AP Biology: Math for Dummies

Big Idea 1 The Process of Evolution Drives the

Diversity and Unity of LifeChange in the genetic makeup of a population over time is evolution.

Organisms are linked by lines of descent from common ancestry.

Life continues to evolve within a changing environment.

The origin of living systems is explained by natural processes.

Hardy-Weinberg EquationsProbabilities

Organisms use feedback mechanisms to regulate growth & reproduction & to maintain dynamic homeostasis.

Growth & dynamic homeostasis f a biological system are influenced by changes in the system’s environment..

Many biological processes involved in growth, reproduction & dynamic homeostasis include temporal regulation & coordination.

Big Idea 2Biological Systems Utilize Free Energy and

Molecular Building Blocks to Grow, Reproduce and Maintain Dynamic Homeostasis

Growth, reproduction & maintenance of the organization of living systems require free energy & matter.

Growth, reproduction & dynamic homeostasis require that cells create & maintain internal environments that are different from their external environments.

Water PotentialGibb’s Free Energy

Big Idea 3Living Systems Store, Retrieve, Transmit and Respond to Information Essential to Life Processes

Heritable information provides for continuity of life. Expression of genetic information involves cellular &

molecular mechanisms.The processing of genetic information is imperfect & is

a source of genetic variation.Cells communicate by generating, transmitting & receiving chemical signals.

Transmission of information results in changes within and between biological systems.

Chi squareGene Linkage

Big Idea 4Biological Systems Interact and These Systems and Their Interactions Possess Complex PropertiesInteractions within biological systems lead to complex properties.Competition & cooperation are important aspects of biological systems.Naturally occurring diversity among & between components within biological systems affect interactions within the systems.

Population Growth

Energy Transfer

Primary Productiv

ity

Phase Number Percent spent in each phase

Interphase 52

Prophase/prometaphase 12

Metaphase 2

Anaphase 5

Telophase 1

1. On the AP Biology Exam, you probably won’t have to do the actual count andidentification; you will be given a data table and asked questions, like-

From the data, calculate the percentage of time a cell spends in interphase

Calculate the percentage of time a cell spends in mitosis

2.

Step 1- count the total # of cells 72Step 2- divide 52/72 x 100 to get % of time spent in interphase 72%Step 3- Subtract 72% from 100% to get time spent in mitosis 28%

A Cellular Biologist wants to double check that statement that cells spend 90 percent of their time in Interphase as compared to the various stages of Mitosis. She grows some Allium in her laboratory. She then takes one of the plants, cuts off the root tips, stains the DNA in the cells so as to be able to see the stages of the cell cycle. Her hypothesis states “If cells spend 90 percent of their time in Interphase, then she should be able to calculate the relative time existing between Interphase and Mitosis based upon the cells counted in her specimen.” She counted 1000 cells from her preserved specimen under the microscope. Her data are shown below. Calculate the X2 to the nearest hundredth.

Stage of the Cell Cycle Number of Cells Observed Number of Cells Expected

Interphase 872 900

Mitosis 128 100

3. To measure the population density of monarch butterflies occupyinga particular park, 100 butterflies are captured, marked with a small dot on a wing and then released. The next day, another 100 butterfliesare captured, including the recapture of 20 marked butterflies. What would you estimate the population to be? NOTE- this is not on your formula sheet!

(100 x 100)/ 20 = 500

4. Study this age structure diagram that shows the human population for India. What percent of the population is under 20?

2 x 2 = 4 6 x 4 = 24

5.

6. The ability to taste PTC is due to a single dominant allele (A). You sampled 215 individuals in biology and determined that 150 could taste PTC and 65 could not. How many individuals in this population show the following genotype?

AA, Aa, aa

65/216 = .3 = q2

q = .55p = .45

(.45)(.45) = .2 = 20%20% x 215 = 43 AA2(.45)(.55) = .495 = 50%50% x 215 = 107 Aa

7. The allele for the hair pattern called “widow’s peak” is dominant over the allele for no “widow’s peak”. In a population of 100 individuals, 64 show the dominant phenotype. What is the frequency of the recessive allele?

.664 show the dominant phenotype, So 36 show the recessive phenotype.Since this is a population of 100, 36% Show the recessive phenotype .36 = q2

.6 = q

8. Atmospheric pressure is the combined partial pressures of all of the gases that make up the atmosphere. At the summit of a high mountain, the atmospheric pressure is 380mm/Hg. The partial pressure of oxygen is 69mm/Hg. What percentageOf the atmosphere is made up of oxygen at this altitude?

18%69/380 = .18 = 18%(the average partial pressure of oxygen at sea level is 21%)

9. Cytosine makes up 12% of the nucleotides in a sample of DNA from an organism. Approximately, what percentage of the nucleotides in this sample will be thymine?

38%

Remember Chargaff’s Law? Amount of A = T and C=G

If 12% is cytosine, then 12% is guanine76% is made up of adenine and thymine; divided equally between them

10. How many oxygen atoms would be in the polysaccharide that results from 100 glucose molecules being linked together by dehydration synthesis?

H2O

HO

HO H

H HHO

100 C6H1206 would give you 600 oxygen atoms99 water molecules would be produced by dehydration synthesisEach water molecule contains one oxygen atom

600 - 99 = 501

1st Law of Thermodynamics- energy cannot be created or destroyed, but it can change form.18,000 energy accumulated as biomass; 12,000 going to the tree layer; 4,400 going to the shrub layer; 1,600 left, to go to the grass layer. 1,600 is 9% of 18,000 (1,600/18,000 x 100)

11.

12.

13. Geneticists working in an agriculture lab wanted to develop a crop that combines the disease resistance of rye grain with the high crop yielding capacity of wheat grain. Rye grain has a diploid chromosome number (2n) of 14 and wheat grain has a diploid chromosome number of 42. The resulting grain is called triticale and is an alloploidy plant. How many chromosomes are found in the pollen grain of triticale?

Alloploidy results when two different plant species combine their diploid genome to make new and unique species. That would mean that this particular species would have 56 chromosomes. The cells in a pollen grain of would be haploid so the resulting number is 28.

14. What is the water potential of a cell with a solute potential of -0.67 MPa and a pressure potential of 0.43 MPa?

-.24MPa

-0.67 + 0.43

How many naked ladies to you see in this picture?

15. You measure the total water potential of a cell and find it to be -0.24 MPa. If the pressure potential of the same cell is 0.46 MPa, what is the solute potential of that cell?

Since water potential is equal to the solute potential + the pressure potential,

-0.24 MPa = 0.46 MPa + X. Solve for x= -0.7

16. A wind blown pollen grain with a solute potential of -3.0 MPa has dried out somewhat after blowing around in the wind. This has caused its turgor pressure to go to zero. It lands on a flower stigma whose cells have a solute potential of -3.0 Mpa and a pressure potential of +1 Mpa.

Which way will water flow? From the pollen grain to the stigma or the stigma to the pollen grain? Show how you deduced your answer.

17. From the Osmosis/Diffusion Lab. Once the molar concentration of a sucrose solution has been determined, how would you determine the water potential of that sucrose solution? Work one to show how. Assume room temperature (22C).

.4

-(1)(.4)(.0831)(295) = -9.8Since this was done in an Open container, your pressure potential is 0, so the water potential is the same as the solute potential

18. Use the Station 1 data to calculate the Primary Productivity of a water sample.

Report your answer in units of mg Carbon fixed/Liter

Station 1

4.2 mg O2/L 0.698 = 2.9 mL O2/L

2.9 mL O2/L 0.536= 1.6 mg Carbon fixed/L

19. What is the mean rate of growth per day between day 5 and day 25? Record your answer to the nearest hundredth of a cm.

If this same rate of growth continues, how tall will the plant be on day 50? Record your answer to the nearest hundredthof a cm.

18-3 = 1515/20 days = .75 cm

.75 x 25 = 18.7518.75 + 18 = 36.75

20. A population of ground squirrels has an annual per capita birth rate of 0.06 and an annual per capital death rate of 0.02. Estimate the number of individuals added to (or lost from) a population of 1,000 individuals in one year.

dN/dt = B-D Change in population size/time = Birth rate – Death rate

0.06 – 0.02 = 0.04 x 1000 = 40 individuals added per year

21. In 2005, the United States had a population of approximately 295,000,000 people. If the birth rate was 13 births for every 1,000 people, approximately how many births occurred in the United States in 2005?

295,000,000/1,000 = 295,000295,000 x 13 = 3,835,000

22. A population of June beetles is estimated to be 3,000. Over the course of a month, there are 400 births and 150 deaths. Estimate r and calculate what the population size is predicted to be at the end of the month. Predict the population at the end of the second month.

Birth rate = 400/3000 = 0.1333 birthsDeath rate = 150/3000 = 0.0500 deaths r = birth rate – death rate = 0.1333 – 0.0500 = 0.0833 3000 x 0.0833 = 250250 + 3000 = 3250

3250 x 0.0833= 271271 + 3250 = 3521

23. A population of frogs exhibits logistic growth. If the carrying capacity is 500 frogs and r = 0.1 individuals (per month), what is the maximum population growth rate for this population? (Maximum population growth rate occurs when N = K/2)

K/2 = 500/2 = 250= N

dN = rN(K-N)dT K

0.1 x 250 (0.5) = 12.5 individuals/month

12.5

24. What is the population standard deviation for the numbers: 75, 83, 96, 100, 121 and 125?

Raise your hand if you need help.

1. Find the mean2. Find the difference between the mean and each number (some will be + and some will be –)3. Square each difference (now they’re all +)4. Add them together5. Take the square root6. Bam! That’s it.

I hate standard deviations.

25. How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?

Aa and Bb and Dd can each make 2. CC and EE can only make 1.

2 x 2 x 1 x 2 x 1 = 8

26. Assuming the genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes?

aabbccdd AaBbCcDd AABBCCDD AaBBccDd AaBBCCdd

AaBbCcDd x AaBbCcDd¼ AA ½ Aa ¼ aa¼ BB ½ Bb ¼ bb¼ CC ½ Cc ¼ cc¼ DD ½ Dd ¼ dd

¼ x ¼ x ¼ x ¼= 1/256

½ x ½ x ½ x ½ = 1/16

¼ x ¼ x ¼ x ¼= 1/256

½ x ¼ x ¼ x ½ = 1/64

½ x ¼ x ¼ x ¼ = 1/128

The End

Science Practices- The student can:1. -use representations & models to communicate

scientific phenomenon & solve scientific problems.2. -use mathematics appropriately.3. -engage in scientific questioning to extend thinking

or to guide investigations within the context of the AP course.

4. -plan & implement data collection strategies appropriateto a scientific question.

5. -perform data analysis & evaluation of evidence. 6. -work with scientific explanations & theories. 7. -is able to connect & relate knowledge across

various scales, concepts & representations in and across domains.

Experimental Design

Pose scientific questions Refine scientific questionsEvaluate scientific questions

Put your enzyme stuff here-

Justify the selection of the kind of data needed to answer a particular scientific question.

Design a plan for collecting data to answer a particular scientific question.

Collect data to answer a particular scientific question.Evaluate sources of data to answer a particular scientific question.

Analyze data to identify patterns or relationships.Refine observations and measurements based on data analysis.Evaluate the evidence provided by data

sets in relation to a particular scientific question.

Science Practices- The student can:1. -use representations & models to communicate

scientific phenomenon & solve scientific problems.2. -use mathematics appropriately.3. -engage in scientific questioning to extend thinking

or to guide investigations within the context of the AP course.

4. -plan & implement data collection strategies appropriateto a scientific question.

5. -perform data analysis & evaluation of evidence. 6. -work with scientific explanations & theories. 7. -is able to connect & relate knowledge across

various scales, concepts & representations in and across domains.

Models & Representations1. Create a model

2. Describe what is taking place

Specifically, describe the relationship between colony growth and sucrose concentration.

3. Refine Model

Using the model, calculate the water potential of the potato cores.

4. Use and ApplyThe purpose of a particular investigation was to see the effects of

varying salt concentrations of nutrient agar and its effect on colony formation. Below are the results Determine the mean for each treatment and graph the results.

 

What might you be asked to do with this graph?

1. Describe the relationship between the salt concentration and colony growth

2. Predict the salt concentration that would allow no colony growth. Justify your answer.

5. Re-express

Signal transduction pathways are regulatory mechanisms in living things. Identify a signaling molecule and the response brought about in a) an animal b) a plant

Extra Mathcause we love math

Forty flies were put into a choice chamber with two chambers. In one chamber there was a cotton ball soak with vinegar. The other chamber had nothing. After 20 minutes the number of flies were counted in both chambers. This was repeated four more times. Perform a chi-square analysis to determine if the difference between in the number of flies found in the two chambers is significant.

76 124

Water Vinegar

100 100

-24 24

576 576

5.76 5.76

11.52

In a certain population of deer on Fire Island, NY, the allele for a black spot behind the eye is dominant to the allele for no spot. After the hunting season, the percent of deer with no black spot is 15% and the population is in Hardy-Weinberg Equilibrium. What is the frequency for the allele for having no black spot, to the hundredths?

15% = .15 = q2

q = .39

In a dihybrid cross between two heterozygotes, if you have 200 offspring, how many should show both dominant phenotypes?

112This is a 9:3:3:1 ratio, with 9/16 showing both dominant phenotypes. 9/16 = .56 = 56%56% of 200 is 112

In this genetic cross, Aa x aa, there are 348 offspring. How many individuals are expected to have the dominant phenotype?

174; this is a 1:1 ratio, so 50% are expected to Have the dominant phenotype

In a typical Mendelian monohybrid cross, two heterozygotes produce 400 offspring. How many individuals are expected to have the recessive phenotype?

100; this is a 3:1 phenotypic ratio

A study was conducted on the island of Daphne Major in the Galapagos Islands by Peter and Rosemary Grant. This study lasted over 20 year s. The study investigated how the type of seeds available to the finches impacted the depth of their beaks. In years when rain and water were plentiful, the available seeds were smaller and easy to crack. In years experiencing drought, fewer seeds were produced, and the finches had to eat the larger, leftover seeds produced from previous years. During years of drought, birds with a greater beak depth had a selective advantage.

Use the data above to determine the increase in the mean of the depth of the beak between the wet and dry years. Give your answer to the nearest hundredth of a millimeter.

In geckos, spots are dominant to the solid color. If the frequency In a population of 700 geckos, what percentage of the geckos would have spots, if the frequency of the recessive allele is 0.2, and the population is in Hardy-Weinberg equilibrium?

96%

q= .2p= .8

Homozygous dominant (p2) = .64 (64%)Heterozygotes (2pq) = .32 (32%)

The formula is easy: it is the square root of the Variance. So now you ask, "What is the Variance?“The Variance is defined as: The average of the squared differences from the Mean.

To calculate the variance follow these steps:1. Work out the Mean (the simple average of the numbers) 2. Then for each number: subtract the Mean and square the result (the squared difference). 3. Then work out the average of those squared differences.

14,000 (a) 35 (c)180 (b) 100 (d)You are starting with 87,400 kJ and simply subtracting to get the answers.

Consider a field plot containing 200 kg of plant material. Approximately how many kg of carnivore production can be supported?a. 200b. 100c. 20d. 2