AP Biology Final FRQ

studied byStudied by 320 people
5.0(1)
Get a hint
Hint

Describe the three structural components of an RNA nucleotide monomer. Explain the role of RNA polymerase during transcription.

1 / 71

flashcard set

Earn XP

Description and Tags

72 Terms

1

Describe the three structural components of an RNA nucleotide monomer. Explain the role of RNA polymerase during transcription.

  • A five-carbon sugar (ribose), phosphate group, and a nitrogenous base (adenine, cytosine, guanine, or uracil).

  • Binds to the promoter to initiate transcription.

  • Synthesizes a new RNA molecule based on a DNA template by matching current DNA with the proper RNA complement

New cards
2

Identify the dependent variable in the experiments. Identify a control group missing from the second experiment. Justify the need for this control group in the second experiment.

  • Dependent variable is the maximum elongation rate of the mRNA

  • Control:

    • Wild strain from the first experiment without amanintin

    • Experimental strain from the first experiment without amanitin

  • The control is needed because the effect of amanitin on the maximum elongation rate cannot be determined without comparison to the maximum elongation rate under the same conditions without amanitin

New cards
3

Describe the effect of amanitin on the maximum elongation rate for the wild-type and modified RNA polymerases. Determine the ratio of the average maximum elongation rate for the modified RNA polymerase compared to the wild strain RNA polymerase in Figure 1.

  • Amanitin decreases the maximum elongation rate for the wild strain and does not affect the rate in the experimental strain.

  • The ratio of avg. max. Elongation rate for modified RNA polymerase compared to the wild strain RNA polymerase is. 1:6

    • Drop in elongation in the wild type, but not significant because the rate was so low, 12:2 6:1 ratio

New cards
4

State the null hypothesis for the experiment in Figure 1. Provide reasoning to justify the claim that the change in the amino acid sequence in the modified RNA polymerase affected the shape of the active site on the enzyme.

  • The modified RNA polymerase would not affect the maximum elongation rate

  • By mutating the amino acid you change the shape, and will no longer being able to perform the function.

New cards
5

Describe the effect on an enzyme of an increase in environmental temperature above the optimum temperature range of the enzyme.

Increase in temperature above optimum range will denature enzyme and reduce the reaction rate, or can alter the shape of the enzyme and reduce the reaction rate.

New cards
6

Using the template in the space provided for your response, construct an appropriately labeled graph to represent the data in Table 1. Determine whether there is a statistical difference in the amount of CO2 fixed by rubisco at 30°C and 4

A bar graph would be used to represent the data in Table 1. There is no statistical difference in the amount of CO2 fixed by rubisco at 30 C and 40 C

New cards
7

Based on the data, identify the temperature range at which the soybean leaves are producing the lowest amount of carbohydrate.

any value above 45˚C and equal to/ below 50˚C

New cards
8

Increased amounts of carbon dioxide in the atmosphere are correlated with increased global surface temperatures. Based on the data, predict how a surface temperature that continues to rise above 40°C most likely affects the amount of energy available to primary consumers in an ecosystem. Provide reasoning to justify your prediction. One model predicts that an increase in greenhouse gases will lead to a 2°C increase in average surface temperature on Earth. Based on the data from the experiment and the prediction of an increase in average surface temperature by the model, predict how the locations of plant species are expected to change over time.

  • Prediction: there will be less energy available

  • Justify: Carbon fixation decreases the temperatures above 40˚C.

  • Predict: As temperatures warm, the ranges were individual plant species are found will move towards coolers areas, toward the poles, and towards higher altitude.

New cards
9

Describe the importance of phenotypic variation in louse body color among individuals in a population of lice.

Phenotypic variation in louse body color is important because it allows natural selection to act upon the lice population. Different body colors provide varying levels of camouflage against different pigeon feathers, increasing survival chances and reproductive success for lice. It also allows for potential evolutionary changes over time.

New cards
10

Justify the decision of the researchers to hold lighting conditions constant in their experiment.

The researchers kept lighting conditions constant to isolate the effect of bird behavior on louse body color. By removing variations in lighting, they ensured that any changes in louse color could be attributed to the pigeons' grooming behavior rather than lighting differences.

New cards
11

Predict the most likely effect of the pigeons cleaning their feathers on the phenotypes of the lice after four years.

Cleaning feathers is likely to increase the frequency of lice with body colors that blend well with their specific host's feathers. Lice that are better camouflaged have a higher chance of surviving and reproducing, leading to an increase in their proportion over time.

New cards
12

Provide reasoning to support your prediction in part (c).

This prediction is supported by natural selection. Pigeons removing lice selectively favors those that are less visible against their feathers. Lice with better camouflage have higher survival and reproduction rates, leading to an increase in their frequency and the evolution of louse coloration over generations.

New cards
13

(a) Describe the factors that influence an individual’s phenotype.

Genetic factors are caused by a mutation that can lead to changes in the phenotype. Different alleles can express different phenotypes; this is called. Environmental factors include changes in the condition that organisms live in.

New cards
14

(b) Explain how the presence of Notonecta chemical cues affects gene expression in the Daphnia.

Chemical cues from Notonecta change the phenotype of Daphnia, affecting gene expression. The modifications happened solely during Daphnia's formative stages and were not passable. In a predator-free environment, the progeny are unlikely to have higher crest sizes and widths.

New cards
15

As a follow‑up experiment, researchers placed the Daphnia that were exposed to the Notonecta chemical cues into a tank without chemical cues. The Daphnia reproduced asexually, and the offspring developed in the tank without chemical cues. Predict the relative size of the crest height and width of offspring raised in the tank without chemical cues as compared to the parent Daphnia.

Growing a crest is energetically costly for Daphnia, so they only develop large crests in the presence of predator cues. This ensures efficient energy allocation when predators are absent.

New cards
16

Provide reasoning to justify your prediction in part (c).

The bar graph in Figure 2 confirms that Daphnia exposed to chemical cues have significantly larger crest height and width compared to the control group. The data supports the hypothesis that the presence of chemical cues leads to the development of larger crests in Daphnia as an adaptive response to predator detection.

New cards
17

Describe the role of mitosis in the growth of a plant.

Mitosis helps the plant grow by creating genetically identical daughter cells. More Cells means more plant growth.

New cards
18

Explain why there are more cells observed in mitosis in the root tip than in the woody stem.

The root tip shows more cells undergoing mitosis because the root tip needs to grow more than the woody stem and the root tip cells divide more frequently.

New cards
19

Refine the model in the space provided for your response by using the letters from four of the cell labels of sample 2 to represent the sequential steps of mitosis.

E is prophase, D is metaphase, B is anaphase, C is telophase.

New cards
20

Explain how the arrangement of cellular components during the step of mitosis portrayed by cell D in sample 2 facilitates the proper distribution of chromosomes to the tow daughter cells.

D is metaphase because the spindle fibers are how they line up, the sister chromatids exact copies and then split up. If a cell passes the spindle checkpoint, it makes sure the chromosomes are lined up and everything is good before splitting.

New cards
21

Identify the rectangular solid that represents the cell with the smallest amount of plasma membrane per volume of cytoplasm.

Rectangular solid 5 represents the cell with the smallest amount of plasma membrane per volume of cytoplasm, as it has the lowest surface area-to-volume ratio.

New cards
22

Describe the relationship between the dimensions of the rectangular solids and their surface area-to-volume ratios.

The dimensions of the rectangular solids and their surface area-to-volume ratios have an inverse relationship. As the dimensions increase, the surface area-to-volume ratio decreases.

New cards
23

(c) Using the data from the model, evaluate the hypothesis that cells shaped like rectangular solids 1 and 2 are better suited for slow metabolism and the long‑term storage of energy‑containing molecules than are cells shaped like rectangular solids 4 and 5.

Cells shaped like rectangular solids 1 and 2, which have smaller surface area-to-volume ratios, are better suited for slow metabolism and long-term storage of energy-containing molecules compared to cells shaped like rectangular solids 4 and 5, which have larger ratios.

New cards
24

(d) Based on the trends shown in the data, explain why some rectangular cells include folds or projections in their plasma membranes.

Rectangular cells may have folds or projections in their plasma membranes to increase the surface area-to-volume ratio. This helps optimize cellular functions by enhancing nutrient uptake, waste removal, and other essential processes.

New cards
25

(a) Describe the pattern of inheritance that is most likely associated with a mutation in the MT-ND5 gene. Explain why individuals are not typically heterozygous with respect to mitochondrial genes.

The pattern of inheritance associated with a mutation in MT-ND5 is maternal inheritance. An explanation that MT-ND5 is a mitochondrial gene and mitochondria only carry a single, unpaired chromosome.

New cards
26

Identify a dependent variable measured in the researcher’s experiment. Identify one control that the researcher could use to improve the validity of the experiment. Justify the researcher analyzing blood samples at many intermediate time points instead of at only the beginning and the end of the 20-week period.

· the concentration of lactic acid in the blood and the concentration of NAD+ in the blood.

· An identification of the control as measuring NAD+ and lactic acid levels in a group of other affected individuals treated with placebo or without the NAD+. \n · The justification that by collecting more data, the researchers will see a more accurate trend.

New cards
27

Describe the relationship between the concentration of NAD+ in the blood and the concentration of lactic acid in the blood during the first 5 weeks of treatment with the vitamin. Based on Figure 2, calculate the average rate of change in blood NAD+ concentrations from week 5 to week 17.

A description that the concentration of NAD+ increased and the concentration of lactic acid decreased. \n · A calculation of the average rate of change in blood NAD+ as ±20 μmol/l/week

New cards
28

The researcher performed a follow-up experiment to measure the rate of oxygen consumption by muscle and brain cells. Predict the effect of the MT-ND5 mutation on the rate of oxygen consumption in muscle and brain cells. Justify your prediction. The researcher had hypothesized that the addition of the vitamin that is similar in structure to NADHNADH would increase the activity of the mutated NADHNADH dehydrogenase enzyme in individuals with the disorder. Explain how the vitamin most likely increased the activity of the enzyme.

· The prediction that the mutation will decrease the rate of oxygen consumption by the electron transport chain. \n · The justification that the mutation likely inhibits the rate of oxygen consumption because the electron transport chain will be less efficient, so less oxygen will be required as a terminal electron acceptor. \n · The explanation that the enzyme had decreased activity but still functioned. Increasing the amount of substrate should increase the amount of product produced. Because the vitamin was similar to NADH, it could bind to the active site of the enzyme and effectively increase the substrate concentration.

New cards
29

(a) Describe how amino acids are categorized by their chemical properties. Explain how a change in the amino acid sequence of the FXR1 protein could decrease the ability of the protein to bind to RNA.

Amino acids can be categorized by their chemical properties, such as polarity, charge, and functional groups. Changes in the amino acid sequence of the FXR1 protein can reduce its ability to bind to RNA by altering the protein's structure and properties.

New cards
30

Using the template in the space provided for your response, construct an appropriately labeled graph that represents the data shown in Table 11. Determine whether there is a statistical difference in the amount of FXR1FXR1 protein produced by the cells after 1616 and 2424 hours in the presence of IL-19.

Conduct a statistical analysis, such as a t-test, to determine if there is a significant difference in FXR1 protein production after 16 and 24 hours.

New cards
31

Based on the data for the 4848-hour period, describe the effect of IL-19 on FXR1 gene expression.

IL-19 appears to have a dynamic effect on FXR1 gene expression. The relative amount of FXR1 protein fluctuates over the 48-hour period, indicating that IL-19 influences the expression of the FXR1 gene.

New cards
32

The researcher hypothesizes that the FXR1 gene codes for a protein that binds to mRNAs that encode some of the proteins that damage arteries. Individuals with a particular mutation of the FXR1 gene tend to have high levels of these proteins. Based on this information, predict how the FXR1FXR1 protein most likely interacts with the mRNAs.

The FXR1 protein likely interacts with mRNA molecules through RNA binding. It may have specific domains or motifs that recognize and bind to mRNA, regulating their function or stability. This interaction could involve physical contact between the protein and mRNA, influencing mRNA activation or degradation.

New cards
33

Describe one outcome that would demonstrate that a given population has evolved.

(a) One outcome demonstrating evolution in a population is the development of a new trait or characteristic not present in the ancestral population. In this case, if the Cit+ bacteria successfully grow using citrate as an energy source, it shows they have evolved this ability.

New cards
34

Identify the dependent variable measured in the experiments.

(b) The dependent variable measured is the growth or density of the bacterial populations, indicating how well they can utilize citrate as an energy source.

New cards
35

Predict the results obtained by the researchers when they grew the Cit+ and Cit−bacteria in the medium that contained only citrate.

(c) The researchers would likely observe significant growth and density of the Cit+ bacteria in the citrate-only medium, indicating their ability to use citrate as an energy source and outcompete the Cit− bacteria.

New cards
36

The researchers claim that the Cit+ mutation increases the fitness of the bacteria. Provide reasoning to support the claim.

(d) The claim that the Cit+ mutation increases fitness is supported by the rapid growth and high density of the Cit+ bacteria in the citrate-only medium. Their ability to use citrate provides an advantage, enabling faster growth and outcompeting the Cit− bacteria. This demonstrates increased fitness in the specific environment of the citrate-only medium.

New cards
37

(a) Describe the process that maintained a stable Tasmanian devil population size before the appearance of DFTD in 1996.

(a) Before the appearance of DFTD, the Tasmanian devil population was kept stable by factors such as the carrying capacity of their habitat and ecological influences like competition and predation.

New cards
38

(b) Explain how the huge reduction of the Tasmanian devil population since 1996 affects the susceptibility of the current population to new diseases in comparison with the susceptibility of the population before 1996.

(b) The significant reduction in the Tasmanian devil population since 1996 has made the current population more susceptible to new diseases compared to the population before 1996. The low genetic diversity makes it harder for them to develop resistance and adapt to new diseases.

New cards
39

(c) Tasmanian devils are top predators and are considered a keystone species in their community. Predict the effect of the rapid reduction of the Tasmanian devil population on the rest of the community.

(c) The rapid decline in the Tasmanian devil population has had a notable impact on the rest of the community. As top predators and keystone species, their decline disrupts the balance of the ecosystem, leading to potential overpopulation of prey species and negative effects on vegetation and other animal species.

New cards
40

(d) Justify the prediction of part (c).

(d) The prediction in part (c) is supported by the important role Tasmanian devils play as top predators and keystone species. Their decline disrupts predator-prey relationships and can have cascading effects on the ecosystem's stability and biodiversity.

New cards
41

(a) A skin cell completes one round of the cell cycle. Describe the products.

(a) When a skin cell completes one round of the cell cycle, it produces two identical daughter cells.

New cards
42

(b) Based on Figure 1, explain how p53 regulates the cell cycle in the presence of damaged DNA.

(b) p53 regulates the cell cycle in the presence of damaged DNA by activating protein kinases, which then activate p53. Activated p53 binds to the damaged DNA and triggers cell cycle arrest and DNA repair.

New cards
43

(c) Draw an X on the template in the space provided for your response to indicate the phase during which the replication of damaged DNA would occur.

(c) The replication of damaged DNA occurs during the S phase of the cell cycle.

New cards
44

(d) Based on Figure 1, explain how a mutation to p53 may lead to an increased risk of cancer.

(d) Mutations in p53 can disrupt its function as a tumor suppressor, leading to an increased risk of cancer. Mutated p53 may fail to properly regulate the cell cycle and DNA repair, allowing damaged cells to replicate and accumulate mutations, potentially leading to uncontrolled cell growth and cancer.

New cards
45

(a) Identify the individual who most likely exhibits symptoms of cystinuria.

(a) Individual 1 is most likely exhibiting symptoms of cystinuria, as they have a significantly higher concentration of cysteine in their urine compared to normal values.

New cards
46

(b) Describe the relationship between the total number of mutant alleles in an individual and the concentration of cysteine in the urine.

(b) The concentration of cysteine in the urine increases with the total number of mutant alleles in an individual. More mutant alleles are associated with higher cysteine levels.

New cards
47

(c) Evaluate the hypothesis that mutations in SLC7A9 have a greater effect on the transport of cysteine across the plasma membrane of kidney cells than do mutations in SLC3A1.

(c) The data provided in Table 1 do not directly compare the effects of mutations in SLC7A9 and SLC3A1 on cysteine transport. Further investigation is needed to evaluate the hypothesis that mutations in SLC7A9 have a greater impact on cysteine transport.

New cards
48

(d) Explain how the data support the claim that cysteine is a large polar molecule.

(d) The data support the claim that cysteine is a large polar molecule. Elevated cysteine levels in the urine of individuals with cystinuria indicate that cysteine struggles to cross the kidney cell membrane, suggesting it is a large polar molecule that requires functional transporter proteins for efficient reabsorption by the kidneys.

New cards
49

(a) Describe how C. parvum obtains the glucose it needs for glycolysis after it has infected another cell. Explain the role of lactate dehydrogenase in enabling C. parvum to continue producing ATP by glycolysis.

(a) After infecting a host cell, C. parvum obtains glucose for glycolysis through various mechanisms. Lactate dehydrogenase plays a crucial role in C. parvum's ATP production by facilitating fermentation after glycolysis.

New cards
50

(b) Identify the independent variable used in the experiment. Identify the difference between the control cells and the experimental cells used in the experiment. Justify the researchers using a different range of concentrations for FX11 than was used for gossypol.

(b) The independent variable in the experiment is the concentration of gossypol or FX11. The control cells were untreated, while the experimental cells were treated with different concentrations of the inhibitors. The different concentration ranges for FX11 and gossypol were likely chosen to determine their optimal effectiveness without harming human cells.

New cards
51

(c) Based on the data in Figure 1, identify the concentration of gossypol that reduced C. parvum growth to 50% of that in control cells.

(c) The concentration of gossypol that reduced C. parvum growth to 50% of the control cells' growth is approximately 14 micromolar.

New cards
52

(d) Researchers discovered a strain of C. parvum that expresses a functional variation of the lactate dehydrogenase gene. A DNA sequence comparison showed that the variant differs from the normal sequence in the region that codes for the enzyme’s allosteric site. Predict the effect of FX11 treatment on C. parvum cells that express this variant of lactase dehydrogenase. Provide reasoning to support your prediction. Explain how gossypol and FX11 might be used as drugs to treat C. parvum infections in humans without negatively affecting human cells.

(d) If C. parvum cells express a lactate dehydrogenase variant with a different allosteric site, the FX11 inhibitor may not effectively bind and inhibit the enzyme, allowing ATP production to continue. Gossypol and FX11 can target C. parvum without harming human cells due to their specific interaction with C. parvum's lactate dehydrogenase, which differs from the corresponding enzyme in human cells. This selective targeting enables the inhibition of C. parvum's energy production while minimizing negative effects on human cells.

New cards
53

(a) Describe the situations in which a normal human cell would enter the cell cycle and undergo mitotic cell division. Explain how spindle fibers help ensure the products of mitosis are two identical cells with a full set of chromosomes.

(a) Normal human cells undergo mitotic cell division during growth, development, tissue repair, and cell replacement. Spindle fibers ensure that chromosomes are properly distributed, attaching to kinetochores and aligning chromosomes during metaphase. They help separate sister chromatids, ensuring each daughter cell receives an identical set of chromosomes.

New cards
54

(b) Using the template in the space provided for your response, construct an appropriately labeled graph that represents the data shown in Table 1. Based on the data, determine the concentration(s) of paclitaxel that is (are) most effective in causing tripolar cell division.

(b) Based on the data in Table 1, higher concentrations of paclitaxel lead to a higher percentage of tripolar cell division. The concentrations of 6 nM, 8 nM, and 10 nM are most effective in causing tripolar cell division.

New cards
55

(c) Based on the data, identify the lowest level of paclitaxel that will allow for at least 50% of the cells to be tripolar. From the start codon through the stop codon, the length of the fully processed AURKA mRNA is 1,212 nucleotides. Calculate the number of amino acids in the polypeptide chain coded for by the mRNA.

(c) The concentration of paclitaxel that allows at least 50% of cells to be tripolar is 4 nM, as shown in Table 1. The number of amino acids in the polypeptide chain coded by the AURKA mRNA cannot be determined without additional information.

New cards
56

(d) Predict the effect of a mutation that prevents the expression of AURKA on a normal (noncancerous) cell.

(d) A mutation preventing AURKA expression in a normal cell would disrupt spindle fiber assembly during mitosis, leading to errors in chromosome alignment and separation. This could cause unequal distribution of chromosomes to daughter cells, genomic instability, and potential cell death. Without functional AURKA, normal cell division would be severely affected.

New cards
57

(a) Describe the way the scientists will determine the evolutionary fitness of the mice in the experiment.

(a) The scientists will determine the mice's evolutionary fitness by comparing their survival and reproductive success in different sand enclosures. They will observe the number of offspring produced and the ability to survive and reproduce. Mice with fur colors that blend well with their surroundings and provide camouflage are expected to have higher fitness.

New cards
58

(b) Identify the independent variable in the scientists' experiment.

(b) The independent variable in the experiment is the color of the sand in the enclosures.

New cards
59

(c) State the null hypothesis of this experiment.

(c) The null hypothesis states that the sand color does not affect the frequency of fur color in the mouse populations.

New cards
60

(d) The scientists claim that the changes in the frequency of fur color were the result of natural selection. Justify the researchers’ claim.

(d) The researchers' claim is justified by the observed correlation between sand color and fur color. The populations in light sand enclosures became lighter, while those in dark sand enclosures became darker. This suggests natural selection favoring mice with better camouflage. The absence of ground predators and wild mice, along with predatory birds' presence, supports this claim.

New cards
61

(a) Describe the relationship between a parasite and its host.

(a) The relationship between a parasite and its host is a symbiotic association where the parasite benefits at the expense of the host.

New cards
62

(b) Explain how producing the enzymes that digest α-pinene is beneficial to the bacterial species living within the nematodes.

(b) Producing enzymes that digest α-pinene is beneficial to the bacteria living within the nematodes because it allows them to utilize α-pinene as a nutrient source and overcome the tree's defense mechanism.

New cards
63

(c) Predict the effect of the antibiotic treatments on the mortality rate of the nematodes when exposed to α-pinene.

(c) The antibiotic treatments are predicted to increase the nematodes' mortality rate when exposed to α-pinene due to the elimination or reduction of the symbiotic bacteria that degrade the defensive chemicals.

New cards
64

(d) Provide reasoning to justify your prediction in part (c).

(d) The prediction is based on the fact that without the protective bacteria, the nematodes would be unable to neutralize α-pinene, making them more vulnerable to its toxic effects and resulting in increased mortality.

New cards
65

(a) Describe the major process that takes place in this eukaryotic organelle.

(a) The chloroplast is involved in photosynthesis, the process of converting light energy into chemical energy in the form of glucose.

New cards
66

(b) Explain the function of the structure labeled with an X in Figure 1.

(b) The structure labeled with an X in Figure 1 is the outer surface of the chloroplast, which acts as a barrier and controls the exchange of molecules and ions in and out of the chloroplast.

New cards
67

(c) On the template in the space provided for your response, represent the location where carbon fixation takes place by writing "CF" and the location of the electron transport proteins by writing "ETP".

(c) Carbon fixation occurs in the stroma, while the electron transport proteins (ETP) are located in the thylakoid membranes.

New cards
68

(d) Explain how the shape and stacking of the thylakoids contributes to the rate of carbon fixation by the chloroplast.

(d) The shape and stacking of the thylakoids optimize light absorption and create a concentration gradient for ATP synthesis, contributing to the efficient rate of carbon fixation in the chloroplast.

New cards
69

(a) Identify the genotypes of the male and female flies used in cross 2.

(a) In cross 2, the male fly used is unaffected (homozygous dominant), and the female fly used is affected (homozygous recessive).

New cards
70

(b) Identify the cross in which the female parent was most likely heterozygous.

(b) Cross 3 is most likely the cross where the female parent was heterozygous for the trait. The offspring from this cross show a percentage of affected male offspring (50%), indicating that the female parent carried one copy of the affected allele.

New cards
71

(c) The researchers hypothesize that crossing any unaffected female and an affected male will result in a 0% chance of producing an affected male offspring. Evaluate the validity of the hypothesis.

(c) The hypothesis suggesting a 0% chance of producing an affected male offspring when crossing an unaffected female with an affected male is not valid. Cross 2 demonstrates that all male offspring from this cross are affected, contradicting the hypothesis. This suggests that the trait is likely X-linked recessive and inherited from the affected male parent.

New cards
72

(d) Explain how the results exclude the possibility that the trait is encoded by a mitochondrial gene.

(d) The results of the crosses rule out the possibility of the trait being encoded by a mitochondrial gene. If the trait were encoded by a mitochondrial gene, all offspring of affected females would show the same phenotype. However, in cross 4, where both parents are affected, all offspring are affected regardless of their sex. This indicates that the trait is not solely inherited from the maternal line and is not associated with mitochondrial genes.

New cards

Explore top notes

note Note
studied byStudied by 101 people
... ago
4.0(1)
note Note
studied byStudied by 27 people
... ago
5.0(1)
note Note
studied byStudied by 1211 people
... ago
4.9(15)
note Note
studied byStudied by 14 people
... ago
5.0(1)
note Note
studied byStudied by 27 people
... ago
4.5(2)
note Note
studied byStudied by 35 people
... ago
5.0(1)
note Note
studied byStudied by 122 people
... ago
5.0(3)
note Note
studied byStudied by 18 people
... ago
5.0(3)

Explore top flashcards

flashcards Flashcard (100)
studied byStudied by 9 people
... ago
5.0(1)
flashcards Flashcard (202)
studied byStudied by 4 people
... ago
5.0(1)
flashcards Flashcard (63)
studied byStudied by 10 people
... ago
5.0(1)
flashcards Flashcard (60)
studied byStudied by 35 people
... ago
5.0(2)
flashcards Flashcard (36)
studied byStudied by 16 people
... ago
5.0(2)
flashcards Flashcard (97)
studied byStudied by 18 people
... ago
5.0(1)
flashcards Flashcard (55)
studied byStudied by 41 people
... ago
5.0(2)
flashcards Flashcard (26)
studied byStudied by 5 people
... ago
5.0(1)
robot