AP Bio Topic Questions Unit 4
What is the expected percent change in the DNA content of a typical eukaryotic cell as it progresses through the cell cycle from the start of the G1phase to the end of the G2phase?
+100%
A group of researchers cultured yeast cells in a nutrient-rich environment and a nutrient-poor environment and observed the duration of the stages of their cell cycles. The results of their study are summarized in Table 1. Table 1. Duration (in minutes) of yeast cell cycle phases in a nutrient-rich environment and a nutrient-poor environment Cell Cycle Stages Nutrient-rich environment Nutrient-poor environment Sand G2 23 33 Mitosis 48 61 G1 20 57 Cytokinesis 5 10 The cell cycle of yeast cells grown in the nutrient-poor environment is approximately what percent of the cell cycle of yeast cells grown in the nutrient-rich environment?
168
Glycogen synthetase kinase 3 beta is a protein kinase that has been implicated in many types of cancer. Depending on the cell type, the gene for glycogen synthetase kinase 3 beta (GSK3β)can act either as an oncogene or as a tumor suppressor. Which of the following best predicts how GSK3βmutations can lead to the development of cancer?
Cells with inactive GSK3β fail to trigger apoptosis.
Figure 1 shows the number of chromosomes observed in an actively dividing human cell at each stage of cell division. Which of the following presents a correct interpretation of the changes in chromosome number depicted in Figure 1 ?
Chromosomes enter metaphase containing two chromatids attached by a centromere. During anaphase, the chromatids are separated, each becoming a chromosome. Cytokinesis distributes the chromosomes into two separate cells.
Notch is a receptor protein displayed on the surface of certain cells in developing fruit fly embryos. Notch's ligand is a membrane-bound protein called Delta that is displayed on the surface of adjacent cells. When Notch is activated by its ligand, the intracellular tail of the Notch protein becomes separated from the rest of the protein. This allows the intracellular tail to move to the cell's nucleus and alter the expression of specific genes. Which of the following statements best explains Delta's role in regulating cell communication through the Notch signaling pathway?
Delta restricts cell communication to short distances within a developing embryo.
The epinephrine signaling pathway plays a role in regulating glucose homeostasis in muscle cells. The signaling pathway is activated by the binding of epinephrine to the beta-2 adrenergic receptor. A simplified model of the epinephrine signaling pathway is represented in Figure 1. A researcher claims that the epinephrine signaling pathway controls a catabolic process in muscle cells. Which of the following statements best helps justify the researcher's claim?
Glycogen phosphorylase catalyzes the conversion of glycogen to glucose-1-phosphate.
The epinephrine signaling pathway plays a role in regulating glucose homeostasis in muscle cells. The signaling pathway is activated by the binding of epinephrine to the beta-2 adrenergic receptor. A simplified model of the epinephrine signaling pathway is represented in Figure 1.
In involves enzymes activating other enzymes
Researchers performed an experiment to determine the effect of certain genetic mutations on mitosis in tropical fruit fly embryos. They determined the percentage of cells in each of four phases of mitosis as shown in Figure 1. Which of the following patterns is shown by the data?
In mutant 3 cells, more time is spent in prophase/prometaphase than in the later stages of mitosis.
The epinephrine signaling pathway plays a role in regulating glucose homeostasis in muscle cells. The signaling pathway is activated by the binding of epinephrine to the beta-2 adrenergic receptor. A simplified model of the epinephrine signaling pathway is represented in Figure 1. Which of the following statements best describes the role of adenylyl cyclase in the epinephrine signaling pathway?
It accelerates the production of a second messenger.
In a certain signal transduction pathway, the binding of an extracellular molecule to a cell-surface protein results in a rapid increase in the concentration of cyclic AMPinside the cell. The cyclic AMPbinds to and activates cytosolic enzymes that then activate other enzymes in the cell. Which of the following statements best describes the role of cyclic AMPin the signal transduction pathway?
It acts as a second messenger that helps relay and amplify the signal within the cell.
Ethylene causes fruits to ripen. In a signaling pathway, receptors activate transcription factors, which ultimately leads to ripening. Which of the following best supports the claim that ethylene initiates the signal transduction pathway that leads to ripening of fruit?
Loss-of-function mutations in ethylene receptors result in changes to the ripening process.
Figure 1 is a proposed model of the feedback system controlling erythrocyte (red blood cell) production. Air is less dense at very high elevations, so less oxygen is available than in the denser air at sea level. Based on the model in Figure 1, if a person travels from sea level to a high elevation location, which of the following correctly predicts the response to the decreased blood oxygen level?
More erythropoietin will be secreted from the kidneys, increasing production of erythrocytes.
The epinephrine signaling pathway plays a role in regulating glucose homeostasis in muscle cells. The signaling pathway is activated by the binding of epinephrine to the beta-2 adrenergic receptor. A simplified model of the epinephrine signaling pathway is represented in Figure 1. Cyclic AMP phosphodiesterase is an enzyme that catalyzes the conversion of cyclic AMP to a different molecule. Which of the following best predicts the effect of inhibiting cyclic AMP phosphodiesterase in a muscle cell stimulated by epinephrine?
Phosphorylase kinase will remain active because protein kinase A will no longer be deactivated.
In flowering plants, plasmodesmata are narrow channels through cell walls that connect the cytoplasm's of adjacent cells. An explanation of how plant cells communicate across cell walls will most likely refer to the diffusion through plasmodesmata of which of the following?
Small, water-soluble molecules
A student claims that the Y chromosome contains the sex-determining region gene, known as the SRY gene, which causes male fetuses to develop testes. Which of the following provides correct information about cell signaling that supports the claim?
The SRY gene produces a protein that binds to specific regions of DNA in certain tissues, which affects the development of these tissues.
The relative amount of DNA in a cell at various stages of the cell cycle is shown in Figure 1 . Which of the following best describes how the amount of DNA in the cell changes during M phase?
The amount of DNADNA is halved as the cell divides into two daughter cells.
Cancer cells behave differently than normal body cells. For example, they ignore signals that tell them to stop dividing. Which of the following conditions will most likely cause a normal body cell to become a cancer cell?
The environment contains mutagens that induce mutations that affect cell-cycle regulator proteins.
Glucocorticoids are steroid hormones that control cellular responses through several different signaling pathways. One of the signaling pathways involves the glucocorticoid receptor, an intracellular protein that is activated by binding to a glucocorticoid molecule. A simplified model of the glucocorticoid receptor signaling pathway is represented in Figure 1. Which of the following statements best predicts the effect of a mutation that results in a loss of the glucocorticoid receptor's ligand binding function?
The glucocorticoid receptor will remain associated with the accessory proteins.
Which of the following outcomes will most likely result from the irreversible binding of GDP to the G protein?
The intracellular concentration of glycogen will increase.
Figure 1 shows a model of a signal transduction cascade, initiated by the binding of a ligand to the transmembrane receptor protein A A DNAmutation changes the shape of the extracellular domain of transmembrane receptor protein Aproduced by the cell. Which of the following predictions is the most likely consequence of the mutation?
The molecule that normally binds to protein A will no longer attach, deactivating the cellular response.
Vertebrate immune responses involve communication over short and long distances. Which of the following statements best helps explain how cell surface proteins, such as MHC proteins and T cell receptors, mediate cell communication over short distances?
The proteins interact directly with proteins on the surfaces of other cells.
The coagulation cascade controls blood clot formation in response to blood vessel injury. Thrombin is an enzyme that plays a key role in regulating the coagulation cascade. A simplified model of thrombin's role in regulating the coagulation cascade is represented in Figure 1. Argatroban is a competitive inhibitor of thrombin. Which of the following effects on the coagulation cascade is most likely to result from inhibiting thrombin activity with argatroban?
The rate of fibrin formation will decrease.
At the start of mitosis, sister chromatids are held together by a complex of proteins. Separase is an enzyme that cleaves the complex, enabling the chromatids to separate during mitosis. Separase is overexpressed in many cancer cells, and scientists hypothesized that they might be able to slow or stop the growth of cancer cells by blocking the activity of separase. The scientists found a compound they named Sepin‑1 that appears to effectively cleave and thus inactivate purified separase protein in vitro (in a test tube). To test whether Sepin‑1 inhibits the growth of cancer cells, the scientists added increasing concentrations of Sepin‑1 to many different types of cancer cell lines growing in culture. A representative sample of the data they obtained is shown in Figure 1. The scientists also proposed to examine whether there is a relationship between the sensitivity of different types of cancer cells to Sepin-1, as measured by the concentration of Sepin-1 that caused 50% of the cells to die, and the relative concentration of separase in the different cell lines. A representative sample of the data is shown in Table 1. (b) Identify an independent variable in the experiment graphed in Figure 1. Based on the data in Figure 1, identify a control that shows that Sepin‑1 rather than something else in the culture medium is inhibiting the growth of the cancer cells.
The response indicates that an independent variable is the cell lines OR the concentrations of Sepin-1.The response indicates that a control showing that Sepin-1 is responsible for inhibiting the growth of the cancer cells is that all cells are 100% viable at a very low concentration (approximately 0.2 μM ) of Sepin-1.
Students in a class are studying patterns of inheritance using genes involved in determining the body color and wing shape of Drosophila flies. Each of the genes has only two alleles, one of which is completely dominant to the other. Each student in the class performed a parental cross between a fly that is true-breeding for ebony body and vestigial wings and a fly that is true-breeding for gray body and long wings. Each student then crossed several pairs of the F1flies and determined the phenotypes of 500 of the resulting F2flies with respect to body color and wing shape. The students in the class averaged their data for the frequencies of the four possible phenotypes (Table 1). The students performed a second cross. The parental cross was between flies that are true-breeding for gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. They crossed pairs of F1flies and determined the phenotypes of the resulting F2flies. The students found an approximate 3:1ratio of flies with the dominant phenotype (gray bodies and long wings) to flies with the recessive phenotype (ebony bodies and curly wings). Only a few of the flies expressed the dominant phenotype of one trait and the recessive phenotype of the other trait. (b) Using the template, construct an appropriately labeled graph, including error bars, to represent the data in Table 1. Based on the data in Table 1, determine whether there is a significant difference between the number of flies in each of the four phenotypes.
The response indicates that based on error bars, the number of flies with the phenotype ebony body and long wings is the same as the number of flies with the phenotype gray body and vestigial wings. Based on the error bars, the numbers of flies with the two other phenotypes are significantly different from each other and from those of the first two phenotypes.
Students in a class are studying patterns of inheritance using genes involved in determining the body color and wing shape of Drosophila flies. Each of the genes has only two alleles, one of which is completely dominant to the other. Each student in the class performed a parental cross between a fly that is true-breeding for ebony body and vestigial wings and a fly that is true-breeding for gray body and long wings. Each student then crossed several pairs of the F1flies and determined the phenotypes of 500 of the resulting F2flies with respect to body color and wing shape. The students in the class averaged their data for the frequencies of the four possible phenotypes (Table 1). The students performed a second cross. The parental cross was between flies that are true-breeding for gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. They crossed pairs of F1flies and determined the phenotypes of the resulting F2flies. The students found an approximate 3:1ratio of flies with the dominant phenotype (gray bodies and long wings) to flies with the recessive phenotype (ebony bodies and curly wings). Only a few of the flies expressed the dominant phenotype of one trait and the recessive phenotype of the other trait. (a) In the first analysis, all of the F1flies from the students' crosses have the identical phenotype with respect to body color and wing shape, but the F2flies have four different phenotypes. Describe how fertilization contributes to this genetic variability.
The response indicates that fertilization joins gametes with different allele combinations of the genes for body color and wing shape.
Students in a class are studying patterns of inheritance using genes involved in determining the body color and wing shape of Drosophila flies. Each of the genes has only two alleles, one of which is completely dominant to the other. Each student in the class performed a parental cross between a fly that is true-breeding for ebony body and vestigial wings and a fly that is true-breeding for gray body and long wings. Each student then crossed several pairs of the F1flies and determined the phenotypes of 500 of the resulting F2flies with respect to body color and wing shape. The students in the class averaged their data for the frequencies of the four possible phenotypes (Table 1). The students performed a second cross. The parental cross was between flies that are true-breeding for gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. They crossed pairs of F1flies and determined the phenotypes of the resulting F2flies. The students found an approximate 3:1ratio of flies with the dominant phenotype (gray bodies and long wings) to flies with the recessive phenotype (ebony bodies and curly wings). Only a few of the flies expressed the dominant phenotype of one trait and the recessive phenotype of the other trait. (c) Based on the data in Table 1, describe why the dominant alleles for body color and wing shape are the alleles that produce a gray body and long wings, respectively. Based on the data, describe why the two genes are most likely on different chromosomes or why they are most likely on the same chromosome. Calculate the probability of producing flies that have gray bodies and vestigial wings if a cross is performed between one of the F1 flies from the first analysis and a fly that is homozygous for a gray body and vestigial wings.
The response indicates that flies in the largest fraction of the F2 generation have these two traits, suggesting that the alleles for these traits are dominant.The response indicates that the ratio of phenotypes is 9:3:3:1, which is characteristic of a dihybrid cross with two genes that are on separate chromosomes.The response indicates that the probability is 0.5. The probability of flies having gray bodies is 1.0, because the gray color is dominant and one of the flies in the cross is homozygous for a gray body. The probability of flies having vestigial wings is 0.5, because vestigial wings are recessive and one of the flies is homozygous for vestigial wings, while the F1 fly is heterozygous for this trait. The probability of flies having the two traits is calculated by multiplying the two individual probabilities together: 1×0.5=0.5 .
At the start of mitosis, sister chromatids are held together by a complex of proteins. Separase is an enzyme that cleaves the complex, enabling the chromatids to separate during mitosis. Separase is overexpressed in many cancer cells, and scientists hypothesized that they might be able to slow or stop the growth of cancer cells by blocking the activity of separase. The scientists found a compound they named Sepin‑1 that appears to effectively cleave and thus inactivate purified separase protein in vitro (in a test tube). To test whether Sepin‑1 inhibits the growth of cancer cells, the scientists added increasing concentrations of Sepin‑1 to many different types of cancer cell lines growing in culture. A representative sample of the data they obtained is shown in Figure 1. The scientists also proposed to examine whether there is a relationship between the sensitivity of different types of cancer cells to Sepin-1, as measured by the concentration of Sepin-1 that caused 50% of the cells to die, and the relative concentration of separase in the different cell lines. A representative sample of the data is shown in Table 1. (d) The scientists hope to test Sepin‑1 in clinical trials with human cancer patients. The scientists claim that one way to test the efficacy of the Sepin‑1 treatment will be to obtain periodic samples of the patients' cancer cells and determine the percent of cells in metaphase. Provide reasoning to support the scientists' claim.
The response indicates that if the compound is effective, the rate at which the cells are cycling should decrease because Sepin-1 should decrease the amount of active separase in the cells. If the cells are cycling less frequently, fewer cells should be in metaphase.
Students in a class are studying patterns of inheritance using genes involved in determining the body color and wing shape of Drosophila flies. Each of the genes has only two alleles, one of which is completely dominant to the other. Each student in the class performed a parental cross between a fly that is true-breeding for ebony body and vestigial wings and a fly that is true-breeding for gray body and long wings. Each student then crossed several pairs of the F1flies and determined the phenotypes of 500 of the resulting F2flies with respect to body color and wing shape. The students in the class averaged their data for the frequencies of the four possible phenotypes (Table 1). The students performed a second cross. The parental cross was between flies that are true-breeding for gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. They crossed pairs of F1flies and determined the phenotypes of the resulting F2flies. The students found an approximate 3:1ratio of flies with the dominant phenotype (gray bodies and long wings) to flies with the recessive phenotype (ebony bodies and curly wings). Only a few of the flies expressed the dominant phenotype of one trait and the recessive phenotype of the other trait (d) Predict the most likely cause of the F2 ratio obtained by the students in the second analysis between parental flies that are true-breeding for gray bodies and long wings and flies that are true-breeding for ebony bodies and curly wings. Provide reasoning to justify your prediction.
The response indicates that in the second analysis the genes for gray or ebony body color and long or curly wings are linked on the same chromosome.The response indicates that if the genes are close together/linked, the combination of parental alleles will remain unchanged except for a small percent of new combinations that result from limited crossing over in the F1 flies as they produce gametes. The F1 flies are heterozygous, with one chromosome that has both dominant alleles and one chromosome that has both recessive alleles. A cross between them is like a monohybrid cross. Approximately 14/ of the F2 flies will be homozygous dominant for both genes,12/ will be heterozygous for the two genes, and 14/ will be homozygous recessive for the two genes. This gives a 3:1 phenotypic ratio of dominant to recessive for both alleles.
At the start of mitosis, sister chromatids are held together by a complex of proteins. Separase is an enzyme that cleaves the complex, enabling the chromatids to separate during mitosis. Separase is overexpressed in many cancer cells, and scientists hypothesized that they might be able to slow or stop the growth of cancer cells by blocking the activity of separase. The scientists found a compound they named Sepin‑1 that appears to effectively cleave and thus inactivate purified separase protein in vitro (in a test tube). To test whether Sepin‑1 inhibits the growth of cancer cells, the scientists added increasing concentrations of Sepin‑1 to many different types of cancer cell lines growing in culture. A representative sample of the data they obtained is shown in Figure 1. The scientists also proposed to examine whether there is a relationship between the sensitivity of different types of cancer cells to Sepin-1, as measured by the concentration of Sepin-1 that caused 50% of the cells to die, and the relative concentration of separase in the different cell lines. A representative sample of the data is shown in Table 1. (a) Identify the point in mitosis at which separase cleaves the protein complex that holds sister chromatid pairs together. In normal cells, separase is kept in an inactive state until it is needed. Explain how the progression of cells past sequential cell cycle checkpoints and the activity of enzymes such as separase is controlled by interactions between two major groups of regulatory proteins.
The response indicates that separase is most active at the metaphase-anaphase transition OR at the start of anaphase. The response indicates that cells form complexes with different combinations of cyclins and cyclin-dependent (cdk) kinases and that the combinations of the two proteins in the complexes induce the cells to move to different stages of the cell cycle.
The compound eyes of arthropods contain photoreceptor cells surrounded by other cells such as cone cells and pigment cells. Scientists studied how interactions between neighboring cells in the developing eyes of larval fruit flies affect the differentiation of R7 precursor cells that normally become the photoreceptors of adult wild-type flies. One of their experiments is illustrated in Figure 1. R7 and R8 cells are both present in a cluster of cells in the developing eye. R8 cells have cell-surface proteins called Boss. R7 precursor cells have cell-surface receptor tyrosine kinases ( RTKs ) called Sev and contain the G protein Ras that is involved in intracellular signaling. RTKs are proteins with an extracellular portion that binds to a ligand and an intracellular portion that transfers phosphate groups from ATP to other proteins. Scientists examined the fate of R7 precursor cells in fly larvae with wild-type (normal) forms of the Boss, Sev , and Ras proteins or in larvae with single mutations in the protein Sev ( Sev− ) or double mutations in both Sev ( Sev− ) and Ras ( RasD ). RasD is a form of Ras that is always active. (b) Describe how the defect in the Single-Mutant ( Sev− ) R7 precursor cell causes the cell to become a cone cell instead of a photoreceptor cell.
The response indicates that the Ras protein cannot bind to the mutated intracellular portion of Sev and become phosphorylated, or activated, so Ras does not induce the intracellular signaling pathways necessary for differentiation of the R7 precursor cell to a photoreceptor cell.
The compound eyes of arthropods contain photoreceptor cells surrounded by other cells such as cone cells and pigment cells. Scientists studied how interactions between neighboring cells in the developing eyes of larval fruit flies affect the differentiation of R7 precursor cells that normally become the photoreceptors of adult wild-type flies. One of their experiments is illustrated in Figure 1. R7 and R8 cells are both present in a cluster of cells in the developing eye. R8 cells have cell-surface proteins called Boss. R7 precursor cells have cell-surface receptor tyrosine kinases ( RTKs ) called Sev and contain the G protein Ras that is involved in intracellular signaling. RTKs are proteins with an extracellular portion that binds to a ligand and an intracellular portion that transfers phosphate groups from ATP to other proteins. Scientists examined the fate of R7 precursor cells in fly larvae with wild-type (normal) forms of the Boss, Sev , and Ras proteins or in larvae with single mutations in the protein Sev ( Sev− ) or double mutations in both Sev ( Sev− ) and Ras ( RasD ). RasD is a form of Ras that is always active. (a) Describe the first interaction that must occur between the R7 precursor cell and the R8 cell for the R7 precursor to differentiate to an R7 photoreceptor cell.
The response indicates that the Sev RTK on the R7 precursor binds with Boss on the R8 cell.
At the start of mitosis, sister chromatids are held together by a complex of proteins. Separase is an enzyme that cleaves the complex, enabling the chromatids to separate during mitosis. Separase is overexpressed in many cancer cells, and scientists hypothesized that they might be able to slow or stop the growth of cancer cells by blocking the activity of separase. The scientists found a compound they named Sepin‑1 that appears to effectively cleave and thus inactivate purified separase protein in vitro (in a test tube). To test whether Sepin‑1 inhibits the growth of cancer cells, the scientists added increasing concentrations of Sepin‑1 to many different types of cancer cell lines growing in culture. A representative sample of the data they obtained is shown in Figure 1. The scientists also proposed to examine whether there is a relationship between the sensitivity of different types of cancer cells to Sepin-1, as measured by the concentration of Sepin-1 that caused 50% of the cells to die, and the relative concentration of separase in the different cell lines. A representative sample of the data is shown in Table 1. (c) Based on the data in Table 1, describe the general relationship between the sensitivity of cells to Sepin‑1 and the concentration of separase in the cells. Use the Figure 1 data to calculate the ratio between the amount of Sepin‑1 required to kill 50% of the cells in the cell line that is most sensitive to the compound and the cell line that is least sensitive to the compound.
The response indicates that the greater the concentration of separase in the cells, the more sensitive the cells are to Sepin-1 OR that less Sepin-1 is required to kill cells with a greater concentration of separase. The response indicates that there is a 1:3 ratio between the amount of Sepin-1 required to kill A, the most sensitive cell line, in which approximately 10 μM kills 50% of the cells and D, the least sensitive cell line, in which approximately 30 μM kills 50% of the cells.
R7 and R8 cells are both present in a cluster of cells in the developing eye. R8 cells have cell-surface proteins called Boss. R7 precursor cells have cell-surface receptor tyrosine kinases ( RTKs ) called Sev and contain the G protein Ras that is involved in intracellular signaling. RTKs are proteins with an extracellular portion that binds to a ligand and an intracellular portion that transfers phosphate groups from ATP to other proteins. Scientists examined the fate of R7 precursor cells in fly larvae with wild-type (normal) forms of the Boss, Sev , and Ras proteins or in larvae with single mutations in the protein Sev ( Sev− ) or double mutations in both Sev ( Sev− ) and Ras ( RasD ). RasD is a form of Ras that is always active. (c) The scientists predict that the presence of activated Ras in the Double-Mutant (Sev−RasD) R7 precursor cell will enable the cell to differentiate into an R7 photoreceptor cell. Based on Figure 1, evaluate the likely accuracy of their prediction.
The response indicates that the prediction is most likely supported by the data. Because the Ras protein is already activated, it does not need to become phosphorylated, or activated, by binding to the intracellular portion of Sev, so Ras can signal and the R7 precursor cell should become an R7 photoreceptor cell.
R7 and R8 cells are both present in a cluster of cells in the developing eye. R8 cells have cell-surface proteins called Boss. R7 precursor cells have cell-surface receptor tyrosine kinases ( RTKs ) called Sev and contain the G protein Ras that is involved in intracellular signaling. RTKs are proteins with an extracellular portion that binds to a ligand and an intracellular portion that transfers phosphate groups from ATP to other proteins. Scientists examined the fate of R7 precursor cells in fly larvae with wild-type (normal) forms of the Boss, Sev , and Ras proteins or in larvae with single mutations in the protein Sev ( Sev− ) or double mutations in both Sev ( Sev− ) and Ras ( RasD ). RasD is a form of Ras that is always active. (d) The Sev RTK is expressed in more cell types than just R7 precursor cells, and it appears to be important in determining the phenotypes of these other cells. Explain how one receptor can induce different phenotypes in different cell types.
The response indicates that the proteins that interact with/are activated by/are phosphorylated by the Sev RTK in different cell types may differ OR that the proteins that are activated later in the intracellular signaling pathway may differ.
Researchers studying cell cycle regulation in budding yeast have observed that a mutation in the CDC15 gene causes cell cycle arrest in telophase when the yeast cells are incubated at an elevated temperature. Which of the following statements best predicts the effect of the cell cycle arrest on proliferating yeast cells?
The yeast cells will replicate their chromosomes but will fail to complete cytokinesis.
