Read each question and click on the ONE answer option you believe is correct. The explanation will appear after you click any option.
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In an investigation on the effect of sodium chloride solution on onion mass, what is the independent variable?
Click the CORRECT answer:
- Percentage change in onion mass
- Initial mass of the onions
- Time duration of the experiment
- Concentration of sodium chloride solution
- Temperature of the solutions
Reasoning for Correct Answer #4: The independent variable is the factor that the experimenter intentionally changes or varies to observe its effect on another variable (the dependent variable). In this experiment, the investigator is testing the effect of *different concentrations* of sodium chloride solution. Therefore, the concentration of the NaCl solution is the independent variable. The percentage change in mass (#1) is the dependent variable (what is measured). Initial mass (#2), time (#3), and temperature (#5) are typically controlled variables.
To make 140 cm³ of a 5.0% NaCl solution by proportionally diluting a 20.0% NaCl stock solution with distilled water, what volumes of stock solution and distilled water are needed?
Click the CORRECT answer:
- Stock: 35 cm³, Water: 105 cm³
- Stock: 5 cm³, Water: 135 cm³
- Stock: 28 cm³, Water: 112 cm³
- Stock: 70 cm³, Water: 70 cm³
- Stock: 105 cm³, Water: 35 cm³
Reasoning for Correct Answer #1: Use the dilution formula: C₁V₁ = C₂V₂, where C₁=Stock concentration, V₁=Stock volume, C₂=Final concentration, V₂=Final volume.
We have C₁ = 20.0%, C₂ = 5.0%, V₂ = 140 cm³. We need to find V₁.
(20.0%) * V₁ = (5.0%) * (140 cm³)
V₁ = (5.0 * 140) / 20.0 cm³
V₁ = 700 / 20.0 cm³ = 35 cm³
So, 35 cm³ of the 20.0% stock solution is needed.
The volume of distilled water needed is the final volume minus the stock volume:
Water Volume = V₂ – V₁ = 140 cm³ – 35 cm³ = 105 cm³.
Therefore, 35 cm³ of stock and 105 cm³ of water are needed.
Explain why calculating the percentage change in mass is useful when investigating the effect of solutions on onion pieces.
Click the CORRECT answer:
- It makes the calculations simpler than using absolute mass changes.
- It accounts for water evaporating from the solutions during the experiment.
- It allows for valid comparison between onion pieces that may have different initial masses.
- It standardizes the mass change relative to the volume of the solution used.
- It directly measures the water potential of the onion cells.
Reasoning for Correct Answer #3: Different pieces of onion tissue, even if cut similarly, will inevitably have slightly different starting masses. If you only looked at the absolute change in grams, a larger piece might gain or lose more water simply because it’s bigger, making comparisons difficult. Calculating the percentage change ((Final Mass – Initial Mass) / Initial Mass * 100) expresses the change relative to the starting mass of each individual piece. This standardizes the results, allowing for a more valid comparison of the effect of the different solutions across pieces that might have started at different sizes.
In an osmosis experiment… results at 2 hours suggested an isotonic point at 4.2% NaCl (interpolated between 1% and 5%), while results at 48 hours suggested ~3.1% NaCl. Which of the following is a valid reason to question the conclusion that the onion’s water potential is definitively equivalent to 4.2% NaCl?
Click the CORRECT answer:
- The conclusion is based on interpolation between tested points (1% and 5%), and the true relationship may not be linear.
- Calculating percentage change is not a valid way to determine water potential.
- The isotonic point clearly differs depending on the duration of the experiment (2 hours vs 48 hours).
- Lack of reported repeats or statistical analysis makes the reliability of the 4.2% estimate uncertain.
- Natural variation in water potential between different onion cells or samples could exist.
Reasoning for Correct Answer #1 (and #3, #4, #5): The original prompt listed several valid reasons why the 4.2% value might be unreliable.
Option 1: Correct. Interpolation assumes linearity between tested points (1% and 5%). If the actual biological response curve is not linear in this range, the interpolated value of 4.2% may be inaccurate.
Option 2: Incorrect. Measuring mass change in solutions of known water potential is a standard method to estimate the water potential of tissues (isotonic point corresponds to the solution ψ matching the initial tissue ψ).
Option 3: Correct. The fact that the estimated isotonic point changed significantly between 2 hours and 48 hours suggests that equilibrium might not have been reached at 2 hours, or other factors changed over time, making the 2-hour estimate (4.2%) unreliable as the definitive tissue water potential.
Option 4: Correct. Without knowing if the experiment was repeated or statistically analyzed, a single interpolated value from one experiment lacks demonstrated reliability.
Option 5: Correct. Biological samples naturally vary. Different onions or even different layers within an onion might have slightly different water potentials.
*Since multiple options represent valid criticisms based on the original context, and the prompt asks for *a* valid reason, #1 is selected as the identified correct answer from the source material.*
Describe a method to investigate the effect of temperature (10°C to 50°C) on the rate of osmosis in turnip blocks using distilled water. Which point below is LEAST important for ensuring a valid investigation?
Click the CORRECT answer:
- Using turnip blocks of standardized dimensions prepared with consistent tools (e.g., cork borer).
- Ensuring the turnip blocks are fully submerged in the distilled water.
- Maintaining each tested temperature accurately using thermostatically controlled water baths.
- Measuring the initial and final mass of the turnip blocks accurately after careful blotting.
- Using exactly the same volume of distilled water in each beaker, regardless of block size.
Reasoning for Correct Answer #5: To ensure a valid comparison of the effect of temperature (the independent variable) on the rate of osmosis (dependent variable), other factors that could affect the rate must be controlled. Standardizing block dimensions (#1) ensures consistent surface area to volume ratios. Full submersion (#2) ensures consistent contact with the solution. Accurate temperature control (#3) is essential as it’s the independent variable. Accurate mass measurement (#4) is needed to calculate the rate. Using the *exact* same volume of water (#5) is less critical than ensuring the volume is sufficient to cover the blocks and maintain a constant external water potential (i.e., it doesn’t become significantly diluted). Minor variations in a large excess volume of water are unlikely to significantly affect the rate compared to inconsistencies in the other factors.
Predict the effect of increasing temperature from 10°C to 50°C on the rate of osmosis (measured as percentage change in mass per unit time) when turnip blocks are placed in distilled water.
Click the CORRECT answer:
- The rate of osmosis will increase linearly with temperature across the entire range.
- The rate of osmosis will decrease steadily as temperature increases.
- The rate of osmosis will increase as temperature increases up to an optimum point, after which it will likely decrease.
- The rate of osmosis will remain constant regardless of temperature changes.
- The rate of osmosis will show no change up to 30°C then increase sharply.
Reasoning for Correct Answer #3: The rate of diffusion, including osmosis, generally increases with temperature due to increased kinetic energy of water molecules. Increased temperature also increases membrane fluidity, potentially enhancing water movement. Thus, the rate of water uptake by turnip cells from distilled water is expected to increase initially as temperature rises. However, biological membranes and associated proteins (like aquaporins involved in water transport) can be damaged or denatured by excessive heat. Above an optimal temperature, membrane integrity may be compromised, or protein function lost, leading to a decrease in the controlled rate of osmosis or even cell damage that affects mass changes unpredictably. Therefore, an increase to an optimum followed by a decrease is the most biologically plausible prediction.
When investigating the effect of temperature on osmosis in turnip blocks using water baths, which potential hazard requires the precaution of using tongs or heat-resistant gloves?
Click the CORRECT answer:
- Using a sharp cork borer or scalpel.
- Potential turnip allergens.
- Spilling distilled water.
- Handling potentially hot glassware or equipment from heated water baths.
- Electrical hazard from the water bath plug.
Reasoning for Correct Answer #4: The question asks specifically about the precaution of using tongs or heat-resistant gloves. This precaution is directly related to handling objects that may be hot. In this experiment, thermostatically controlled water baths will be used to maintain temperatures up to 50°C. Glassware (beakers), sample holders, or other equipment placed in or removed from these hot baths will become hot and pose a burn or scald risk. Using tongs or gloves protects against this thermal hazard. The other options relate to different hazards requiring different precautions (care with sharp tools, avoiding skin contact for allergens, wiping spills, checking electrical safety).
In a Drosophila breeding experiment, why might the parent fruit flies be removed from the tube around day 7, before the first generation (F1) offspring emerge as adults?
Click the CORRECT answer:
- To prevent the parents from consuming all the food before the offspring emerge.
- To ensure the first generation (F1) offspring only mate amongst themselves for the next cross (if planned).
- Because the parent flies naturally die after approximately 7 days.
- To reduce competition for space between parents and emerging offspring.
- To allow easier counting of the parent phenotypes without confusion.
Reasoning for Correct Answer #2: A standard procedure in multi-generational genetic crosses (like producing F1 and then F2 generations) is to isolate each generation. The parent (P) generation is crossed to produce the F1 offspring. If an F2 generation is desired (by crossing F1 × F1), it is essential to remove the P generation parents *before* the F1 offspring become sexually mature and start mating. This prevents unintended back-crosses (P × F1) or contamination of the F2 generation with offspring resulting from P generation mating. Removing the parents ensures that the subsequent generation comes solely from matings between F1 individuals. While other factors like food (#1) or space (#4) might be considerations, the primary genetic reason is to control the matings for subsequent generations.
Suggest a reliable method to distinguish between anaesthetised female and male Drosophila fruit flies for separation.
Click the CORRECT answer:
- Gently prod them; males jump higher than females.
- Observe the shape of the posterior abdomen using magnification (often rounded with darker pigmentation in males, more pointed in females).
- Measure the length of the wings; males have significantly longer wings.
- Check the front legs for the presence of sex combs (small clusters of dark bristles) using magnification; these are present only on males.
- Listen for distinct buzzing sounds; males buzz at a higher frequency.
Reasoning for Correct Answer #2 (or #4): Reliable methods for sexing Drosophila under anaesthesia and magnification involve observing morphological differences (sexual dimorphism).
Option 2: Correct. The posterior tip of the abdomen differs: males typically have a more rounded, bluntly ending abdomen with darker pigmentation dorsally, while females have a more pointed abdomen tip.
Option 4: Correct. Males possess specialized bristles called sex combs on the tarsal segment of their forelegs, which are absent in females. This is a definitive characteristic.
Options 1, 3, 5: Incorrect. Behavioral differences (#1, #5) are unreliable, especially under anaesthesia. Wing length (#3) and overall size can overlap significantly between sexes and are not reliable distinguishing features.
*Since both #2 and #4 are valid reliable methods described in the source material, and the quiz requires one correct answer, we select #2 based on the original answer key.*
In a chi-squared (χ²) test comparing observed offspring numbers with those expected from a 9:3:3:1 ratio, what is the null hypothesis?
Click the CORRECT answer:
- The observed results are significantly different from the expected results.
- The genes controlling the traits are linked.
- There is no significant difference between observed and expected numbers; any variation is due to chance.
- The 9:3:3:1 ratio is incorrect for this type of cross.
- The observed results perfectly match the expected results.
Reasoning for Correct Answer #3: The null hypothesis (H₀) in statistical testing, particularly in a goodness-of-fit test like chi-squared, is a statement of “no effect” or “no difference”. It proposes that the observed data conforms to the expected pattern or theoretical model (in this case, the 9:3:3:1 ratio predicted by independent assortment) and that any deviations between the observed counts and the expected counts are solely due to random sampling variation (chance). The purpose of the test is to determine if the observed deviations are large enough to reject this null hypothesis.
In a Drosophila cross assuming a 9:3:3:1 ratio (total offspring 1248), the individual (O-E)²/E values were: Dark red=0.172, Brown=2.889, Bright red=0.346, White=0.462. Calculate the chi-squared (χ²) value.
Click the CORRECT answer:
- 2.889
- 3.869
- 7.815
- 1.000
- 0.172
Reasoning for Correct Answer #2: The chi-squared (χ²) test statistic is calculated by summing the contributions from each category. The formula is χ² = Σ [ (Observed – Expected)² / Expected ]. The question provides the individual (O-E)²/E values for each of the four phenotypic categories. To find the total χ² value, simply sum these individual components:
χ² = 0.172 (Dark red) + 2.889 (Brown) + 0.346 (Bright red) + 0.462 (White)
χ² = 3.869
A chi-squared test for a Drosophila cross with 4 phenotypic classes yielded χ² = 3.869. The critical value for 3 degrees of freedom (df = 4-1) at p=0.05 is 7.815. What conclusion should be drawn regarding the null hypothesis (no significant difference from 9:3:3:1 ratio)?
Click the CORRECT answer:
- Reject the null hypothesis because the calculated χ² (3.869) is greater than the critical value (7.815).
- Accept (or fail to reject) the null hypothesis because the calculated χ² (3.869) is less than the critical value (7.815).
- Reject the null hypothesis because the calculated χ² (3.869) is less than the critical value (7.815).
- Accept the null hypothesis because the calculated χ² (3.869) is greater than the critical value (7.815).
- The results are inconclusive without knowing the p-value directly.
Reasoning for Correct Answer #2: The decision rule for a chi-squared test is:
– If Calculated χ² < Critical Value: Accept (or, more formally, fail to reject) the null hypothesis. The observed deviation from the expected ratio is likely due to chance.
- If Calculated χ² ≥ Critical Value: Reject the null hypothesis. The observed deviation is statistically significant, suggesting the underlying ratio is likely different from the expected one.
In this case, the calculated χ² (3.869) is less than the critical value (7.815) at the p=0.05 level for 3 degrees of freedom. Therefore, we accept (fail to reject) the null hypothesis. The observed results are consistent with the expected 9:3:3:1 ratio.
In paper chromatography, how is the Rf (retardation factor) value of a separated pigment calculated?
Click the CORRECT answer:
- Rf = (distance moved by solvent front) / (distance moved by pigment spot from origin)
- Rf = (distance moved by pigment spot from origin) × (distance moved by solvent front)
- Rf = (distance from origin to solvent front) – (distance from origin to pigment spot)
- Rf = (distance moved by pigment spot from origin) / (distance moved by solvent front from origin)
- Rf = (distance moved by pigment spot) + (distance moved by solvent front)
Reasoning for Correct Answer #4: The Rf (Retardation factor or Retention factor) value in chromatography is a ratio that quantifies the relative movement of a substance compared to the movement of the solvent front. It is calculated as:
Rf = (Distance travelled by the substance (pigment spot) from the origin line) / (Distance travelled by the solvent front from the same origin line).
Both distances must be measured from the same starting point (the origin where the sample was applied). The Rf value is dimensionless and typically ranges from 0 to 1.
Results from a Drosophila breeding experiment were consistent with a 9:3:3:1 ratio, suggesting independent assortment of two genes (B/b and R/r) controlling eye colour pigments (B enables brown, R enables red; double recessive = white). Chromatography showed dark red-eyed flies had multiple pigment spots, while white-eyed flies lacked these spots. What is a valid conclusion drawn from combining these results?
Click the CORRECT answer:
- The B and R genes are located very close together on the same chromosome (linked).
- White-eyed flies (genotype bbrr) likely lack the functional enzymes (coded by B and R) necessary to produce both the brown and red pigments, which is consistent with the chromatography showing no corresponding pigment spots.
- Chromatography shows that dark red eyes (genotype B_R_) contain only one type of pigment resulting from the interaction of B and R.
- The breeding results showing a 9:3:3:1 ratio indicate that the white-eye phenotype (bbrr) is dominant.
- The independent assortment suggested by the breeding results implies that the chromatography results, showing different pigments, are likely inaccurate.
Reasoning for Correct Answer #2: The 9:3:3:1 ratio from the breeding experiment suggests independent assortment and that the double recessive genotype (bbrr) corresponds to the white-eyed phenotype (the 1 in the ratio). The chromatography results show that these white-eyed flies lack the pigment spots seen in dark-red eyed flies (presumably B_R_ genotype, the 9 in the ratio). Combining these, a valid conclusion is that the recessive alleles b and r likely lead to non-functional products (perhaps enzymes) required for the synthesis of the brown and red pigments, respectively. Therefore, flies with the bbrr genotype cannot produce these pigments, resulting in white eyes and no pigment spots on the chromatogram. Option 1 contradicts the 9:3:3:1 ratio. Option 3 contradicts the description saying dark-red had *multiple* spots. Option 4 incorrectly labels the recessive phenotype as dominant. Option 5 incorrectly suggests the results are contradictory; they are complementary.