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AP Biology Math Review
- You may use an APPROVED calculator and formula sheet.
- You will solve each problem and grid in the answer. Tips for using grid sheet:
- Grid LEFT to right
- Use the formula sheet
- Don’t round until the end
- Look at HOW the answer should be given: “Round to nearest…” Example:. The 1 is in the tenths place The 2 is in the hundreds place The 3 is in the thousandths place Q1: Chi Square A hetero red eyed female was crossed with a red eyed male. The results are shown below. Red eyes are sex-‐linked dominant to white, determine the chi square value. Round to the nearest hundredth. Phenotype # flies observed Red Eyes 134 White Eyes 66 Chi Square Strategy: a. Given = observed b. Calculate the expected, usually do a Punnett square to figure this out how many phenotypes c. Plug in +
Q1: Chi Square Answer Observed = 134 red eyes, 66 white eyes Q2: Surface Area and Volume What is the SA/V for this cell? Round your answer to the nearest hundredths.
Q3: Water Potential and Solution Potential Answer
Q4: Hardy Weinberg p2 + 2pq + q2 = 1
A census of birds nesting on a Galapagos Island revealed that 24 of them show a rare recessive condition that affected beak formation. The other 63 birds in this population show no beak defect. If this population is in HW equilibrium, what is the frequency of the dominant allele? Give your answer to the nearest hundredth Hardy Weinberg Strategy: a. figure out what you are given: allele (p or q) or genotypes (p2, 2pq, q2) b. figure out what you are solving for (allele frequency, number in population) c. manipulate formulas to go from given to solving for d. always give answers in decimals
Q4: Hardy Weinberg Answer
Q5: Rate Hydrogen peroxide is broken down to water and oxygen by the enzyme catalase. The following data were taken over 5 minutes. What is the rate of enzymatic reaction in mL/min from 2 to 4 minutes? Round to the nearest hundreds. Time (mins) Amount of O 2 produced (mL) 1 2. 2 3. 3 4. 4 5. 5 5. Q6: Laws of Probability Calculate the probability of tossing three coins simultaneously and obtaining three heads. Express in fraction form. Q6: Laws of Probability
Q7: Population Growth Answer Q8: Net Productivity The net annual primary productivity of a particular wetland ecosystem is found to be 8000 kcal/m^2. If respiration by the aquatic producers is 12,000 kcal/m^2 per year, what is the gross annual primary productivity for this ecosystem in kcal/m^2 per year? Round to the nearest whole number. Q8: Net Productivity Q9: Q 10 Data taken to determine the effect of temperature on the rate of respiration in a goldfish is given in the table below. Calculate Q 10 for this data. Round to the nearest whole number. Temperature (C) Respiration Rate (Min) 16 16 21 22
Q9: Q 10
Q8: Net Productivity Answer Q9: Q 10 Answer Q10: Standard Deviation Grasshoppers in Madagascar show variation in their back-‐leg length. Given the following data, determine the standard deviation for this data. Round the answer to the nearest hundredth. Length(cm): 2.0, 2.2, 2.2, 2.1, 2.0, 2.4 and 2.
Q11: Dilution Answer
Q12: pH log Answer
AP Biology Math Review – Part II
Q13:
Initial mass of pumpkin cores was measured in grams. What is the average initial mass for the pumpkin cores? 29.15, 28.45, 30.92, 29.25, 32.09, 31.67. Round to nearest hundredths. Q14: χ 2 = ∑ In pea plants, smooth seeds are dominant to wrinkled, and purple flowers are dominant to white. In a dihybrid cross where a 9:3:3:1 ratio is expected, the following data was collected: Smooth and Purple = 223 Smooth and White = 84 Wrinkled and Purple = 89 Wrinkled and White = 33 Determine the chi-‐square value. Round to nearest hundredths. Q15: χ^2 = ∑ Two Wisconsin fast plants are crossed. One has the recessive dwarf trait, but the normal pigment anthocyanin, while the other has the recessive anthocyaninless trait, but is on normal height. Their offspring consist of: 89 plants of normal height and pigment 93 anthocyaninless plants and normal height 96 dwarf plants and normal pigment 94 anthocyaninless, dwarf plants A student proposes that the parent plants' genotype must have been ddAa for the dwarf parent and Ddaa for the anthocyaninless parent. Calculate the chi square value that would be used to confirm this hypothesis. Round to nearest hundredths.
Q 20 : total water potential = pressure potential + solute potential ψ total = ψ p + ψ s Scientists are trying to determine under what conditions a plant can survive. They collect the following data and would like to know the water potential of the plant cell. The solute potential is -‐0.6 MPa and the pressure potential is -‐1.0 MPa. What is the water potential? Round to nearest hundredths. Q2 1 : growth rate = dN/dt = rN, r = b-‐d; dN/dt = rmaxN( 1 -‐(N/K)) A hypothetical population has a carrying capacity of 1,500 individuals and rmax is 1.0. a. Fill out the following table: Population size Population growth rate 1, 1, 2, b. What is happening to this population? Why? Q2 2 : pH = -‐log [H+] According to the Acid Rain Monitoring Project at the University of Mass, the pH measured at King Phillip Brook on April 10, 2012, was near 5, which the pH measured at Robbins Pond on that same date was near 9. Determine to the nearest whole number how many times greater the hydrogen ion concentration was at King Phillip Brook. Q23: growth rate = dN/dt = rN, r = b-‐d; dN/dt = (b-‐d)N In 2009, the US had a population of about 307 million people. If there were 14 births and 8 deaths per 1000 people, what was the country's net population growth that year (ignore immigration and emigration)? Round to nearest thousandths.
Q13 Answer: average initial mass = total mass/total number = 181.53/6 = 30.26 grams Q14 Answer: Observed Expected (9:3:3:1) Smooth and Purple = 223 (9/16) 429 = 241 Smooth and White = 84 (3/16) 429 = 81 Wrinkled and Purple = 89 (3/16) 429 = 81 Wrinkled and White = 33 (1/16) 429 = 27 Total = 429 Total = 429 χ^2 = ∑ = ( 241 - 22 3)^2 + (81-84)^2 + (81-89)^2 + (27-33)^2 241 81 81 27 = 1.34 + 0.11 + 0.79 + 1.33 = 3. Q15 Answer: ddAa x Ddaa = 1:1:1: Observed Expected (1:1:1:1) plants of normal height and pigment = 89 (1/4)^372 = 93 anthocyaninless plants and normal height = 93 (1/4) 372 = 93 dwarf plants and normal pigment = 96 (1/4) 372 = 93 anthocyaninless, dwarf plants = 94 (1/4) 372 = 93 Total = 372 Total = 372 χ 2 = ∑ = ( 93 - 89 ) 2
- ( 93 - 93 ) 2
- ( 93 - 96 ) 2
- ( 93 - 94 ) 2 93 93 93 93 = 0.17 + 0 + 0.10 + .01 = 0. Q16 Answer: 250/1000 = 0.25 = aa = q^2 q = 0.5, p = 1 -‐ q = 0. 2pq = 0.5; heterozygous = 0.5 x 1000 = 500
Q21 Answer: logistic growth rate = dN/dt = rN, r = b-‐d; dN/dt = rmaxN(1 -‐(N/K)) population number (N) birth (b) death (d) carrying capacity (K) rate (r) at N = 1600: dN/dT = 1.0 x 1600 (1 -‐ (1600/1500)) = 1600(1 -‐ 1 .06) = -‐ 96 at N = 1750: dN/dT = 1.0 x 1750 (1 -‐ (1750/1500)) = 1750(1 – 1.17) = -‐ 298 at N = 2000: dN/dT = 1.0 x 2000 (1 -‐ (2000/1500)) = 2000 (1 – 1.33) = -‐ 660 Population size Population growth rate 1,600 -‐ 96 1,750 -‐ 298 2,000 -‐ 660 shrinking because over carrying capacity Q22 Answer: pH 5: [H+] = 1 x 10 -‐^5 pH 9: [H+] = 1 x 10 -‐^9 1 x 10 -‐^9 = 10000 1 x 10 -‐^5 Q23 Answer: logistic growth rate = dN/dt = rN, r = b-‐d; dN/dt = (b-‐d)N N = 307 million b = 14/ d = 8/ dN/dt = (b-‐d)N = (14/1000 – 8/1000) 307 million = 1.842 million
