Multi-Station Math Review 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:
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
(Recombinant offspring)
(Total offspring)
(158+162) / (778+785+158+162)
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
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