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ANSWERS Station 1: A) Chi-squared B) pH calculation Compared with a basic solution at pH 8, the same volume of an acidic solution of pH 5 has how many times more hydrogen ions (H+)? Give your answer to the nearest whole number? Station 2: A) Hardy-Weinberg B) Dilution Station 3: A) Rate Calculation from Graphed Data After an enzyme is mixed with its substrate, the amount of product formed is determined at 10-second intervals for 1 minute. Determine the rate of change for the amount of product formed during the first 30 seconds of data collection. Give your answer to the nearest hundredth. dY/dt = (0.7-0)/(30-0) = 0.7/30 = 0.02 B) Cladogram (Part 1) B) Cladogram Analysis (Part 2) Station 4: A) SEM Glaucoma problem See Mrs. Kravitz & Ms. Albu for answers B) Population growth A population of pandas exhibits a logistic growth. The carrying capacity of the population is 174 pandas and the rmax, the maximum per capita growth rate, of the population is 0.16 (Pandas/ [Pandas x month]. Calculate the maximum population growth rate for the population if the maximum growth occurs when N= K/2. Give your answer to the nearest tenth. N= K/2 = 174/2 = 87 dN/dT= rmaxN ([K-N]/K) dN/dT= (0.16) (87) ([174-87]/174) 7.0 Pandas/month Station 5: A) Restriction Mapping See Mrs. Kravitz & Ms. Albu for answers B) Primary Productivity The biomass of a rainforest in Costa Rica is 73% carbon. Additionally, the biomass increases annually at a rate of 3.2 x 105 kg/hectare. Calculate the mass of carbon accumulated and stored in 1.0 hectar of this fabulous forest in one year. Give your answer to the nearest hundred thousand (105) kilograms. 1 ha x (3.2 x 105 kg/hectare) x .73 = 2.34 x 105 kg Station 6: Linkage Mapping: A wild-type fruit fly (heterozygous for gray body color and normal wings) was mated with a black fly with vestigial wings. The offspring has the following phenotypic distribution: • Wild type: 778 • Black-vestigial: 785 • Black normal: 158 • Gray vestigial: 162 What is the recombination frequency between these genes for body color and wing type? (Recombinant offspring) (Total offspring) (158+162) / (778+785+158+162) 320/1873 0.171 = 17.1% B) Branching method In a trihybrid cross involving three traits, where the pigments are both AaBbCc, what is the probability of producing an offspring with the phenotype A-bbC-? ALL recessive = (¼)(1/4) (1/4) = 1/64 A-bbC- (3/4) (1/4) (¾) = 9/64 Station 7: A) Water Potential B) Free Energy Students would need to have previous knowledge that 68F=293 Kelvin