans share a common ancestor from which both the present-day humans and the present-day chimpanzees evolved.

T OR F: An individual organism cannot evolve over the course of its lifetime.

TRUE: is variational, not transformational. Usually, traits that an organism acquires over the course of its lifespan are not passed onto the next generation. 

T or F: A bacterium plated in an antibiotic medium is equally as  likely to develop a mutation that confers antibiotic resistance than bacteria plated on nutrient agar. 

TRUE: Remember that MUTATION IS RANDOM, therefore all bacterium plated in an antibiotic medium is EQUALLy likely to develop the mutation

T OR F: Evolution does not result in organisms that are perfectly suited to their environment.

True

Which of the following statements is NOT part of the theory of evolution?

A. Over their lifetimes, individuals evolve to become better suited to their environment.

B. Over generations, populations evolve to become better suited to their environment.

C. In an environment with tall trees, mutations in giraffe genomes that result in longer necks will occur just as frequently as those that result in shorter necks. 

D. All life on earth is related through descent from a common ancestor.

The correct answer is A.

1. A is NOT part of the theory of evolution (and is therefore correct). As we discussed, evolution occurs in populations, not individuals. Individuals do not evolve over their lifetimes.

2. Populations (groups of a single type of organism living in a given environment) do indeed evolve over several generations.

3. This may seem enticing, but it is actually correct. Mutations are random. It is how selection acts on the mutations that causes evolution in a given direction.

4. Recall LUCA (see previous slide).

A school of fish becomes trapped in a dark underwater cave. Many generations later, the descendants of these fish still live in the cave, but the eyes of the descendants are much smaller than those of their ancestors.

Which of the following explanations most likely accounts for the scenario above?

A. The cave contained a chemical pollutant that increased the mutation rate at a gene influencing eye size. Over time, individual fish accumulated mutations and everyone’s eyes became smaller.

B. Fish in the cave no longer needed to use their eyes, because the cave was so dark. Over time, due to lack of use, eyes became smaller.

C. In the original school of fish, individuals differed in their eye size. In the cave, having large eyes was a waste of energy and small-eyed individuals produced more offspring. Mutations that reduced eye size were favoured by natural selection, and the frequency of small-eyed individuals increased.

D. In the cave, having large eyes was a waste of energy, so mutations occurred that reduced eye size were more likely to occur.

1. Mutations are random. Therefore, there is no reason to believe that a mutation-causing chemical would favour mutations that reduced eye size. More importantly, evolution would occur in this case even if there were no chemical in the cave! This is a good distractor and you need to be careful not to get caught up in options like this one.

2. Sounds right, but is in fact incorrect. While the trend of things not being used and gradually becoming less functional can be observed in nature, this does not lay the logical groundwork that explains the scenario. The question asks us to identify a mechanism for the scenario, and to suggest that body parts become less functional if we need them less is to imply that evolution “wants” to take place (see first point on previous slide), which is not true.

3. Correct because it explains the concepts of variability and natural selection. Some individuals already likely had smaller eyes, which became more favourable to have in terms of survivability. These particular individuals had higher survival rates, and were able to pass on their “small-eyed” genes onto future generations. Eventually, the frequency of small-eyed individuals, given that this phenotype gives them an environmental advantage, should increase.

4. Incorrect, as this statement suggests that mutation occurs BECAUSE it was desired. Variation in traits (like a reduced eye size) is not any more likely than another mutation, it’s just selected for in the population by the environment. Do not confuse mutation with evolution!

Which of the following statements about prokaryotic and eukaryotic cell cycle is correct?

1. Eukaryotic cells undergo binary fission

2. The DNA of prokaryotic cells is found in the nucleoid

3. The prokaryotic cell cycle consists of the G1, S, G2, and M phases

4. Eukaryotic cells replicate through mitosis and meiosis

1. Eukaryotic cells undergo mitosis and meiosis, not binary fission

2. This statement is correct, prokaryotic DNA is located in the nucleoid

3. This is describing the eukaryotic cell cycle, not prokaryotic

4. This statement is correct, eukaryotic cells replicate through mitosis and meiosis

The correct answer is c)

which of the following about the SA:V ratio of the cell is correct?

1. It is preferable for cells to have a low SA:V ratio to ensure cells can transport nutrients efficiently

2. The SA:V ratio in a cell has no effect on keeping up with the cell’s demands

3. Cells prefer a high SA:V ratio to maintain a larger volume

4. Cells require a high SA:V ratio to minimize the diffusion distance

1. It is actually not preferable for cells to have a low SA:V ratio, cells want to have a high SA:V ratio which is achieved by cell division.

2. The SA:V ratio has a significant effect on keeping up with the cell’s demands, as having a shorter diffusion distance allows for more efficient transport of nutrients

3. Although the first part of this statement is correct, a larger volume is actually contradictory, as it would increase the diffusion distance, making the cell less efficient with keeping up with demands

4. This statement correctly identifies the ideal SA:V ratio in cells, and the purpose of the high ratio is to minimize the diffusion distance to allow maximum efficiency of nutrient transport

The correct answer is d)

1 and 3. p53 detects DNA damage in the G1/S checkpoint and G2/M checkpoint, not in the Mitotic Spindle Checkpoint (#2), so p53 would not occur. 

2. Incorrect. The mitotic spindle checkpoint is to check that sister chromatids are getting separated properly. It does not check for DNA damage.

4. Incorrect. The only three checkpoints that are needed to be known for this course are in G1/S, G2M, and mitotic spindle checkpoint. No checkpoint at the end of S.

Therefore 1 & 3 (b) are correct

What role do cell cycle checkpoints play in the “inheritance of sameness”?

1. Checkpoints ensure that DNA polymerase does not read the wrong template strand

2. Checkpoints ensure that all kinetochores are attached to spindles before proceeding to anaphase

3. Checkpoints ensure that organelles are properly replicated

4. Checkpoints ensure that damage is repaired before DNA is replicated

Basically asking what role to cell cycle checkpoints have in DNA replication as well as DNA segregation into its respective daughter cells.

1. Incorrect because never mentioned in the lecture.

2. Correct because it describes the mitotic spindle checkpoint.

3. Incorrect because never mentioned in the lecture that there’s a checkpoint to check if organelles are properly replicated.

4. Correct because it describes the G1S checkpoint.

Therefore 2 & 4 (c) are correct

Sources of genetic variation in mitosis include ______:

Recombination

1. Random alignment of chromosomes in metaphase

2. Independent assortment

3. A & B

4. None of the above

There is no genetic variation in mitosis because both daughter cells are identical.

Therefore e is correct

Imagine you look into a microscope and see a cell that is in the S phase of the cell cycle. Which of the following statements could be correct about the cell?

1. The cell is undergoing anaphase

2. The cell has passed the G₂/M checkpoint

3. The cell is receiving signals to differentiate

4. The cell is replicating its DNA

1. Incorrect because anaphase is during M phase not S.

2. Incorrect because G2M checkpoint occurs at the end of G2 (which is after S).

3. Incorrect because differentiation occurs in the gap phases, not S.

4. Correct. S stands for synthesis of DNA, which is when DNA replication occurs.

Therefore D is correct.

One key idea from the lecture on cell cycling is that not all cells divide all the time. Which of the following statements correctly describes the conditions when a cell might be in G₀? 

Cells that enter G₀ have different probabilities of re-entering the cell cycle, depending on what type of cell they are.

1. Cells enter G₀ only if they have DNA damage. 

2. Cells in G₀ cannot function.

3. Cells in G₀ most likely have high turnover rates (replicate often).

1. Correct. G0 is a rest + functioning phase, but it can re-enter the cell cycle if needed (depending on cell type). Cells such as neurons do not divide so they are in G0 always.

2. Incorrect because G0 is a rest + functioning phase. If a cell has DNA damage, they will be stopped at one of the checkpoints of the cell cycle.

3. Incorrect because G0 is for resting and functioning.

4. Incorrect because a cell will enter G0 when it does not need to cycle/divide, which means it probably has low turnover rates.

Therefore a is correct

Which of the following incorrectly describes the function of a cell cycle checkpoint? 

G1/S : checks for cell size, DNA damage, and growth factor needed for S phase

1. G2/M : confirms that nuclear membrane has dissolved, and that spindle poles have formed

2. M : ensures spindles have correctly attached and chromosomes have lined up evenly on metaphase plate

3. All of the above are correct descriptions

G1/S Checkpoint: Checks if the cell has the proper nutrients to go through S phase and checks for DNA damage because we don’t want to be replicating mutations in S phase.

G2/M Checkpoint: Checks if the cell has the proper nutrients to go through M phase.

Mitotic Spindle Checkpoint: Check if the spindles are attached to the kinetochores which are the cores of the chromosomes. Otherwise, you have aneuploidy.

Therefore, the right answer is b

Which of the following statements are CORRECT regarding meiosis and mitosis?

a. Meiosis results in four haploid daughter cells, while mitosis results in two diploid daughter cells. 

b. Meiosis occurs in somatic cells, while mitosis occurs in gametes.

c. Meiosis involves only one round of cell division, while mitosis involves two rounds of cell division.

d. Meiosis is responsible for growth and tissue repair

A

D

Which one of the following statements about metaphase in mitosis or meiosis is correct?

Homologous chromosomes are pulled to opposite poles in metaphase I of meiosis

1. Single chromosomes are lined at the the metaphase plate in metaphase of mitosis

2. Metaphase in mitosis is the same as metaphase I in meiosis

3. There are two stages of metaphase in mitosis

4. None of the statements are correct

B

Which one of the following accurately describes the number of chromosomes in a human cell at various stages?

1. Germ cell -> 23, Gamete -> 23, Zygote -> 46

2. Germ cell -> 46, Gamete -> 23, Zygote -> 23

3. Germ cell -> 23, Gamete -> 23, Zygote -> 23

4. Germ cell -> 46, Gamete -> 23, Zygote -> 46

5. None of the above are correct

D

Alleles A and B are separated by a very short distance on one homolog of Chromosome 1, while allele C is found on the other homolog. Assuming Meiosis occurs without nondisjunction, and no crossing over occurs, what is the likelihood a gamete will inherit all three alleles?

1. 100%

2. Less than 100% but greater than 75%.

3. Less than 75% but greater than 50%

4. Less than 50% but greater than 0%

5. 0%

0%

XYY syndrome is a genetic condition in which a male has an extra Y chromosome.

Assuming the XYY 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 mother

2. Meiosis I in the father

3. Meiosis II in the mother

4. Meiosis II in the father

4

Which of the following statements are CORRECT regarding non-disjunction events?

1. A non-disjunction event in the 1st meiotic division will result in all gametes exhibiting aneuploidy. 

2. Trisomy 21 occurs from a meiotic non-disjunction event. 

3. A non-disjunction event in the 2nd meiotic division will result in 50% of the produced gametes having a normal amount of chromosomes. 

4. Non-disjunction events occur due to a malfunction in prophase I or II. 

1, 2, 3 

which of the following orders of the stages of meiosis are incorrect?

1. Meiosis I - equational, Meiosis II - reductional

2. Meiosis I - reductional, Meiosis II - equational

3. Prophase I, Metaphase I, Anaphase I, Telophase I

4. Prophase II, Metaphase II, Anaphase II, Telophase II

1

which of the following is not a method of genetic variation?

1. Independent assortment

2. Homologous recombination

3. Random natural selection

4. Random fertilization

3

Consider a tetraploid organism with 6 chromosomes per set. During prophase, what is:

1. the coefficient of n?

2. the coefficient of C?

Tetraploid means that this organism is 4n → so that means we have 4 sets of the organism’s unique chromosomes

1. the coefficient n = 4 the whole way through the cell cycle + mitosis

2. DNA replicates before prophase. Initially, 4C. After replication, 8 C

→  C across cell cycle: 4C, 8C, 8C, 8C, 8C, 8C, 8C, 4C

Scroll down for correct answer

• Homologous group of chromosome 1: coefficient of C 

  • 4C - correct 

  • Copies of the genome - how many sister chromatids in a homologous grouping 

• Karyotype is taken during metaphase: telomeres is at the end of each chromosome 

  • Count how many tips of sister chromatids we have = number of telomeres 

  • 4 per grouping 

• N value: how many UNIQUE chromosomes (# of type of chromosomes)

(a)

1. The organism is a diploid (2n) → 2 sets of 8 unique chromosomes. Each chromosome has 2 copies of the genome. Thus, the karyotype has 2 x 2 = 4 copies of the genome

2. Each chromosome has 4 telomeres. Since each unique chromosome has two sets of chromosomes, each set has 8 telomeres. 8 unique chromosomes x 8 telomeres = 64 telomeres

3. There are 8 unique chromosomes

4. The karyotype does NOT contain 2 copies of the genome

Answer: C

May be initially difficult to tell if this insect is diploid or triploid. However, the two chromosomes on the far right are distinct in shape from the others. Since there are 2 chromosomes with the unique “y” shape, we can reasonably conclude that the insect is diploid with 2n=6. 

To determine the coefficient of C, count the number of chromatids per homologous pair. There are 4 chromatids per homologous pair, therefore this cell is 4C. 

Scroll down for correct answer

Answer: A

• Plant is 6n → this means it has 6 sets of unique chromosomes and each set has 7 unique chromosomes

• Therefore, the plant before the S phase has 6 sets x 7 chromosomes = 42 chromosomes

• Karyotype is taken during metaphase, therefore, the 42 chromosomes are replicated and now we have 42 chromosomes x 2 chromatids/chromosome = 84 chromatids → thus 1 is correct

• The plant has 6 sets of unique single-stranded chromosomes (6C). During S phase, each chromosome becomes double-stranded, therefore the coefficient of C likewise doubles (12 C) → thus 2 is correct

• Karyotype is made in metaphase. For a cell to have reached that point, it would have passed the G2/M checkpoint → thus 3 is correct

• The cell is not undergoing apoptosis since it has passed all the checkpoints → thus 4 is incorrect

C

A

First have to identify that the cell is undergoing MITOSIS, meaning that the cell will start and end as diploid, not haploid as it would it during meiosis

1. CORRECT: the cell begins as 2C in the G1 phase, then during S phase, duplicates to 4C as the cell undergoes 

2. INCORRECT: the endpoint of the cell is 0C, which is not a possible outcome of mitosis 

3. INCORRECT: the cell does not duplicate its DNA during S phase 

4. INCORRECT: the endpoint of the cell is 0C, which is not a possible outcome of mitosis 

D - n coefficient (ploidy) doesn’t change and only one reduction occurs

Answer: A

• DNA replication is semiconservative so parent strand would not have aminopurine → not C or D

  • If chicken is fed with aminopurine during S phase, it will incorporate aminopurine into the new strand only

• In option B, one chromosome is shown to have 2 new strands and the other to have 2 parent strands. This does not exhibit semi-conservative replication

  • So the only correct option is A

The correct answer is A because it shows aminopurine in only one strand of each double helix, which would be the correct image of semi-conservative replication if aminopurine is replacing adenine in the newly synthesized strands.

(aka DNA replication will create strands with aminopurine, the original and  synthesized strand should not have aminopurine, and the aminopurine should not be random)

C is wrong because we are synthesizing the new strands with aminopurine

7:36

Answer: A because the continuous leading strand is away from the fork but it should be toward the fork. Continuous replication occurs in the 5’ to 3’ direction towards the fork.

-   B is incorrect because the diagram is correct in having the Okazaki fragments on the lagging strand.

-   C is incorrect because the blue strands are the template strands and therefore the “parent DNA”. The red strands are the newly synthesized strands.

-   D is incorrect because only one strand (lagging) should have Okazaki fragments, due to discontinuous replication.

A

When the RNA primer is replaced with DNA, uracil is replaced with thymine.

Answer : D) 4 only because Pol III extends from RNA primers and Pol I removes RNA primers and replaces it with DNA.

• I is incorrect because all DNA polymerases read the template strand 3’ to 5’ and synthesize the new strand 5’ to 3’.

• II is incorrect because DNA polymerase I is not only active in lagging strand synthesis, it can act elsewhere when needed (i.e. removing RNA primers)

• III is incorrect because Pol III is responsible for synthesizing the Okazaki fragments and Pol I replaces the RNA primers with DNA after the synthesis of Okazaki fragments.

The correct answer is D. 1, 2, 3, 4

Which of the following statements are true?

1. In a replication bubble, there is both RNA and DNA

2. The replication fork will open strands in a 5’ to 3’ direction

3. DNA replication occurs in both directions from the Ori

4. Okazaki fragments are synthesized on the leading strand

1. True, because of the RNA primers and DNA nucleotides

2. False, because the strands are antiparallel, the fork is not only on one strand like polymerase is; therefore it does not move in a 5’-3’ or 3’-5’ direction

3. True, it opens like a bubble, so both sides have their own lagging and leading strand

4. False, Okazaki fragments are synthesized on the lagging strand

Telomerase contains:

1. A built-in DNA template

2. A built-in RNA template

3. No template; it copies from the DNA strand directly

4. Only protein domains

B—It is an RNA dependent DNA polymerase (uses RNA as a template to synthesize DNA on the other strand).

Which of the following statements about telomerase is false?

1. It builds nucleotides in the 5’ to 3’ direction

2. It is not expressed in most of our cells

3. It can increase a cell’s Hayflick limit

4. It extends the 5’ end of the template DNA strand

D - Telomerase extends the 3’ end of the template (parent) DNA strand. This is because during replication, it is the 5’ end of the daughter strand that is shortened. In order to extend this, we would need to extend the 3’ end of the parental strand, as the parent and daughter strands are complementary to each other. 

Answer: A

• Telomerase extends the 3’ end of the template strand (position B)

• DNA is antiparallel, so position A must be the 5’ end of the DNA

• Telomerase uses an RNA template → therefore C and D are not correct

• WHY NOT d

  • Orange is RNA and not DNA

1. Suppose a population of bacteria, called “population A”, ARE exposed to an antibiotic that disrupts the cell wall structure, causing the bacteria to lyse (die). Another population of bacteria, called “population B”, ARE NOT exposed to any antibiotics. Over time, a small sub-population of population A is observed to survive the antibiotic. When researchers examine the sub-population, they find an antibiotic-resistant gene, responsible for their survival. However, when they investigate population B, no such antibiotic-resistant gene is found. 

How would modern evolution theory explain the difference in resistance?

1. When population A is exposed to the antibiotic, it causes the population to develop a helpful gene mutation in order to survive. Since Population B was not exposed to the antibiotic, no such gene was needed and therefore was not found.

2. Mutations that offer resistance to antibiotics have likely occurred in both populations (A & B). Individuals in population A with the mutation conferring resistance have a survival advantage compared to those without the mutation, whereas those in population B with the mutation conferring resistance do not have a survival advantage when compared to those without the mutation. Therefore, the antibiotic resistant gene is found in population A but not population B.

3. It was possible for both populations (A & B) to develop the mutation. But since population A was exposed to the antibiotic, it increased the likelihood of developing a helpful antibiotic resistant gene, whereas population B was not exposed to the antibiotic, so the probability of developing an antibiotic resistant gene did not change. Due to the difference in probabilities, population A developed the resistant gene, whereas population B did not.

4. The antibiotic resistant gene should be found in Population B NOT Population A, so a mistake must have occurred in the experimental process.

Source: Made by BMP 2025/26 

1. B

As you have learned, mutations are constantly happening at random, and therefore are occurring in both populations. Since mutations are constantly happening, and at a high rate in the case of bacteria, due to the fact that they reproduce so fast, an antibiotic resistant mutation has likely occurred in both populations. However, since population B is not exposed to the antibiotic, there is no survival advantage for individuals with the mutation, and the mutation will likely not make its way through the population (i.e it won’t be found when researchers test for it). On the other hand, population A was exposed to the antibiotic, and therefore individuals with the mutation had a survival advantage and were able to reproduce, whereas the individuals without the mutation could not survive and reproduce. This lead to subsequent generations having the antibiotic resistant gene, and the gene being found by researchers.

1. Unlike other types of infections, viral infections are particularly difficult to treat with vaccines, since variants of the virus develop so easily and quickly. One example of this is with COVID-19 and the many variants including Omicron which had many mutations in certain proteins, making it more easily transmittable, even with vaccinated individuals.

What about viruses, such as HIV, make them so hard to treat or to develop vaccines for?

1. They are unlike human cells

2. They do not contain genes

3. They use forward transcription

4. They have high rates of mutation

1. 1, 2, and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3, and 4 are correct

Source:  Made by BMP 2025/26

1. D (4 only)

1. Incorrect. While it is true that viruses are unlike human cells, this is not the reason that they are hard to develop a vaccine for. Take for example bacteria, (which are also unlike human cells) we are much more successful at developing antibiotic drugs than antiviral ones.

2. Incorrect. Viruses do in fact have a genome, so this answer must be incorrect.

3. Incorrect. While some viruses use forward transcription, HIV was the main focus of the class, and HIV uses reverse transcription. Reverse transcription does make developing vaccines harder because it increases the mutation rate of the virus, while forward transcription does not. Therefore, forward transcription does not make developing vaccines for viruses more challenging.

4. Correct. One key characteristic of viruses, especially retroviruses (like HIV), is that they have very high rates of mutations due to transcription (reverse transcription for retroviruses). Due to the high mutation rates of viruses, they are often able to evade the effects of vaccines, making developing vaccines a challenge.

What structure or process in cells is NOT also a characteristic of viruses?

1. Nucleic acid genome

2. Cytoplasm

3. Reproduction or replication

4. Ribosomes 

1. 1, 2, and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3, and 4 are correct

Source: Made by BMP 2025/26 

1. C (2 and 4)

1. Incorrect. Like cells, viruses also have a nucleic acid genome (either DNA or RNA).

2. Correct. Unlike cells, cytoplasm. This is because viruses are acellular (no cellular structures) and only consist of genetic material, a protein coat (capsid), and some have a lipid bilayer overtop of the capsid, called an envelope.

3. Incorrect. Like cells, viruses also reproduce or replicate. Although they need a host cell to do so, one they have attached to a host, viruses replicate many copies of themselves using the host cell machinery. 

4. Correct. Unlike cells, viruses do not have ribosomes. This is because they don’t perform protein synthesis themselves, and instead have to rely on the host cell and the host cell’s ribosomes. 

Which of the following statements is correct in describing how genome mutations contribute to evolution?

A. Mutations can generate errors that reduce fitness.

B. Mutations are the source of variation.

C. Mutations occur randomly.

D. Mutations are heritable.

E. All of the options are true.

4. E (All of the Above)

1. Correct. Mutations are random and can cause an increase, decrease or no change in fitness

2. Correct. You learned in lecture that mutations are in fact the source of variation, since they can result in new variations of genes

3. Correct. You learned in lecture that mutations are random.

4. Correct. Since mutations change the “code” of the DNA, they are heritable. 

5. Correct. All of the above answers are correct.

 A friend, who sadly is not taking BIOLOGY 1000A, tells you that: “Animals have evolved

adaptations that are ideally matched with their environment and their habitat. For example, camels need a way to keep sand out of their eyes, so they evolved long eyelashes and a third eyelid. Camels with these adaptations are more likely to survive and reproduce in the deserts where they live, so these adaptations are an example of evolution by natural selection.” Your friend has misunderstood some aspects of evolution. Which of these statements corrects a misunderstanding?

1.  Adaptations do not evolve with a pre-determined purpose, such as evolving a way to keep sand out of the eyes of camels.

2. It is possible for long eyelashes and a third eyelid to evolve if camels did not live in desert sands.

3. Evolution does not result in the best or ideal traits for a particular environment.

4. Natural selection is not based on better success in reproducing.

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.

1. Correct. Adaption happens as a result of random mutations that confer a survival advantage to the individual in the given environment, over generations.They therefore do not have a predetermined purpose.

2. Correct. It is possible for long eye-lashes and a third eyelid to evolve if camels lived in an environment where those traits were advantageous to survival and reproduction.

3. Correct. Evolution does not result in the best or ideal traits for a given environment, it results in ones that happen to be advantageous (not perfect) for survival and reproduction.  

4. Incorrect. Natural selection is in fact based on what traits will help an individual better survive and reproduce 

6. If viruses are considered to be alive, which of the following traits is shared by all life, which is also evidence that life shares a common ancestry and evolves?

A. Genetic information is contained within a protein shell.

B. Genes are transcribed from a RNA code to a DNA message.

C. DNA or RNA genomes are copied without error.

D. Copies of DNA or RNA genomes are passed on (inherited) to offspring.

 D

1. Incorret. Not all living organisms have a protein sheath surrounding their genetic information, for example human cells.

2. Incorrect. Many organisms don’t transcribe RNA to DNA.

3. Incorrect. Oftentimes, an organism's genetic information is copied with unintentional errors. This is what leads to variation in populations

4. Correct. In all forms of life, genetic information is in fact passed on to offspring. This also suggests that all forms of life did evolve from a distant ancestor.

. 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.

. B (1 and 3)

1. Correct. One of the main components of evolutionary theory is that all life is related through a common ancestor, called LUCA (Last Universal Common Ancestor).

2. Incorrect. Adaptation does not occur when individuals respond to changes in their environment. It occurs when certain traits are more advantageous than others in a certain environment, and selected for.

3. Correct. In order for a population to evolve, the variations must be heritable, otherwise they would be lost from one generation to another.

4. Incorrect. No individual or species is more highly evolved than another, since we have all had the same amount of time to evolve since LUCA.

8. 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

8. C

1. Incorrect. Making predictions is a fundamental part of the scientific method.

2. Incorrect. Trying to falsify your predictions is part of the scientific method, since predictions (or hypotheses) must be falsifiable to be considered scientific.

3. Correct. Trying to prove that your theory is true is not part of the scientific method, instead we try to test if the theory is strong and can be backed by evidence.

4. Incorrect. Revising current theories when new evidence comes to light is part of the scientific method since theories should be as up-to-date as possible (even if evidence falsifies them). 

Cave salamanders have eyes that are considered vestigial because these salamanders are blind. Which of the following statements best explains the existence of cave salamander eyes based on the theory of evolution?

1. In caves, there is little light to use for sight. So, the goal of salamanders living in caves was to reduce the use of their eyes.

2. Cave salamanders are perfectly adapted to their environment, so their blind eyes must have an advantageous function that we have not aware of.

3. The ancestor of cave salamanders must have had eyes used for seeing, but since good eyesight does not provide an advantage in caves, salamanders with poor eyesight became more common.

4. Salamanders living in caves did not use their eyes. So, as individuals lived in darkness, they lost their ability to see and this loss of eyesight was inherited by their offspring.

9.  C

1. Incorrect. The goal of the salamander was not to reduce their eyes since they lived in a dark environment…, they cannot control what traits such as eyes they have or don't have.

2. Incorrect.  According to evolution theory, no organism is perfectly adapted to their environment.

3. Correct. Since eyes are a vestigial structure, a distant relative must have had them, and because eyes offered no advantages to living and reproducing, salamanders without eyes became more prevalent (in other words, there was no selective pressure selecting for having eyes)

4. Incorrect. According to evolutionary theory, individuals do not lose or gain traits in response to their environment, and then pass that on to their offspring, so this is incorrect. 

. Evolution is a scientific theory, with notions of this theory first emerging in ancient times. Which of the following statements accurately describes a scientific theory? 

1. A scientific theory is an explanation about how something in nature works that cannot be disputed 

2. A scientific theory is intended to be rigorously tested, which may strengthen or weaken the theory 

3. A scientific theory is a set of ideas that are accepted as truth when initially proposed 

4. A scientific theory is modifiable based on evidence from experiments and observations 

1. 1, 2, and 3 

2. 1 and 3 

3. 2 and 4 

4. 4 only 

5. All of 1, 2, 3, and 4 are correct 

10. C (2 and 4) 

1. Incorrect. Scientific theories can be disputed and should be falsifiable (claimed false if contradicting evidence occurs). 

2. Correct. Theories should be testable and can be revised.

3. Incorrect: experiments can and should be run on theories to test them 

4. Correct: explained above

Evolution is gradual but can occur quickly. For example, HIV evolves resistance to the drug AZT in only a few months. On the other hand, phenotypic changes take hundreds of years in a population of elephants. 

Which factor would contribute to a faster rate of evolution in HIV than in elephants? 

1. If elephants live in environments where individuals rarely compete for resources 

2. If HIV produces new generations of offspring more quickly than elephants. 

3. If elephants have a higher mutation rate (more mutations per genome copy) than HIV. 

4. If HIV evolution is variational, but elephant evolution is transformational. 

11. B 

1. Incorrect. Competition can drive evolution, however whether elephants experience competition or not does not explain the difference in speed between elephants and HIV. The question is asking for a reason why HIV evolves faster. 

2. Correct! Generation time is crucial to evolution. Since HIV reproduces faster, more mutations occur, leading to rapid evolution. 

3. Incorrect. Even if elephants did have a higher mutation rate than HIV, this would lead to more rapid evolution for elephants, not HIV. 

4. Incorrect. Evolution is never transformational and always variational; individuals do not evolve but populations do. 

12. The genome of the human immunodeficiency virus (HIV) is approximately 9500 nucleotides long and encodes genes for 15 proteins that are used by HIV in its life cycle. Currently, standard treatment for an HIV infection requires a patient to take several antiviral drugs together that interfere with proteins functioning at multiple points of the HIV life cycle. Under this drug cocktail regime, HIV will continue to evolve. Where will mutations most likely appear in the genomes of HIV collected from a patient taking a drug cocktail?

1. In the reverse transcriptase gene

2. In genes where a mutation will be beneficial 

3. In genes that are targeted by antiviral drugs 

4. In any gene, but not in non-coding regions 

5. Anywhere in the genome 

12. E 

1. Incorrect. The reverse transcriptase enzyme makes mutations, but the reverse transcriptase gene itself is equally as likely as anywhere else in teh genome to be mutated. 

2. Incorrect. Mutations have an equal likelihood of occurring anywhere in the genome. They have a higher chance of being selected for if they are beneficial to the HIV (eg in genes targeted by the antiviral drugs but they are equally likely to be made anywhere. 

3. Incorrect. See explanation to B 

4. Incorrect. Mutations can and do occur in noncoding regions. 

5. Correct! See explanations to A, B, D 

13. 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.

1. 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.

2. 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.

3. 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.

13. A

Recall that the transformational view of evolution by Lamarck proposes that individuals in a species change over their lifetime in response to their environment and then pass these acquired traits to their offspring, leading to the transformation of the species. This is different from the Darwinian theory of evolution, which is variational and proposes that populations evolve, not individuals. Of the 3 answer options, only A states that individual virions change during their lifetime and that the changes are passed down to their offspring. B aligns more with the Darwinian theory of evolution and C states that the traits are not passed down to their offspring, which does not alight with the transformational view of evolution.

14. 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.

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct.

14. D (4 only)

1. Incorrect. While evolution can lead to adaptation, adaptation is not guaranteed and is not always directional.

2. Incorrect. Mutations are not purposefully beneficial or harmful and occur randomly. Thus, it is incorrect to say that beneficial mutations occur more often than harmful mutations.

3. Incorrect. Similar to 2, mutations occur randomly and it is not guaranteed that a mutation that appears would improve the fitness of the population.

4. Correct. Beneficial mutations tend to increase in frequency over generations because individuals with them are more likely to survive and reproduce. Thus, these mutations are MORE LIKELY (but not guaranteed) to spread through the population

15. When people who are infected with HIV take the anti-retroviral drug AZT, the viral population evolves to become more and more AZT-resistant. But what happens to levels of AZT-resistance in an HIV viral population if the human host is not taking AZT? And why?

1. AZT-resistance does not increase: the virus does not need to mutate, so no variation in resistance is generated.

2. AZT-resistance does not increase: mutations still occur, but AZT-resistant mutations do not have an advantage over AZT-susceptible forms.

3. AZT-resistance increases, but more slowly than if the human host were taking AZT. Without AZT, HIV’s reverse transcriptase makes fewer errors, so mutations improving AZT-resistance do not happen very frequently. But when these mutations do occur, they have an advantage and spread throughout the viral population.

4. AZT-resistance increases: mutations still occur, and even though AZT-resistant mutations are not currently useful to the virus, they will be useful in the future if the human host begins taking the drug.

15. B

Mutations are constantly occurring randomly and the rate of mutations does not change due to the absence of a specific selective pressure (making A and C incorrect). Since the individual is not taking AZT, there is no selective pressure that makes AZT-resistant mutations more advantageous compared to AZT-susceptible mutations. Moreover, they will not be “kept” in the population in anticipation of the future when the individual takes the drug (making D incorrect). Thus, AZT-resistance does not increase or change.

Imagine that a malfunctioning X-ray machine accidentally exposes living cells on your hand to excessive ionizing irradiation.

Which of the following consequences is most likely?

1. Mutations will likely result from repair of double-strand breaks in chromosomes.

2. The cell will arrest at the metaphase cell cycle checkpoint.

3. A thymine dimer will occur

4. Transposable elements will randomly start inserting into various coding regions.

The correct answer is A. We learned that ionizing radiation can lead to double-stranded breaks in DNA. These are repaired by NHEJ, which often leads to mutation since the cell sloppily tries to reassemble the broken DNA strands without a template

• B is wrong, because if there was DNA damage, the cell would be checking for this at the G1/S checkpoint, and arrest the cell there so that the damage can be repaired before replication

• C is wrong because UV light is what leads to thymine dimers, not X-ray radiation

• D is wrong because TEs are unrelated to DNA damage

Which of the following is true about reactive oxygen species?

1. Aged cells show an accumulation of ROS. 

2. ROS can lead to double stranded breaks in the DNA.

3. Antioxidants help lower the levels of ROS in the cell. 

4. ROS are only created by damaged mitochondria.

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct

The correct answer is A! 1, 2 and 3 are all correct statements regarding ROS.

Statement 4 is incorrect, as ROS can be byproducts even in healthy cells, particularly during the electron transport chain in the mitochondria. 

Which of the following is false regarding mismatch repair?

1. Repair enzymes cut out the damage using endonuclease activity

2. It occurs during G1/S phase and S

3. It’s used when DNA polymerase proofreading doesn’t fix the damage

4. DNA polymerase filling in proper nucleotides is the last step in the mechanism

D is the answer!

A, B, and C are all true for mismatch repair, however DNA ligase sealing the nicks is the last step of the mechanism, not DNA polymerase filling in the proper nucleotides.

One of the mechanisms that leads to DNA mutations is a process known as DNA polymerase slippage which occurs during DNA replication. Which of the following statements about DNA polymerase slippage is correct?

1. Backward slippage will lead to an insertion mutation.

2. DNA polymerase slippage often leads to a point mutation.

3. Forward slippage will lead to DNA polymerase reading more base pairs than it should.

4. None of the above.

The correct answer is A! 

1. Correct! When Polymerase slips backwards, this leads to an insertion mutation on the synthesized strand as it doesn’t realize that the bases have already been copied.

2. Not correct. DNA Polymerase slippage leads to either an insertion or deletion mutation (InDel), which are double stranded changes in the sequence. Point mutations are single base pair changes.

3. Not correct. Forward slippage leads to a deletion mutation on the template strand, resulting in reading less base pairs than it should. 

4. Not correct because A is correct

B is correct!

Recall previous slides about backward and forward DNA polymerase slippage causing insertion and deletion.

Exposure to radiation causes a double stranded break in the DNA. After repairing the damage using Non-Homologous End Joining, which of the following scenarios could occur in the repaired DNA?

1. An insertion mutation

2. A deletion mutation

3. An inversion mutation

4. The wild type sequence

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

The answer is E, all of 1,2,3 and 4 are correct.

Any of these scenarios could occur. Although it would be unlikely for the NHEJ to produce the wild type sequence, it is possible.

Imagine that a malfunctioning X-ray machine accidentally exposes living cells on your hand to excessive ionizing irradiation.

Which of the following consequences is most likely?

1. Mutations will likely result from repair of double-strand breaks in chromosomes.

2. The cell will arrest at the metaphase cell cycle checkpoint.

3. A thymine dimer will occur

4. Transposable elements will randomly start inserting into various coding regions.

The correct answer is A! 1, 2 and 3 are all correct statements regarding ROS.

Statement 4 is incorrect, as ROS can be byproducts even in healthy cells, particularly during the electron transport chain in the mitochondria. 

Imagine that a malfunctioning X-ray machine accidentally exposes living cells on your hand to excessive ionizing irradiation.

Which of the following consequences is most likely?

1. Mutations will likely result from repair of double-strand breaks in chromosomes.

2. The cell will arrest at the metaphase cell cycle checkpoint.

3. A thymine dimer will occur

4. Transposable elements will randomly start inserting into various coding regions.

The correct answer is A! 1, 2 and 3 are all correct statements regarding ROS.

Statement 4 is incorrect, as ROS can be byproducts even in healthy cells, particularly during the electron transport chain in the mitochondria. 

Which of the following is false regarding mismatch repair?

1. Restriction enzymes cut out the damage using endonuclease activity

2. It occurs during G1/S phase and S

3. It’s used when DNA polymerase proofreading doesn’t fix the damage

4. DNA polymerase filling in proper nucleotides is the last step in the mechanism

D is the answer!

A, B, and C are all true for mismatch repair, however DNA ligase sealing the nicks is the last step of the mechanism, not DNA polymerase filling in the proper nucleotides.

One of the mechanisms that leads to DNA mutations is a process known as DNA polymerase slippage which occurs during DNA replication. Which of the following statements about DNA polymerase slippage is correct?

1. Backward slippage will lead to an insertion mutation.

2. DNA polymerase slippage often leads to a point mutation.

3. Forward slippage will lead to DNA polymerase reading more base pairs than it should.

4. None of the above.

The correct answer is A! 

1. Correct! When Polymerase slips backwards, this leads to an insertion mutation on the synthesized strand as it doesn’t realize that the bases have already been copied.

2. Not correct. DNA Polymerase slippage leads to either an insertion or deletion mutation (InDel), which are double stranded changes in the sequence. Point mutations are single base pair changes.

3. Not correct. Forward slippage leads to a deletion mutation on the template strand, resulting in reading less base pairs than it should. 

4. Not correct because A is correct

B is correct!

Recall previous slides about backward and forward DNA polymerase slippage causing insertion and deletion.

1. Which surface area-to-volume ratio would enable a cell to be most efficient at exchanging materials with its environment?

1. 6:1

2. 3:1

3. 1.5:1

4. 0.75:1

1. A

Out of the options, 6:1 is the highest surface area-to-volume ratio. The surface area-to-volume ratio determines how efficiently a cell can exchange materials. A higher surface area-to-volume ratio means more surface area is available per unit of volume, allowing for faster and more efficient exchange.

1. Which cell cycle checkpoint does NOT directly check for DNA mutation?

1. G1/S Checkpoint

2. G2/M Checkpoint

3. M Checkpoint

4. None. All checkpoints directly check for DNA mutation.

1. C

The G1/S checkpoint checks for DNA mutations before the cell undergoes DNA synthesis. The G2/M checkpoint checks for DNA mutations again after DNA synthesis in case it got past G1/S. The M checkpoint checks for proper spindle attachment to the chromosomes during metaphase. M checkpoint is the only checkpoint that does not directly check for DNA mutation. 

1. A drug inhibiting p21 activity is administered. Which of the following may happen? 

1. p53 can no longer detect DNA damage. 

2. The cyclin-CDK complex is inactivated.

3. p21 production will decrease

4. Tumours will form.

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct.

1. D 4 only

1. p53 can no longer detect DNA damage. Incorrect. The drug only affects p21’s activity. p21 is affected BY p53, but does not affect p53, so p53 should function as normal.  

2. The cyclin-CDK complex is inactivated. Incorrect. Inhibition of p21’s activity will prevent the inhibition of the activated cyclin-CDK complex, allowing the cyclin-CDK complex to stay activated. 

3. p21 production will decrease. Incorrect. The drug inhibits p21 activity, not its production. p21 is produced in response to p53 activation following DNA damage. Since the drug acts on p21 after it is made, it does not interfere with the transcription or translation of p21. Therefore, p21 levels may remain unchanged, even though its function/activity is blocked.

4. Tumours will form. Correct. By inhibiting p21 activity, the drug prevents an important brake on the cell cycle from functioning. The inhibition of p21 activity will prevent its inhibition of the activated cyclin-CDK complex. This unchecked complex allows the cell cycle to proceed without stopping, even if the cell has DNA damage or is not ready for cell division. This will lead to uncontrolled cell growth, which is characteristic of tumours.

1. What do early embryonic cells and mature nerve cells have in common?

1. Both cell types have already differentiated 

2. Both cell types receive and respond to signals to divide 

3. Both cell types have exited G1

4. Both cell types are metabolically active 

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct.

1. D 4 only

1. Both cell types have already differentiated. Incorrect. Early embryonic cells have not differentiated. They lack gap phases such as G1 which would enable them to differentiate. Instead, they cycle between S phase and Mitosis. On the other hand, mature nerve cells have already differentiated because they are now nerve cells.

2. Both cell types receive and respond to signals to divide. Incorrect. Mature nerve cells remain in G0 and do not divide. They don't respond to division signals like early embryonic cells do.

3. Both cell types have exited G1. Incorrect. Exiting G1 suggests the cell rests in G0. Only mature nerve cells are in G0, while early embryonic cells are not. 

4. Both cell types are metabolically active. Correct. Even though nerve cells don’t divide, they are very metabolically active. Embryonic cells are also metabolically active to support rapid division and growth.

1. One key idea from the lecture on cell cycling is that not all cells divide all the time. Which of the following statements correctly describes the conditions when a cell might be in G0?

1. Cells that enter G0 have different probabilities of re-entering the cell cycle, depending on what type of cell they are.

2. Cells enter G0 only if they have DNA damage.

3. Cells in G0 cannot function.

4. Cells in G0 most likely have high turnover rates (replicate often).

Source: Taken from BIO1001A Practice Midterm 2022

1. A

Some cells, like liver cells, can re-enter the cycle when stimulated. Others, like neurons, enter G0 permanently (terminally differentiated).

1. The cell cycle plays an important role in controlling cell division. Which of the following can lead to increased cell division?

1. Loss of p53.

2. Loss of CDK.

3. Loss of p21.

4. Loss of cyclins.

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct.

1. B 1 and 3

1. Loss of p53. Correct. p53 is a negative regulator so it stops the cell from continuing to the next stage. Losing p53 would allow the cell to continue onto the next stage without stopping, leading to increased cell division. 

2. Loss of CDK. Incorrect. CDK is a positive regulator so it signals the cell to continue to the next stage. Losing CDK would arrest the cell and lead to decreased cell division. 

3. Loss of p21. Correct. p21 is a negative regulator so it stops the cell from continuing to the next stage. Losing p53 would allow the cell to continue onto the next stage without stopping, leading to increased cell division. 

4. Loss of cyclins. Incorrect. Cyclin is a positive regulator so it signals the cell to continue to the next stage. Losing CDK would arrest the cell and lead to decreased cell division. 

1. Studies of frog oocytes led to the important discovery of a regulatory protein involved in cell cycling. Using the frog, Rana pipiens, scientists learned that if oocytes are arrested in prophase of meiosis I and then injected with cytoplasm containing a specific protein, meiosis progression can be induced (activated). Which of the following proteins were the oocytes most likely injected with?

1. A protein that activates p53.

2. A protein that triggers cells to enter G0.

3. A protein that activates p21.

4. A cyclin-dependent kinase (CDK).

1. D

p53 is a tumor suppressor that halts the cell cycle in response to DNA damage, so it would inhibit progression instead. If a cell is triggered to enter G0, it would not proliferate and would not enter meiosis. p21 inhibits CDK which would inhibit cell cycle progression. 

1. What do meiosis II and mitosis have in common?

1. Crossing over occurs.

2. Ploidy remains the same.

3. Homologous chromosomes separate.

4. Sister chromatids separate.

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct.

1. C 2 and 4

1. Crossing over occurs. Incorrect. Crossing over is when homologous chromosomes exchange genetic material. It happens during prophase I of meiosis I, not during either meiosis II or mitosis.

2. Ploidy remains the same. Correct. In mitosis, cells remain diploid (2n → 2n). In meiosis II, cells are already haploid, and remain haploid (n → n). So, ploidy does not change in either process.

3. Homologous chromosomes separate. Incorrect. This occurs during meiosis I, not in meiosis II or mitosis. In meiosis II and mitosis, sister chromatids separate. 

4. Sister chromatids separate. Correct. In anaphase II (meiosis) and anaphase (mitosis), sister chromatids separate.

1. In a diploid cell, the pair of Chromosome 14 contains genes A, B, C, and D. The following are the chances that alleles of two genes are inherited together (not separated by crossing over):

A and B = 80%

B and C = 30%

C and D = 40%

Which pair of genes are most likely linked genes?

1. Gene A and Gene B

2. Gene A and Gene C

3. Gene C and Gene D

4. Gene B and Gene D

1. A

Linked genes are genes that are located close together on the same chromosome, so they are inherited together more often because crossing over between them is less likely. The higher the probability that two alleles are inherited together, the closer they likely are on the chromosome. Genes A and B had the highest chance of being inherited together (80%) which means they are most likely linked.

1. Which of the following can you see on a karyotype?

1. Homologous chromosomes

2. Chromosomes in mitotic anaphase 

3. Chromosomes at mitotic metaphase 

4. Chromatin

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct.

1. B 1 and 3

1. Homologous chromosomes. Correct. Scientists will arrange homologous chromosomes together, aligned by size and gene pattern. 

2. Incorrect. Chromosomes are in mitotic metaphase when karyotyping

3. Chromosomes undergoing mitotic metaphase. Correct. Karyotypes are derived from cells stuck in metaphase, when the chromosomes are most condensed and visible under a microscope.

4. Chromatin. Incorrect. Chromatin is uncondensed DNA found during interphase, not metaphase. Hence, it’s not visible in a karyotype.

1. Trisomy 18 is a chromosomal disorder. Who is more likely to give birth to a baby with Trisomy 18?  

1. 55 year old woman

2. 45 year old woman

3. 35 year old woman

4. 25 year old woman

1. A

The risk of chromosomal disorders like Trisomy 18 increases exponentially with maternal age, due to more mutations accumulated throughout life and a higher chance of nondisjunction during oocyte cell division. The 55-year-old woman is the oldest woman among the options listed, and hence has the highest risk.

1. What are some consequences of a nonfunctional M Checkpoint?

1. Tumor growth

2. Aneuploidy

3. Premature entry to anaphase 

4. Cell arrest

1. 1, 2 and 3

2. 1 and 3

3. 2 and 4

4. 4 only

5. All of 1, 2, 3 and 4 are correct.

1. A 1, 2 and 3

1. Tumor growth. Correct. If the M checkpoint is nonfunctional, it can no longer monitor cell progression through the cell cycle and can contribute to cancer development. 

2. Aneuploidy. Correct. A nonfunctional M checkpoint allows chromosomes to segregate incorrectly, resulting in abnormal chromosome numbers in daughter cells otherwise known as aneuploidy.

3. Premature entry to anaphase. Correct. The M checkpoint ensures that all chromosomes are properly attached in metaphase before anaphase. If it’s nonfunctional, the cell may enter anaphase too early.

4. Cell arrest. Incorrect. Cell arrest is a function of a working checkpoint. If the M checkpoint is nonfunctional, the cell will fail to arrest, even when there are errors.

1. Which life cycle has gametes that arise by meiosis?

1. Animal Life Cycle

2. Plant and most Fungi Life Cycle

3. Some Fungi and Algae Life Cycle

4. None. All listed life cycles have gametes that arise by mitosis.

1. A

Only in the animal life cycle do the gametes arise by meiosis. In both Plant and most Fungi Life Cycle and Some Fungi and Algae Life Cycle, their gametes arise by mitosis. 

The best way to study these life cycles is to draw them out and be sure to know when it’s mitosis and meiosis, so you know how the ploidy changes!

1. Which one of the following statements about metaphase in mitosis or meiosis is correct?

1. Homologous chromosomes are pulled to opposite poles in metaphase I of meiosis

2. Single chromosomes are lined at the the metaphase plate in metaphase of mitosis

3. Metaphase in mitosis is the same as metaphase I in meiosis

4. There are two stages of metaphase in mitosis

1. B

In the metaphase of either mitosis and meiosis, there are chromosomes lining up on the metaphase plate. The difference is in mitotic metaphase and metaphase II, the sister chromatids are lined up while in metaphase I, the homologous chromosomes are lined up.

B

C

Which of the following statements regarding transposable elements is/are true?

1. An individual with a c value of 9000 Mb likely has more transposable elements than an individual with a c value of 550 Mb

2. All transposon jumping will cause deleterious effects such as hemophilia

3. All organisms possess transposable elements 

4. Transposable elements have evolved to insert into exons (coding regions) in the genome

A. 1, 2, and 3

B. 1 and 3

C. 2 and 4

D. 4 only

B

Which of the following are types of genomic variation in humans?

1. inDel mutations from DNA polymerase slippage

2. Non-Homologous end joining (NHEJ)

3. Tautomeric shifts

4. Transposable elements

1. Only II

2. I and III

3. II and IV

4. All of the Above

4