Di.ulled H2O

Pa.:m.11 et Iodide kit:tier

04,,ssur:Itarc, Aquect, Sciutic,u

20

25 111101111111110 101111 11

1111111111 III

1.0 M 0.0 M

02 IA

0 4 M 0.6M

0.8 M

Sucrose Morality

Rns ■NwiK Key

AP Biology Graphing & Calculations

Diffusion & Osmosis

Students in an AP Biology class explored the diffusion of different molecules through dialysis tubing, a semipermeable membrane.

They used glucose test strips to check for the presence of glucose in an aqueous solution and potassium-iodide (IKI) to test for the

presence of starch in the same aqueous solution. IKI reacts with starch to give a dark blue, almost black color. When IKI reacts with

starch, it becomes part of the starch molecule and is removed from solution. The design of the experiment is shown below and the following data were obtained:

Figure 2

sample tattle 1 SoIntir n Color Glucose Test Results

Location Solution Initial Final Initial Final

Dialysis Bag G,.,11::°„7:1 Clear Dark Blue + +

Cup IKI Amber Amber ..4-

1. What does the data tell you about the sizes of the molecules relative to the pore size of the dialysis tubing?

no\f cuAts oar ---S--- cA-4 °O\uwse, a're- 5wVatv one (_>st w■AAt 5-k u I ch is off .

2. Does this activity account for the diffusion of all the molecules found in the initial set up? If not, what data could have been collected to show the net direction of diffusion of this molecule or molecules?

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wader.

ANCAk COAk t(\&A VNACKSS tS bads or

so.k \ickk out\ci --■(\01 Waft 1(\ \00.(bS CO

3c\oAlc Vetin okfuouted

Students in an AP Biology class designed an experiment to investigate the influence (if any) of solute concentration on the net

movement of water molecules through a semipermeable membrane. The solute they used was sucrose (cane or table sugar) in the following molar concentrations: 0.0 M (distilled water); 0.2 M; 0.4 M; 0.6 M; 0.8 M; and 1.0 M.

3. Sketch and label the probable experimental setup used by the students: t be _ _ .

ev-s •„„,

1-k-1.0 4(A4. fS

ieN'Aiat a-dA, -V‘ry,3 v)ags.

W1\\ NCI-(3

6:\reA" 41:t the '(\n6\0,f svc rose soVA0A 5. What is the Independent variable?

V\6\6,.(‘A1xp s u1/4crost 544.L.V1kX1

6. What is the dependent variable?

• 6nt (-1 1,A0,% a.,031s‘5 \00,.zs 1 .

7. What constants should have been used by the students?

4 MC\G--\ V\C•SS 0;- ‘0C.Z5 k`en)troAxlft. ty%

y)( S‘f The students collected their data and graphed it — the following results were obtained:

Title: Percent Changer in Mass vs. Sucrose Mnlarity

NAASt,f'S

4. What would have been an appropriate hypothesis for this experiment?

The Qerc~ef‘k e‘-‘a,n9€. ■1\ ASS 6k VC.

*!-‘e ‘01:15..

10

15

\log,c‘l ros-e

tiNAo_r%-tieS

Enzyme Catalysis

In a particular experiment, students observed some characteristics of enzyme action. The specific reaction they investigated was the

decomposition of hydrogen peroxide by the enzyme, catalase. At room temperature, hydrogen peroxide very slowly decomposes

into water and oxygen. The addition of catalase lowers the activation energy of the reaction until it proceeds swiftly at room

temperature. At the end of the reaction, the catalase is unchanged and is available to catalyze the reaction of more hydrogen

peroxide. Catalase, like most enzymes, is a protein. Its ability to form a complex with hydrogen peroxide is based on its molecular

shape. Any factor that can alter the shape of a protein molecule can be expected to impact the ability of catalase to facilitate the

decomposition of hydrogen peroxide. The following graph was produced by the students:

Title: Catalyzed jinx Decompositian 4 i4,02

08

0 6

0.5

0.4

0.2

0.1

0 20 40 60 80 00 120 lax 160 180

Time tsecan.40

What was the independent variable in this experiment? What was the dependent variable?

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I ■rie :-- rake 0-C- Aecomp.

1.

2. What does the line on the graph represent?

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Zygote

• (41pliodi

Fertility, -

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DJdlet No crgtol owr

<di ,

it it

• •

• • 444,

1 11 dl

Ave

IMPIZ; ..1 Filornen13

i 1 0 0 •

0 0 # •tie

Four black ascospores In • run uext to four tau atco.,poles lu a cue indicatei that cro, lug wet Cu \ OT occurred. Airy other artouseintut of ..rosport,ludlcales that tio,int user has hih.ru place.

''Ascospores (hoploul)

Bleck 5pro (AoplOid)

1]

8. On the basis of the results, write a statement that expresses the relationship of solute concentration and direction of net

movement of water molecules in osmosis.

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9. In which, if any, of the experimental setups were the solutions in the bag and outside of the bag isotonic to each other?

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10. When you drink a glass of water, most of it is absorbed by osmosis through cells lining your small intestine. Drinking seawater

can actually dehydrate the body. How?

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Mitosis & Meiosis

Environmental factors can affect the rate of mitosis in plant cells. Ina recent issue of Biology Weekly scientists reported that a

fungal pathogen may negatively affect the growth of soybeans (Glycine max). Soybean growth decreased during three years of high

rainfall, and the soybean roots were poorly developed. Close relatives of the R. anaerobis are plant pathogens and grow in the soil.

A lectin-like protein was found in the soil around the soybean roots. This protein may have been secreted by the fungus. Lectins

induce mitosis in some root apical meristem tissues. In many instances, rapid cell divisions weaken plant tissues.

Suppose you were asked to investigate whither the fungal pahtogen lectin affects the number of cells undergoing mitosis in a

different plant, using onion root tips. You hypothesize that treating onion root tip cells with lectin will induce mitosis in the cells, as

was the case with the soybean plant. After designing and conducting your experiment, the following data are reported:

1' S 1\411 1-eckIn uattl

49

obsemi,ho.kst NO

yaws - e Cc eck- 0110-K6-

Number of Cells

Group Interphase Mitosis Total

Control/Non-Treated 1550 200 1750

Experimental/Treated

w/ Lectin 500 1250 1750

OF= \

Cr .\-\-\ ki0.1u1f s kn3 19

Sordaria fimicola is a common species of ascomycete fungi that grows on the dung of herbivores. Eight spores (ascospores) are

produced in an ascus. Many asci are grouped together within a vase-shaped structure called a perithecium. Two nuclei within a

developing ascus fuse to produce a diploid (2n) nucleus. This diploid nucleus then undergoes meiosis, followed by a mitotic division

to produce eight ascospores in a linear series within the ascus. When the growing filaments of two haploid strains of Sordaria that

produce spores of different colors meet, fertilization occurs and zygotes form. The figure below shows spore formation in Sordaria:

3. What is the rate of the enzyme-catalyzed reaction? What does the graph tell you about the rate of the reaction over time?

nes

1. Are deviations from the expected value due to chance alone? Calculate

the chi-square value for this data

Is

your hypothesis

supported or refuted? Does lectin induce m i tosis in onion root cells?

72, (500 15 5()Z. ( 12-co- + 5 5 \ (.9 311

ISSO 0200

X-- (X L va\u‘e.? crAico_k vo,W. Null \oykes-ks is

- \edit.) DOGS

O 10 15

—17(we, (MN) (

•

Sample Table :t• Percent Light Transmittance of Chloroplast Suspensions

Time

CIIVf3tte 0 D min 5 mln 10 min IS min

2. UnhoilediDark 35 1 35.4 34 y , 34 o

3. Unhoiledilight 36.3 40.6 44.8 48.9

4 Boiled/Light 33.7 34.0 34.0 34.0

5. No ChInropiosts 35.0 35.3 35.4 35.3

The number of map units between two genes or between a gene and the centromere is calculated by determining the percentage of

recombinants that result from crossing over. The greater the frequency of crossing over, the greater the map distance. Calculate

the percent of crossovers by dividing the number of crossover asci (spores arranged 2:2:2:2 or 2:4:2) by the total number of asci x

100. To calculate map distance, divide the percent of crossover asci by 2. The percent of crossover asci is divided by 2 because only

half of the spores in each ascus are the result of crossing over.

A group of asci formed from crossing light-spored Sordaria with dark-spored Sordaria produced the following results:

Number of Ascl Counted Spore Arrangement

7 4 light / 4 dark spores

8 4 dark / 4 light spores

3 2 light / 2 dark / 2 light / 2 dark spores

4 2 dark / 2 light / 2 dark / 2 light spores

1 2 dark / 4 light / 2 dark spores

2 2 light / 4 dark / 2 light spores

2. How many of these asci contain a spore arrangement that resulted from crossing over?

t 0,-(\ 0.4 6-v‘w wt.* lks■m ack6c._ 4e, r QS o- CrosSi0 over

3. From this small sample, calculate the map distance between the gene and centromere.

c )0 tA6.,4i• Ma? (Waive c x co An eltio4eS t.lt /A5 X1001, k iCd 63 a CA SC

Plant Pigments & Photosynthesis

In the light reactions of photosynthesis, light energy is absorbed by chlorophyll and used to excite electrons. The excited electrons

then enter one of two electron transport chains. One chain converts ADP + P to ATP; and the other converts NADP + H to NADPH. In

a particular experiment, students added a solution of DPIP (a blue dye) to a suspension of chloroplasts. The DPIP substituted for NADP in the light reactions: DPIP + H DPIPH. DPIPH is colorless, so as the light reactions took place, the blue color of the solution

diminished. The students used this color change as an indication that the light reactions were taking place and they used the rate at

which the color change took place as a measure of the rate of the light reactions.

The students used a spectrophotometer to measure loss of color by DPIP. They prepared a sample by adding chloroplast suspension,

DPIP, and a buffer to water in a tube or vial called a cuvette. The instrument works by shining a light of known intensity into one side

of the cuvette. On the opposite side of the test chamber is a photocell. DPIP will absorb some of the light that enters the cuvette,

thus, the photocell will "see" less light. As the light reactions take place, there will be less DPIP and the photocell will "see" more

light. Since DPIP absorbs light most strongly at orange-red wavelengths, they set the spectrophotometer to read the amount of light

transmitted in that part of the spectrum.

The following were the content of the cuvettes and data obtained from their experiment:

Contents of the Cutrettes

C.Pt1VY

CfAlt,11. 1

RI4nlo IllsMam/I

2

1.1m114.41, Dv k

3

U4MmIel: Light

4

BoileM 1,10

S 5,

1:1110,,IasI 4 41141* Putter I mt

' ' I ml I ml I rot

Dist Red .120 4 mt 1 ml 3 mt 3 ,,,,.

3 mt. p.st, II,.

Pi'4' — I mt. I mt I ml I mt. UMmf14,1 „...,,,,,,p,mi, ) drIms 1 Mem I t14,4, 4

P,414411 ChlmorMosts

3 drops

1. Write a hypothesis that this experiment was designed to test:

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"k- L)r 44"- ‘ic\k* cook sc C`bilir4kfs\%• 2. What was the independent variable? What was the dependent variable?

-b oile. by: '10 Is\,\A- iro,t1sroAk-ance

3. Which cuvette served as a control for this experiment? Explain your answer.

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. \) \"b rki1 /41%,c Cue*eS Can cooToseikb rc 4. What variables are tested in this experiment?

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UmoAvot e■AsTv,e5 (WA Vs4 (0,6:■AS

6. Plot the data from sample table 3 in the space provided below.

rm. C,6qe in 1-ictAst411\61ve -1-uwf

5. What affect did boiling have on the chloroplast suspension?

Cell Respiration

In a particular experiment, students in an AP Biology class used a respirometer to measure the rate of respiration of germinating and

nongerminating pea seeds at two different temperatures. The respirometer consists of a vial that contains the peas and a volume of

air. The mouth of the vial is sealed with a rubber 1-hole stopper that has a pipet inserted in it. The respirometer is submerged in

water. If the peas are respiring, they will use oxygen and release carbon dioxide. Since 1 mole of carbon dioxide is released for each

mole of oxygen consumed, there is no change in the volume of gas in the respirometer. (Avogadro's Law: At constant temperature

and pressure, 1 mole of any gas has the same volume as 1 mole of any other gas.) The students altered this equilibrium by placing a

solution of potassium hydroxide (KOH) in the vial. Potassium hydroxide reacts with carbon dioxide to form potassium carbonate,

which is a solid (CO2 + 2 KOH --4K2CO3 + H20). Since the carbon dioxide produced is removed by reaction with potassium

hydroxide, as oxygen is used by cellular respiration the volume of gas in the respirometer will decrease. As the volume of gas

decreases, water will move into the pipet. Students used this decrease of volume, as read from the scale printed on the pipet, as a

measure of the rate of cellular respiration. The following data were obtained from the experiment:

Oxygen Consumption of Peas

0.50

0.40

0.30

Ciafrninating peas at morn temperature X Cry peas at roam iernparatura • Germinating peas ai lin

A Dry peas at 16,C., 0.20

0

ls

20

Time (Minutes)

1. Write 2 hypotheses that this experiment was designed to test.

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2. Which respirometer served as the control? Explain.

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(

3. What constants were important in this experiment?

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5. From the graph, calculate the rate of oxygen consumption for each treatment.

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6. If you used the same experimental design to compare the rates of respiration of a 25g reptile and a 25g mammal at 10.C, what

results would you expect? Sketch a graph of your expected results and explain your reasoning.

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0.10

0

4. Using the graph, summarize your findings, comparing results from respirometers 1 and 2, and results obtained at room

temperature vs. results at the colder temperature. Speculate as to the cause(s) of any differences between the treatments.

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low-ts A-evyy Alcce&s.ect 01/4-‘1 has— )0-4' jA2.6 atagalex, ki,31k01 late.Si 4ein? .

Plate III LB agar

no plasmid added

Plate IV LB agar with kenamccin

no plesmld added

c(Aompit5

Restriction Enzyme Analysis of DNA

Gel electrophoresis is a procedure that separates molecules on the basis of their rate of movement through a gel under the

influence of an electrical field. The direction of movement is affected by the charge of the molecules, and the rate of movement is

affected by their size and shape, the density of the gel, and the strength of the electrical field. DNA is negatively charged, so it will

move toward the positive pole of the gel when current is applied. When DNA has been cut by restriction enzymes, the different-

sized fragments will migrate at different rates. Smaller fragments move more quickly, therefore they will migrate the fastest

through the gel. The length of each fragment is measured in number of DNA base pairs.

The figure below shows the results of electrophoresis performed by a group of AP Biology students. Semilog paper was used to plot

the results of the Hindilll digest. Since its fragments sizes are known, this is the standard curve. It was used to determine the other fragement sizes from DNA I and DNA II samples.

1. How many base pairs is the fragment circled in Figure 6.4?

3500

DNA

H*4iii I1 it I li t

DNA!) I _fl

DWI I

Figure BA

Below is a plasmid with restriction sites for BamHl and EcoRl. Several restriction digests were done using these two enzymes either

alone or in combination. Use this data to answer the questions below. HINT: Begin by determining the number and size of the

fragments produced with each enzyme. "kb" stands for kilobases, or thousands of base pairs.

2. In the samples below, which lane shows a digest with BamHl only? •

• °' 3 resificiion stl-es leAs'ol',A so kjoi VCJOLt( 3 -cyri ("yrt

3. Which lane shows a digest with both BamHl and EcoRl? HINT: Begin by determining the number and size of the fragments

produced with each enzyme. "kb" stands for kilobases, or thousands of base pairs.

UM?, • qtr{ at 1-1 r esiYichon51-{-es too*k

6(Nsyt■IS- c)o L\ANufe v-M■ ‘oc T--fackt4,h

Plasmic)

Barn HI

21 ti

Bern Hl -

4 it,

Eco RI

Transformation In Bacteria

Genetic transformation occurs when a host organism takes in foreign DNA and expresses a foreign gene. Bacterial cells have a single

main chromosome and circular DNA molecules called plasmids, which carry additional genetic information. Plasmids are circular

pieces of DNA that exist outside the main bacterial chromosome and carry their own genes for specialized functions including

resistance to specific drugs. In genetic engineering, plasmids are one means used to introduce foreign genes into a bacterial cell.

In a molecular biology laboratory, a student obtained competent E. coli cells and used a common transformation procedure to

introduce the uptake of plasmid DNA with a gene for resistance to the antibiotic kanamycin. Competent cells are cells that are most

likely to take up extracellular DNA. The obtained results are shown below:

If there is no antibiotic in the agar (growth medium for bacteria), then bacteria will cover the plate with so many cells that a "lawn"

growth is produced. Only transformed cells can grow on agar with antibiotic. Since only some of the cells exposed to the kanamycin

plasmid will actually take them in, only some cells will be transformed, and individual bacterial colonies are produced on the agar

plate. If none of the E. coli cells have been transformed, nothing will grow on the agar plate containing antibiotic.

1. On which petri dish do only transformed cells grow?

2. Which of the plates is used as a control to show that nontransformed E. coil will not grow in the presence of kanamycin?

3. If a student wants to verify that transformation has cocurred, what procedure might he/she use?

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oseir kaurao,y in and mcv1,01/44-e —

o. \°'-u)(\ cff ovikif\ S\r■back rasal-b

4. During the course of an E. coil transformation laboratory, a student forgot to mark the culture tube that received the

kanamycin-resistant plasmids. The student proceeds with the laboratory because he thinks that he will be able to determine

from his results which culture tube contained cells that may have undergone transformation. Sketch and label a plate below

that would most likely indicate transformed cells.

?\0-4- € uSti'h 1004 Co\OAifs ov)v\ on LA

ckv - \/341-\ 4(\mn'i".

lot A- 11.6.A

Gel lanes I II III IV V

, f .1 IMP Me •IM

4•10

■111•0

INVO• AMR

6 rb —

3 k b MID ONO OM.

Plate II.

LB agar with kanamycin +ken plesmld

Plate I LB agar

+ken plasmic'

(A 0 Li CO, SY' 190.5

57'1- 571.0 571.5

lib -190 .5 190.5

63.5)a' 63.s-

3 Se7o ualtAt -

Population Genetics - Hardy-Weinberg Calculations

1. In the Sample Table 2, are the Generation 5 values

for p and q different from their initial values?

MC) 1-40-AA CU■S- 1-k-9

2. Is your answer to 1, above, consistent with the

alleles being at equilibrium? If not, why not?

GeN. coo9 s+e,21 cArtaii■co Se4NAk-

vat10.' 9'v, wNct.A. °cued", tyti skstASO r 3i co

3. Review the five conditions that must be et for allele frequencies to remain constant. Which, if any, of these conditions might

4

c

In ter I I

Au 4 =O.s = 1.5

p = OR = 22

Pu,tit (ego-) Pp SJ " twit

not have been met in this simulation? &V■.- s ( 2 e -r1,/ pet.ul ts 4(.9-0

Sample Table 1: Selection

(Jim .4-141._ d;cVeitA,A.3-.

eJe_eti a4t. kw, ARAA.41).LJ out \,(9-rnozu\c‘&u..S AQ-Cg4844 Utd

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In the Sample Table 3, are the Generation 5 values for p and q

different from their initial values? Explain why the values did or did

not change.

5. Compare the Generation 5 values for p and q calculated in Activity B with t

calculated for Activity C. Describe your findings.

-rktcdteleuituez Gem C.-- A 41 3 -rt,404.631t Go, 4 4- c cutAL.

ne_Ai (4) In 6. What would you expect t appen to the frequencies of p and q if you ran the simulation for another e gene Lions? 0,40

17)., j_tA)(3-14.1fAyt9od) CAMe.61:64V"-tgle0 4.13 z e

7. This simulation involved very high selection against the recessive allele. Did selection eliminate the allele from the population?

If not, why did the allele persist?

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1-L.t Lizi(Aozw-le_

Genetics of Organisms -Chi Square Analysis:

Consider the following data for F 2 grains of corn from a dihybrid cross of corn with grains that are purple and starchy and corn with

grains that are yellow and sweet (purple is dominant to yellow and starchy is dominant to sweet). The data are used to determine

whether or not corn color and sweetness assort independently or are linked on the same chromosome.

1 2 3 4 Total

Phenotype Purple, Starchy l.'iiiple, Swoon YelitlW, Starchy Y.,iit1W, Swart 4

Number of Grains

lActual Count;

577 204 176 59 1016

Expected Number of Grains 571.5 190.6 190.5 63.5 1015 •

Jul i . What is the correct null hypothesis for this exp e obse c-RA

Sk- It C61-k-o+ 034115 sv

LLotiaiwitio Ca1104". ■ A 2. Calculate x2 for the actual count above. Show all work and state whether the results support the null hypothesis in favor of the

expected ratio or cause you to reject it.

3. If rejected, identify possible sources of error that could cause this.

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Sample Table 2: Hardy-Weinberg Equilibrium

Initial Clie. Frequenci..-5

= 0.5 ' .1 = , 4

Initi.d lienutype

A.

21,1;en,41Pe ill, r,,,,i,

1 s. A .■

L,1.11.r.111011 2 11.11frowv) 12

0,11..r.ilion 3 ,,,,li,,,,,0 12

Gent:ration 4 II Iii I on, 17

(..ilvf.111o. 1 15

generation 5 Class ji - 0.5 1 ,

Frt.:Neu:it,

r) Ai k6f(-14-4.19`"' Generation 5 values for p and q th yo just

Cole 1,..quemit •

, _ , a

3

Initial Ga. i. •

=11111 MESIMMI:=111

121=1111123121•1111MMIIIIIMIll 0=1111111:111211M11111=ME•11 1=111EMEICIMIIMI=11•1111= 11111=111111=111IMINIM= INIMMIZIEMEMEINIE=INIMEN

Stc Genotwe

Dafc‘ 14-t4xi gatt. 00f- ciAiu2A"

Troi,esei a_4i f5k wo/tA 149-1.6 LA-A'

/

. .

. . .

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4. Interpret the graph above - explain how each environmental factor affects the transpiration rate in plants:

30

I 30c 710

2 LID

?ID

15.0

I ;0

eto

b0

"SO

A-y) e)./A oitAttA,,, L

t.St•vti) "iniec4ekx.4 1A- 0 -4 CLA

ecAL

rat' Ag.4.0.■

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Th tNa ro-t7 ;

Transpiration in Plants

Activit A: Nivosuring Trampinition With a Potometer

lirvatnwral A, ambient cOtaifirrrts

Sample Table 1: Potorneter Reading,.

,,mt fovol, , 1•, ,:, „11 2; V

..:', :,:,,, :V 0.“ :.±1., :V i,245 0,42 e3:

Wet., CL' .':4 OA; ' C.: ...,4 21! 034

11 Mc!

,,,,,e en , :ii ,i.s.: S,,,,t. ,N.6, 114t 1:54,

Tidal lealsorlace area 5. 00271

Sample Table 2 Class Averages. Ire Cumulative Water leas leetim,.

le.,a3twra 30 mil

A. Normal fl,,m k. ..1,,i 11. V.itiiiii

i Mottng Aii,

II f Iles Iturni,,,,,

tAk tra fwo :k,110,t11 v4.41. ktew ,nagan.4 oung odtAnc,frk.

Activity C: Heart Rate c•f• Dolmia

Sample Table 3: Heart Rate of Daphnia

1 omp,ra, u re -4: /-1,ArrIvarsil V 6ec 11,11.■ T3I, ill N'iti, min qi,oril,,t,'10 ..,/,‘ x f •.!

, 25 1 111'. 12 30 2 .... ,:i I , 5.1 'II, 16 61 SW 24 72 432

Graphing Heart Rate of Water Flea

1. Graph the temperature and heart rate data of Daphnia.

Give the graph and axes appropriate titles and labels.

1. Graph the class averages from Table 2. Give the graph and axes appropriate titles and labels.

2. Interpret the graph above - explain how temperature affects the heart rate of Daphnia.

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r e ce IT this a

r had been performed with an endotherm, Mw ght the resu hav differed? Sketch a g ph of your expected results

and explain.

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Sample Tidal. 1: Temperature and Dissolved Oxygen

temperature Class Average IX)

Class Average % Saturation

VC 6.50 mg/l- 53%

20°C 5.60 ini,11 60%

30'C 4.80 awl, 63%

Temperature and Dissolved Oxygen 1. Plot the Class Averages for DO from Table 1. Give the graph and axes

appropriate titles and labels.

i.

mumun 11113

IM 1 ,

.0 00.5

ID

3

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Animal Behavior Activity A2: (.)ricnttititsi Behavior of Ivopods

Sample Data Tel le 1

Time 140*V.K.}

---

t.k 00

ui 40

h 00

t: 8)

2e 48!

2: Ii.)

1: 99

4: NJ

4 00

4: 34.,

8: 00

5: ik,

to; OP

0:. )0

;: 00

7: 59

tt 90

tic: 39

9: 99

9: )0

448

2))

At On

..........._.......

i 0 * 8 8 8 9 7 8 8 7 9 7 8 8 10 10 10 10 10 10

* Oa

Jr, 4,1v i v t 4 1 2 0 0 2) tsitred ox

Talc 1. What hypothesis was tested in this experiment?

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SO 2. Sci-1 , a4v42_ slat o C. ctiv.ut-eA e(te,te., 6 , eke_ ,

v b()) -x' 3. Sketch a graph the data from Table 1 for both the wet side and the dry side of t e chamber. Give the graph and axes

2. What variables should have been held constant during/his experiment?

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4-PA.A-peAceitve

appropriate titles and labels. Interpret the results of the experiment.

L--a)4'tsi

Tv, -p (°C) 2. Interpret the graph above - explain how dissolved oxygen varies according to temperature changes.

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&-v2. vvvsvz_ pO,C_A 3.Predict what would happen to the dissolved oxygen rer± -10°C a a 40°C.

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c. Plug in

Chi Square Strategy:

a. Given = observed

b. Calculate the expected, usually do a Punnett square

to figure this out how many phenotypes

x 2 =

sEe. x sset bitter/non-explosive - 88 bitter/explosive - 68

non-bitter/non-explosive - 62 non-bitter/explosive - 81

V = 5V--1 (DO

Additional Mathematical Calculations

A plant geneticist is investing the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Citrullus lanatus). Explosive rind is recessive and causes watermelons to burst when cut. Non-bitter watermelons are a result of the recessive genotype (susu). The geneticist wishes to determine if the genes assort independently. She performs a testcross between a bitter/non-explosive hybrid and a plant homozygous recessive for both traits. The following offspring are produced:

Calculate the chi-squared value for the null hypothesis that the two genes assort independently. Give your answer to the nearest hundredth.

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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 A eli,o(il Red Eyes 134 1 5°

50 White Eyes 66

72-2 (131-141-* (U6 "r

What is the SA/1.7 for this cell? Round your answer to the nearest hundredths.

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= t53 . 33

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a-1 ,k a15 = 300V, -----------

6) 6) 60

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4

• Solute potential= -iCRT • i = The number of particles the molecule will make in water; for NaCI this would

be 2; for sucrose or glucose, this number is 1 • C = Molar concentration (from your experimental data) • R = Pressure constant = 0.0831 liter bar/mole K • T = Temperature in degrees Kelvin = 273 + °C of solution

The molar concentration of a sugar solution in an open beaker has been determined to be 0.3M. Calculate the solute potential at 27 degrees Celsius. Round your answer to the nearest tenths.

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

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 02 produced (mL)

1 2.3

2 3.6

4.2

4 5.5

5 5.9

b? 5.(0 I.1 LI" --7 - 2 ._

Calculate the probability of tossing three coins simultaneously and obtaining three heads. Express in fraction form.

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N—total number in pop r—rate of growth Data taken to determine the effect of temperature on the rate of respiration in a goldfish is given in the table below. Calculate Q.1.0 for this data. Round to the nearest whole number.

Te;npeiat,,n ,,i, Respiration 01 Rate (Min)

16 pi,

2 1 22

Q,0 -; (adlico)10(a\-‘u)

— (I, 515)z-

There are 2000 mice living in a field. If 1000 mice are born each month and 200 mice die each month, what is the per capita growth rate of mice over a month? Round to the nearest tenths.

N. d.000

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(.) (1)

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(1)

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3

0

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The net annual primary productivity of a particular wetland ecosystem is found to be 8000 kcal/m2. If respiration by the aquatic producers is 12,000 kcal/m2 per year, what is the gross annual primary productivity for this ecosystem in kcal/m2 per year? Round to the nearest whole number.

\ 0

Grass 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 2.5

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Joe has a 2 g/L solution. He dilutes it and creates 3 L of a 1 g/L solution. How much of the original solution did he dilute? Round to the nearest tenths.

Vi = C2V2

We ace lo6kunl v%

What is the hydrogen ion concentration of a solution of pH 8? Round to the nearest whole number

pH = - log [1-1+1

OA= Y.0

• D-\k]

• C_\'-'1\i= \O

Four blocks of phenolphthalein agar are placed in a vinegar solution. The pH indicator solution changes to pink when in contact with an acidic solution.

Block A: 2 cm x 4 cm x 4 cm Block B: 2 cm x 8 cm x 4 cm Block C: 1 cm x 8 cm x 8 cm Block D: 1 cm x 1 cm x 64 cm

a. Which block would the vinegar solution penetrateiinoi,st thoroughly into after ten minutes? D cyeck r 5(4 14 cAs4 e dk.C7-1As■ ,,,o

b'. Which block would have the greatest volume of pink phenolphthalein at the( end of ten

5,-,,ckweo-- volume wv\oce cYS.(--usel %(\§,cle

Justify your answer mathematically and relate your predicted results to the surface area of your blocks.

minutes?

W total 11)

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.

a. Fill out the following table:

Population size Population sr6 wth rate

1,600 I - 9 1,750 - ,,,a1c1) as 2,000 _.-ACi U0 __—

A hypothetical population has a carrying capacity of 1,500 individuals and I-. is 1.0.

b. What is happening to this population? Why?

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