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**AP Biology: Math for Dummies**

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**Science Practices- The student can:**

-use representations & models to communicate scientific phenomenon & solve scientific problems. -use mathematics appropriately. -engage in scientific questioning to extend thinking or to guide investigations within the context of the AP course. -plan & implement data collection strategies appropriate to 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.

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**The Process of Evolution Drives the Diversity and Unity of Life**

Big Idea 1 The Process of Evolution Drives the Diversity and Unity of Life Change 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 Equations Probabilities

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**Water Potential Gibb’s Free Energy Big Idea 2**

Biological 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. 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. Water Potential Gibb’s Free Energy

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**Chi square Gene Linkage Big Idea 3**

Living 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 square Gene Linkage

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**Population Growth Energy Transfer Primary Productivity Big Idea 4**

Biological Systems Interact and These Systems and Their Interactions Possess Complex Properties Interactions 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 Productivity

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**100; this is a 3:1 phenotypic ratio**

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

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**In this genetic cross, Aa x aa, there are 348 offspring**

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

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**This is a 9:3:3:1 ratio, with 9/16 showing both dominant phenotypes. **

In a dihybrid cross between two heterozygotes, if you have 200 offspring, how many should show both dominant phenotypes? 112 This is a 9:3:3:1 ratio, with 9/16 showing both dominant phenotypes. 9/16 = .56 = 56% 56% of 200 is 112

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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? .6 64 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

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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

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**The ability to taste PTC is due to a single dominant allele (A)**

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

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**is 0.2, and the population is in Hardy-Weinberg equilibrium?**

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= .2 p= .8 Homozygous dominant (p2) = .64 (64%) Heterozygotes (2pq) = .32 (32%)

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**Number of Cells Observed Number of Cells Expected**

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

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**Forty flies were put into a choice chamber with two chambers**

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.

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Water Vinegar 11.52

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

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**Need help with standard deviations?**

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**The formula is easy: it is the square root of the Variance**

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: Work out the Mean (the simple average of the numbers) Then for each number: subtract the Mean and square the result (the squared difference). 3. Then work out the average of those squared differences.

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**To measure the population density of monarch butterflies occupying**

a particular park, 100 butterflies are captured, marked with a small dot on a wing and then released. The next day, another 100 butterflies are captured, including the recapture of 20 marked butterflies. One would estimate the population to be: a. 200 b. 500 c. 1,000 d. 10,000 (100 x 100)/ 20 = 500

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

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**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)

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**(the average partial pressure of oxygen at sea level is 21%)**

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 percentage Of 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%)

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**Sometimes we concentrate on the complicated math formulas, and we**

have students who can’t calculate percentages. Phase Number Percent spent in each phase Interphase Prophase/prometaphase Metaphase Anaphase Telophase

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The 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. Trial No Treatment 1% salt 3% salt 5% salt 7% salt 9% salt 1 47 41 25 28 24 5 2 46 42 32 23 21 6 3 34 18 4 57 44 17 39 27

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2 x 2 = 4 6 x 4 = 24

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**What is the mean rate of growth per day between day 5 and day 25**

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 hundredth of a cm. 18-3 = 15 15/20 days = .75 cm .75 x 25 = 18.75 = 36.75

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**What is the water potential of a cell with a solute potential of -0**

What is the water potential of a cell with a solute potential of MPa and a pressure potential of 0.43 MPa? -.24MPa

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**You measure the total water potential of a cell and find it to be -0**

You measure the total water potential of a cell and find it to be 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

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**A population of ground squirrels has an annual per capita birth rate**

of 0.06 and an annual per capital death rate of Estimate the number of individuals added to (or lost from) a population of 1,000 individuals in one year. 120 individuals added 40 individuals added 20 individuals added 20 individuals lost 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

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A small population of white-footed mice has the same intrinsic rate of increase (r) as a large population. If everything else is equal, a. the large population will add more individuals per unit time b. the small population will add more individuals per unit time c. the two populations will add equal number of individuals per unit time d. the J-shaped growth curves will look identical

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**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? a. 3, 800 b. 38,000 c. 380,000 d. 3,800,000 295,000,000/1,000 = 295,000 295,000 x 13 = 3,835,000

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**As N approaches K for a certain population, which of the**

following is predicted by the logistic equation? a. the growth rate will not change b. the growth rate will approach zero c. the population will increase exponentially d. the carrying capacity of the environment will increase

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**Which of the following might be expected in the logistic model of**

population growth? a. as N approaches K, b increases b. as N approaches K, r increases c. as N approaches K, d increases d. both a and b are true

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**According to the logistic growth equation:**

a. the number of individuals added per unit time is greatest when N is close to zero b. the per capita growth rate (r) increases as N approaches K c. population growth is zero when N equals K d. the population grows exponentially when K is small

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An experiment determined that when a protein unfolds to its denatured (D) state from the original folded (F) state, the change in Enthalpy is ΔH = H(D) – H(F) = 46,000 joules/mol. Also the change in Entropy is ΔS = S(D) – S(F) = 178 joules/mol. At a temperature of 20⁰C, calculate the change in Free Energy ΔG, in j/mol, when the protein unfolds from its folded state. The correct answer is joules/mol. ΔG = ΔH – TΔS ΔG = (46,000 joules/mol) – 293 K (178 joules /mol) ΔG = 46,000 joules/mol – 52,154 joules/mol = joules/mol

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

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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.

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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.

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How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE? a. 4 b. 8 c. 16 d. 32

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Given the parents AABBCc x AabbCc, assume simple dominance and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent? a. 1/4 b. 1/8 c. 3/4 d. 3/8

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**Consider a field plot containing 200 kg of plant material**

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

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Cytosine makes up 38% of the nucleotides in a sample of DNA from an organism. Approximately, what percentage of the nucleotides in this sample will be thymine? a. 12 b. 24 c. 31 d. 38

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