lab 7 review mitosis instructor’s...

Lab 7 Review

Mitosis

Instructor’s Material

Appendix A A1

Appendix A

AP BIOLOGY EQUATIONS AND FORMULAS

StAtiSticAl AnAlySiS And ProbAbility s = sample standard deviation (i.e., the sample based estimate of the standard deviation of the population)

x = meann = size of the sampleo = observed individuals with observed genotypee = expected individuals with observed genotype

Degrees of freedom equals the number of distinct possible outcomes minus one.

Standard Error Mean

Standard Deviation Chi-Square

chi-SquAre tAble

Degrees of Freedomp 1 2 3 4 5 6 7 80.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.510.01 6.64 9.32 11.34 13.28 15.09 16.81 18.48 20.09

lAwS of ProbAbilityIf A and B are mutually exclusive, then P (A or B) = P(A) + P(B)If A and B are independent, then P (A and B) = P(A) x P(B)

hArdy-weinberg equAtionSp2 + 2pq + q2 = 1 p = frequency of the dominant

allele in a populationp + q = 1 q = frequency of the recessive

allele in a population

Metric PrefixeS

Factor Prefix Symbol109 giga G106 mega M103 kilo k10-2 centi c10-3 milli m10-6 micro μ10-9 nano n10-12 pico p

Mode = value that occurs most frequently in a data setMedian = middle value that separates the greater and lesser halves of a data setMean = sum of all data points divided by number of data pointsRange = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum)

A2 Appendix A

rATe And growTh water Potential (Ψ)Ψ = Ψp + ΨsΨp = pressure potentialΨs = solute potentialThe water potential will be equal to the solute potential of a solution in an open container, since the pressure potential of the solution in an open container is zero.The Solute Potential of the SolutionΨs = – iCRTi = ionization constant (For sucrose

this is 1.0 because sucrose does not ionize in water.)

C = molar concentrationR = pressure constant (R = 0.0831 liter

bars/mole K)T = temperature in Kelvin (273 + ºC)

ratedY/dtPopulation growthdN/dt=B-Dexponential growth

logistic growth

dY= amount of changet = timeB = birth rateD = death rateN = population sizeK = carrying capacityrmax = maximum per capita growth rate

of population

Temperature Coefficient q10

Primary Productivity Calculationmg O2/L x 0.698 = mL O2 /LmL O2/L x 0.536 = mg carbon fixed/L

t2 = higher temperaturet1 = lower temperaturek2 = metabolic rate at t2

k1 = metabolic rate at t1

Q10 = the factor by which the reaction rate increases when the temperature is raised by ten degrees

SurFACe AreA And VoluMe dilution – used to create a dilute solution from a concentrated stock solutionCiVi = CfVf

i = initial (starting)C = concentration of solutef = final (desired)V = volume of solution

Volume of a SphereV = 4/3 π r3

Volume of a Cube (or Square Column)V = l w hVolume of a ColumnV = π r2 hSurface Area of a SphereA = 4 π r2

Surface Area of a CubeA = 6 aSurface Area of a rectangular SolidA = Σ (surface area of each side)

r = radiusl = lengthh = heightw = widthA = surface areaV = volumeΣ = Sum of alla = surface area of one side of the cube gibbs Free energy

ΔG = ΔH – TΔSΔG = change in Gibbs free energyΔS = change in entropyΔH = change in enthalpyT= absolute temperature (in Kelvin)ph = – log [H+]

Lab 7 Mitosis

1. Above is photograph of an onion root tip squash observed under a microscope. In the data table below, count the number of cells in each phase of mitosis. Combine prophase and prometaphase.

Phase Number of Cells

Percent spent in each phase Minutes in Each Phase

Interphase Prophase/prometaphase Metaphase Anaphase Telophase Total Number of Cells Number of Cells in Mitosis

2. Determine the percentage of cells spent in interphase versus mitosis and record in the table shown.

3. If the life cycle of the onion root tip cell is 24 hours, how many minutes will the average cell spend

in each phase of mitosis? Record the data in the table shown.

4. This is a root tip squash (shown above) that has been treated with a 1 M concentration of caffeine. Determine the percentage of cell that is actively involved in a mitotic division. Record this information in the data table below.

5. Does it appear that caffeine has an effect on the number of cells involved in mitosis? Justify your answer. ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Root tip (Data from Part I) Root tip treated with caffeine Phase Count Phase Count Interphase Interphase Mitosis Mitosis Total Cell Count Total Cell Count % Cells in Mitosis % Cell in Mitosis

6. What is the null hypothesis for this investigation? ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

7. Do a Chi-square analysis to determine if soaking onion bulbs in a 1 M concentration of caffeine solution significantly affects mitosis of the root tips. Use the null hypothesis to aid in determining what is expected in the caffeine trial.

Mitosis Interphase Observed Caffeine Trial Expected Caffeine Trial Difference

Chi square calculation

8. Is this value significant?

Part II.

The roots of several onion bulbs were removed. Then the remaining severed root mass of onions bulbs were submerged in a beaker of distilled water, and others were submerged in a beaker of water seeped with bark from a weeping willow tree. It has been reported that a substance found in the bark of the weeping willow tree bark is a root stimulator. The roots were allowed to regenerate for five days. After five days the roots were measured and recorded in the data table below. To determine if the there is a substance in the weeping willow bark that is a root stimulator, statistical analysis must be done. Determine the mean, standard deviation and SEM for each set of data.

Untreated Roots Treated Roots with Willow Bark

Root # Length (mm)

Di fference from the

mean

(Di fference from the mean)2 Root #

Length (mm)


mean

(Di fference from the mean)2

1 5.7 1 9.1 2 3 2 6.1 3 2.2 3 5.5 4 2.2 4 7.1 5 4.9 5 5.9

Mean

Sum Mean

Sum

Standard deviation

Standard deviation

SEM

SEM

2 SEM

2 SEM

1. On the axis provided, create an appropriately labeled graph to illustrated the means for each group to within 95% confidence (i.e. sample means + 2 SEM). Remember that the means of this data was derived on the onion had been treated with the “Weeping Willow Tree”. This is discreet data, and it is better to make a bar graph.

2. Does the data support that a 1M caffeine solution can have a negative effect on the mitotic division root tip cells?The error bars do not overlap. Qualitatively, it support that a 1M caffeine solution can have a negative effect on the mitotic division root tip cells and is significantly different from the untreated root tips. This data does not support the null hypothesis that the caffeine has no effect on the mitotic division of onion root tips.

Lab 7 Mitosis

1. Above is an onion root tip observed under a microscope. In the data table shown below, count thenumber of cells in each phase of mitosis. Combine prophase and prometaphase.

Phase Number of Cells

Percent spent in each phase Minutes in Each Phase

Interphase 53 75.71% 1090 Prophase/prometaphase 8 11.43% 165 Metaphase 3 4.29% 62 Anaphase 2 2.86% 41 Telophase 4 5.71% 82 Total Number of Cells 70 Number of Cells in Mitosis

17 24.29%

2. Determine the percentage of cells spent in interphase versus mitosis and record in the table shown.

3. If the life cycle of the onion root tip cell is 24 hours, how many minutes will the average cell spendin each phase of mitosis? Record the data in the table shown.

This is a root tip that has been treated with a 1 M concentration of caffeine. Determine the percentage of cell involved in mitosis.

4. This is a root tip squash (shown above) that has been treated with a 1 M concentration of caffeine. Determine the percentage of cell that is actively involved in a mitotic division. Record this information in the data table below.

5. Does it appear that caffeine has an effect on the number of cells involved in mitosis? Justify youranswer.

Yes, the percentage of cells involved in mitosis decreases almost by half. There may be some minor variation due to interpretation on whether a cell was in the process of mitosis or not.

Root tip (Data from Part I) Root tip treated with caffeine Phase Count Phase Count Interphase 53 Interphase 72 Mitosis 17 Mitosis 11 Total Cell Count 70 Total Cell Count 83 % Cells in Mitosis 24.29% % Cell in Mitosis 13.25%

6. What is the null hypothesis for this investigation?Caffeine has no effect upon the rate of mitosis.

7. Do a Chi-square analysis to determine if soaking onion bulbs in a 1 M concentration of caffeinesolution significantly affects mitosis of the root tips. Use the null hypothesis to aid in determining what is expected in the caffeine trial.

Chi square calculation

Percentage in mitosis untreated = 24%

Percentage in interphase untreated=76%

If caffeine had not effect on mitosis, then the number of cells engaged in mitosis should be similar to the untreated root tip squash.

Mitosis = .24 x 83 cells caffeine squash= 20 Interphase = .76 X 83 cells caffeine squash = 63

Mitosis Interphase Observed 11 72 Expected 20 63 Difference -9 9 Chi square calculation

2 ( ) 81 81 = ( ) +( )= 5.3420 63

o eXe−

= ∑

8. Is this value significant?

With one degree of freedom, a value of 5.34 is greater than 3. 84. So there is a high probability that the difference between the treated and nontreated root tips is significant.

Part II.

The roots of several onion bulbs were removed. Then the remaining severed root mass of onions bulbs were submerged in a beaker of distilled water, and others were submerged in a beaker of water seeped with bark from a weeping willow tree. It has been reported that a substance found in the bark of the weeping willow tree bark is a root stimulator. The roots were allowed to regenerate for five days. After five days the roots were measured and recorded in the data table below. To determine if the there is a substance in the weeping willow bark that is a root stimulator, statistical analysis must be done. Determine the mean, standard deviation and SEM for each set of data.

Untreated Roots Treated Roots

Root # Length (mm)


mean

(Di fference from the mean)2 Root #

Length (mm)


mean

(Di fference from the mean)2

1 5.7 2.1 4.41 1 9.1 2.36 5.57 2 3 -0.6 0.36 2 6.1 -0.64 0.41 3 2.2 -1.4 1.96 3 5.5 -1.24 1.54 4 2.2 -1.4 1.96 4 7.1 0.36 0.13 5 4.9 1.3 1.69 5 5.9 -0.84 0.71

Mean 3.6 Sum 10.38 Mean 6.74 Sum 8.35

Standard deviation 1.61

Standard deviation 1.29

SEM 0.72 SEM 0.58

2 SEM 1.44

2 SEM 1.16

1. On the axis provided, create an appropriately labeled graph to illustrated the means for each group to within 95% confidence (i.e. sample means + 2 SEM). Remember that the means of this data was derived on the onion had been treated with the “Weeping Willow Tree”. This is discreet data, and itis better to make a bar graph.

2. Does the data support that a 1M caffeine solution can have a negative effect on the mitotic division root tip cells?

0

1

2

3

4

5

6

7

8

9

Untreated Onion Treated Onion

Mea

n Le

ngth

of R

oot

Tips

(mm

)

Effect of 1M Caffeine on Mitotic Division of Onion Root Tips

Series1

The error bars do not overlap. Qualitatively, it support that a 1M caffeine solution can have a negative effect on the mitotic division root tip cells and is significantly different from the untreated root tips. This data does not support the null hypothesis that the caffeine has no effect on the mitotic division of onion root tips.

AP® BIOLOGY 2011 SCORING GUIDELINES (Form B)

© 2011 The College Board. Visit the College Board on the Web: www.collegeboard.org.

Question 1

The cell cycle is fundamental to the reproduction of eukaryotic cells.

(a) Describe the phases of the cell cycle. (6 points maximum)

Correct order of cycle phases (1 point for entire correct list) Interphase → Prophase → (Prometaphase) → Metaphase → Anaphase → Telophase → Cytokinesis

OR G1 → S → G2 → M

Correct description of at least one important structural or molecular characteristic of each phase (1 point each; 5 points maximum) • Interphase (including, if specified, G1, S, G2 subphases, correctly ordered): Chromatin dispersed in

nucleus; nuclear envelope and nucleoli are intact and functional; DNA is replicated here. • G1, G2: Cell growth.• S: DNA replication.• Mitosis: Nuclear division.• Prophase: Chromosomes begin to condense from chromatin; spindle apparatus assembled.• (Prometaphase): Nuclear envelope disperses, nucleoli disperse, chromosomes connect to spindle

apparatus fibers and begin to show motility.• Metaphase: Chromosomes reach maximum condensation and align on metaphase plate/plane.• Anaphase: Two-chromatid chromosomes split into two daughter (one-chromatid) chromosomes;

chromosomes move to opposite poles of the spindle apparatus.• Telophase: Chromosomes disperse back to chromatin form, nuclear envelope reassembles, nucleoli

reassemble.• Cytokinesis: If this occurs, it is normally coordinated with telophase; cell division.

(b) Explain the role of THREE of the following in mitosis or cytokinesis. (3 points maximum)

• Kinetochores• Microtubules• Motor proteins• Actin filaments

Correct explanation of function (1 point each; if all four are chosen, only the first three are scored) • Kinetochores: Located in centromeres of condensed chromosomes; microtubule attachment sites

necessary for chromosome positioning and movement. • Microtubules: Fundamental structural element of the spindle apparatus; framework on which

chromosome motility is generated; define axis of division and cytokinesis. • Motor proteins (correct location and function must be specified): In kinetochores, move

chromosomes during mitosis, including anaphase separation; involves kinesins and dyneins. OR

In animal cell cleavage furrow, generate force to pinch cell in two; involves myosins. • Actin filaments: Assemble under the membrane at the cytokinesis site; interact with myosin motor

proteins to generate force to pinch cell in two; also interact with astral microtubules of the spindle to position the spindle apparatus in the cell.



Question 1 The cell cycle is fundamental to the reproduction of eukaryotic cells. (a) Describe the phases of the cell cycle.

(6 points maximum)

Correct order of cycle phases (1 point for entire correct list) Interphase → Prophase → (Prometaphase) → Metaphase → Anaphase → Telophase → Cytokinesis

OR G1 → S → G2 → M Correct description of at least one important structural or molecular characteristic of each phase (1 point each; 5 points maximum) • Interphase (including, if specified, G1, S, G2 subphases, correctly ordered): Chromatin dispersed in

nucleus; nuclear envelope and nucleoli are intact and functional; DNA is replicated here. • G1, G2: Cell growth. • S: DNA replication. • Mitosis: Nuclear division. • Prophase: Chromosomes begin to condense from chromatin; spindle apparatus assembled. • (Prometaphase): Nuclear envelope disperses, nucleoli disperse, chromosomes connect to spindle

apparatus fibers and begin to show motility. • Metaphase: Chromosomes reach maximum condensation and align on metaphase plate/plane. • Anaphase: Two-chromatid chromosomes split into two daughter (one-chromatid) chromosomes;

chromosomes move to opposite poles of the spindle apparatus. • Telophase: Chromosomes disperse back to chromatin form, nuclear envelope reassembles, nucleoli

reassemble. • Cytokinesis: If this occurs, it is normally coordinated with telophase; cell division.

(b) Explain the role of THREE of the following in mitosis or cytokinesis. (3 points maximum)

• Kinetochores • Microtubules • Motor proteins • Actin filaments

Correct explanation of function (1 point each; if all four are chosen, only the first three are scored) • Kinetochores: Located in centromeres of condensed chromosomes; microtubule attachment sites

necessary for chromosome positioning and movement. • Microtubules: Fundamental structural element of the spindle apparatus; framework on which

chromosome motility is generated; define axis of division and cytokinesis. • Motor proteins (correct location and function must be specified): In kinetochores, move

chromosomes during mitosis, including anaphase separation; involves kinesins and dyneins. OR

In animal cell cleavage furrow, generate force to pinch cell in two; involves myosins. • Actin filaments: Assemble under the membrane at the cytokinesis site; interact with myosin motor

proteins to generate force to pinch cell in two; also interact with astral microtubules of the spindle to position the spindle apparatus in the cell.



Question 1 (continued)

(c) Describe how the cell cycle is regulated and discuss ONE consequence of abnormal regulation. (3 points maximum) Regulation: Correct description of checkpoints, which block cell cycle progress unless specific molecular and/or physical conditions are satisfied (1 point each; 2 points maximum) • Action of MPF and CDKs in checkpoint regulation • Contact inhibition of mitosis • Hormones; growth factor control of cell cycle activity Correct discussion of the consequences of abnormal cell cycle regulation (1 point maximum) • Uncontrolled cell proliferation, as in cancer • Apoptosis • Non-disjunction/aneuploidy/broken chromosomes from abnormal spindle events

AP® BIOLOGY 2011 SCORING COMMENTARY (Form B)


Question 1 Sample: 1A Score: 9 The response earned the maximum of 6 points in part (a). One point was earned for correctly identifying the cell cycle stages in correct order and combining the synthetic events with the mitotic events. The response earned another point for correctly identifying S phase as the phase in which DNA is replicated. The remaining four points were earned for correctly describing the events of the remaining stages of mitosis. The student could have earned more points in this section, but the 6-point maximum had already been reached. In part (b) the response earned 1 point for correctly identifying microtubules as spindle fiber components. No point was earned for the description of actin filaments because they are incorrectly identified as spindle apparatus components. No point was earned for the description of motor proteins because it is too vague. In part (c) the response earned 2 points. One point was earned for the correct presentation of the concept of cell cycle checkpoints and for providing an example of a substrate-specific checkpoint. Another point was earned for discussing how cancer cells escape this checkpoint regulation because they lose anchorage dependency. Sample: 1B Score: 6 In part (a) the response earned 1 point for correctly identifying the cell cycle stages in correct order, although most of the answer specifies only the mitotic cycle. Another point was earned for correctly specifying interphase as the phase in which DNA is replicated. The response earned 3 points for describing the events of the remaining phases of mitosis. (The response does not adequately distinguish telophase and cytokinesis, so only 1 point was awarded for that portion of the response.) No points were earned in part (b). No adequate or correct descriptions are given in this section. In part (c) 1 point was earned for linking contact inhibition and the escape of cancer cells from this inhibition. No points were earned for the incomplete description of checkpoints as regulatory control periods of the cycle, along with a vague description of contact inhibition. Sample: 1C Score: 3 In part (a) the response earned 1 point for correctly listing the phases of the mitotic cycle in correct order. One point was earned for describing metaphase as the stage at which chromosomes “line up at the 50- yard line.” The response earned 1 more point for the description of chromatid separation at anaphase. No other statements are correct or sufficiently precise to earn points. No points were earned in part (b) because the statement that motor proteins “aid in energy” is too vague. No points were earned in part (c) because no correct or precise statements are given in this section, nor is an example of a substrate-specific checkpoint presented.



Question 1 (continued)

(c) Describe how the cell cycle is regulated and discuss ONE consequence of abnormal regulation. (3 points maximum) Regulation: Correct description of checkpoints, which block cell cycle progress unless specific molecular and/or physical conditions are satisfied (1 point each; 2 points maximum) • Action of MPF and CDKs in checkpoint regulation • Contact inhibition of mitosis • Hormones; growth factor control of cell cycle activity Correct discussion of the consequences of abnormal cell cycle regulation (1 point maximum) • Uncontrolled cell proliferation, as in cancer • Apoptosis • Non-disjunction/aneuploidy/broken chromosomes from abnormal spindle events

AP® BIOLOGY 2011 SCORING COMMENTARY (Form B)


Question 1 Sample: 1A Score: 9 The response earned the maximum of 6 points in part (a). One point was earned for correctly identifying the cell cycle stages in correct order and combining the synthetic events with the mitotic events. The response earned another point for correctly identifying S phase as the phase in which DNA is replicated. The remaining four points were earned for correctly describing the events of the remaining stages of mitosis. The student could have earned more points in this section, but the 6-point maximum had already been reached. In part (b) the response earned 1 point for correctly identifying microtubules as spindle fiber components. No point was earned for the description of actin filaments because they are incorrectly identified as spindle apparatus components. No point was earned for the description of motor proteins because it is too vague. In part (c) the response earned 2 points. One point was earned for the correct presentation of the concept of cell cycle checkpoints and for providing an example of a substrate-specific checkpoint. Another point was earned for discussing how cancer cells escape this checkpoint regulation because they lose anchorage dependency. Sample: 1B Score: 6 In part (a) the response earned 1 point for correctly identifying the cell cycle stages in correct order, although most of the answer specifies only the mitotic cycle. Another point was earned for correctly specifying interphase as the phase in which DNA is replicated. The response earned 3 points for describing the events of the remaining phases of mitosis. (The response does not adequately distinguish telophase and cytokinesis, so only 1 point was awarded for that portion of the response.) No points were earned in part (b). No adequate or correct descriptions are given in this section. In part (c) 1 point was earned for linking contact inhibition and the escape of cancer cells from this inhibition. No points were earned for the incomplete description of checkpoints as regulatory control periods of the cycle, along with a vague description of contact inhibition. Sample: 1C Score: 3 In part (a) the response earned 1 point for correctly listing the phases of the mitotic cycle in correct order. One point was earned for describing metaphase as the stage at which chromosomes “line up at the 50- yard line.” The response earned 1 more point for the description of chromatid separation at anaphase. No other statements are correct or sufficiently precise to earn points. No points were earned in part (b) because the statement that motor proteins “aid in energy” is too vague. No points were earned in part (c) because no correct or precise statements are given in this section, nor is an example of a substrate-specific checkpoint presented.

Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved.AP is a registered trademark of the College Entrance Examination Board.

-3-

20003. Information transfer is fundamental to all living organisms. For two of the following examples, explain in detail

how the transfer of information is accomplished.

a) The genetic material in one eukaryotic cell is copied and distributed to two identical daughter cells

b) A gene in a eukaryotic cell is transcribed and translated to produce a protein

c) The genetic material from one bacterial cell enters another via transformation, transduction, or conjugation

4. To survive, organisms must be capable of avoiding, and/or defending against, various types of environmentalthreats. Respond to each of the following.

a) Describe how adaptive coloration, mimicry, or behavior function as animal defenses against predation.Include two examples in your answer.

b) Describe how bacteria or plants protect themselves against environmental threats. Include two examples inyour answer.

c) Compare the human primary immune response with the secondary immune response to the same antigen.

END OF EXAMINATION

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AP® Biology 2000 ─ Scoring Standards


Question 3 Scoring Guide

a) The genetic material in one eukaryotic cell is copied and distributed to two identical daughtercells. The maximum for part a is 6 points

Part a is asking for “copy” and “distribute”, there is an internal maximum of four points for each. a part 1 DNA Replication (max 4 points) 1 – when DNA is copied – interphase, S phase of cell cycle 1 - recognition of origin site on DNA 1 - concept of unwinding enzyme 1 - RNA primer 1 - DNA polymerase – functional definition 1 - Concept of complementary relationship among bases – semiconservative 1 - Discontinuous/continuous or lagging /leading or Okazaki fragments (due to antiparallel backbones and 5’ to 3’ generation of new segments) 1 - DNA ligase – functional definition 1 - Other/Elaboration – telomere replication, proofing by DNA polymerase, expanded details a part 2 Mitosis (max 4 points) 1 - concept of chromatid pairs or ‘doubled chromosomes’ 1 - prophase – condensation, spindle formation 1 - metaphase – alignment of chromosomes 1 - anaphase – separation of chromatids or equivalent statement 1 - telophase or origin of cytokinesis – nuclear membrane reforms, cell plate or cell furrow 1 - Other or Elaboration – cell cycle control, cell surface area/volume ratio and mitosis, MTOC(microtubule organizing center), centromere or kinetochore attachment b) The gene in a eukaryotic cell is transcribed and translated to produce a protein. There is a

maximum of 6 points for this part.

This part asks for transcription and translation, there is an internal maximum of four points each.

b part 1 Transcription (max 4 points)

1 - Functional definition: DNA sequence to RNA sequence1 - Promoter Recognition1 - RNA polymerase – function1 - Complementarity relationships (T to U)1 - 5’ to 3’ – growth of new strand1 - Start site/ Termination Sequences1 - Introns/Exons – with general explanation1 - Caps/ Tails – with general explanation1 - Other or Elaboration: sense and antisense , transcription factors, spliceosomes, multiple RNA’s, enhancers, conserved segments

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b part 2 Translation (max 4 points)

1 - Functional Definition: base sequences to aa sequences

Initiation

1 - Sequence of events – complex( m-RNA, small unit of ribosome, first t-RNA)1 - Structure of ribosomes – complete description – two subunits , 2 or more action sites, r-RNA and proteins

Elongation

1 - t-RNA structure – amino-acyl site and anticodon1 - Complementarity – codons to anticodons, m-RNA base sequence to t-RNA base sequence

Translocation –with basic description1 - Peptide Formation – amino acids joined by peptide bonds to form polypeptide.

Termination

1 - Stop codon + release polypeptide + release ribosomes (must have 2 of 3)

1 - Other or Elaboration – triplet code, recognition segments, wobble (redundancy)

One point can be granted to either section of part b for describing the movement of RNA from thenucleus to the cytoplasm.

c) The genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation. There is a 6 point maximum on this part. Choose one only!

Transformation (max 6 points)

1 - Functional definition1 - Competency – cell membrane permeable to fragments1 - How to make competent – calcium chloride, heat shock, cold for stability, gene gun, electroporation1- Parameters for individual DNA segments – size, double helix1 - Description of Griffiths/Avery – information transfer emphasis2 - Other or Elaboration – recognition of transfer – plasmid description, gene technology, antibiotic resistance

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Transduction (max 6 points)

1 - Functional definition – viral vector1 - Lytic Cycle – describe1 - Virus transfers bacterial DNA1 - Lysogenic Cycle – describe1 - Introduce viral DNA into bacteria1 - Excision2 - Other or Elaboration – prophage, oncogene, gene technology

Conjugation (max 6 points)

1 Functional definition (contact and one-way)1 - Pili – describe1 - F+ factor – donor(+), recipient(-)1 - Hfr Cells1 - Replication of transferred segment2 - Other or Elaboration – antibiotic resistance, plasmid, gene technology

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lab 7 review mitosis instructor’s...

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