AP Lecture Guide 15 – The Chromosomal Basis of Inheritance

AP Biology: CHAPTER 15

THE CHROMOSOMAL BASIS OF INHERITANCE

 

1. Describe some of the pieces of information that scientists discovered that contributed to the

“Chromosome Theory of Inheritance”?

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2. Summarize the Chromosomal Theory of Inheritance.

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3. Why was Thomas Hunt Morgan’s choice of the fruit fly a good model organism?

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4. Describe Morgan’s first mutant. Why was it so significant from the wild type?

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5. Show the cross P, F1, F2 for the white-eyed male mutant.

6. What happens when we trace the inheritance of traits found on the same chromosome?

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7. Use the diagram to trace the body color and wing shape in this linked two trait cross.

 

8. What is recombination and when does it occur? ___________________________________

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9. How is recombination frequency calculated? _____________________________________

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10. What determines sex in humans? ______________________________________________

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11. In what ways are sex-linked traits distinct from autosomal traits? _____________________

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12. Why are sex-linked recessive traits more common in human males than females?

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13. How many X chromosomes are typically expressed in humans and cats?

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14. What happens to X chromosomes that are inactivated? _____________________________

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15. How many Barr bodies would be found in a person with: XXY_____ XO_____ XXX_____.

16. Define each term & and indicate when each occurs.

a. aneuploidy _____________________________________________________________

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b. polyploidy ______________________________________________________________

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17. Identify the each of the alterations of chromosome structure.

18. List and describe a few specific examples of non-disjunctions that occur in humans.

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19. Describe genomic imprinting and give an example.

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AP Lab 5 Sample 7

 

Cellular Respiration

Blake Lockwood

Introduction:

 

The human body has to have energy in order to perform the functions that allow life. This energy comes from the process of cellular respiration. Cellular respiration releases energy that the body can use in the form of ATP from carbohydrates by using oxygen. Cellular respiration is not just one singular reaction, it is a metabolic pathway made up of several reactions that are enzyme mediated. This process begins with glycolysis in the cytosol of the cell. In glycolysis, glucose is split into two three-carbon compounds called pyruvate, producing a small amount of ATP The final two steps of cellular respiration occur in the mitochondria. These final two steps are the electron transport system and the Krebs Cycle. The overall equation for cellular respiration is

C6H12O6 + 6O2 -> 6CO2 + 6H2O + 686 kilocalories of energy per mole of glucose oxidized.

There are three ways to measure the rate of cellular respiration. These three ways are by measuring the consumption of oxygen gas, by measuring the production of carbon dioxide, or by measuring the release of energy during cellular respiration. In order to measure the gases, the general gas law must be understood. The general gas law state: PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of molecules of gas, R is the gas constant, and T is the temperature of the gas (in K). The gas law also shows concepts about gases. If temperature and pressure are kept constant, then the volume of the gas is directly proportional to the number of molecules of the gas. If the temperature and volume remain constant, then the pressure of the gas changes in direct proportion to the number of molecules of gas present. If the number of gas molecules and the temperature remain constant, then the pressure is inversely proportional to the volume. If the temperature changes and the number of gas molecules is kept constant, then either pressure of volume will change in direct proportion to the temperature.

In this experiment, the rate of cellular respiration will be measured by measuring the oxygen gas consumption by using a respirometer in water. This experiment measures the consumption of oxygen by germinating and non-germinating at room temperature and at ice water temperature. The carbon dioxide produced in cellular respiration will be removed by potassium hydroxide (KOH). As a result of the carbon dioxide being removed, the change in the volume of gas in the respirometer will be directly related to the amount of oxygen consumed. The respirometer with glass beads alone will show any changes in volume due to atmospheric pressure changes or temperature changes.

 

Hypothesis:

 

The germinating peas will have a higher rate of respiration, than the beads and non-germinating peas.

 

Materials:

 

This lab requires two thermometers, two water baths, beads, germinating and non-germinating peas, beads, six vials, twelve pipettes, 100 mL graduated cylinder, scotch tape, tap water, ice, KOH, absorbent and non-absorbent cotton, six washers, six rubber stoppers, scotch tape, and a one mL dropper.

 

Methods:

 

Start the experiment by setting up two water baths, one at room temperature and the other at 10 degrees Celsius. Then, find the volume of twenty-five germinating peas. Next, put 50 mL of water in a graduated cylinder and put twenty-five non-germinating peas in it. Then, add beads until the volume is the same as twenty-five germinating peas. Next, pour our the peas and beads, refill the graduated cylinder with 50 mL of water, and add only beads until the volume is the same as the twenty-five germinating peas. Repeat these steps for another set of peas and beads. Also, put together the six respirometers by gluing a pipette to a stopper and taping another pipette to the pipette for all six respirometers. Then, put two absorbent cotton balls, several drops of KOH, and half of a piece of non-absorbent cotton into all six vials. Next, add the peas and beads to the appropriate respirometers. Place one set of respirometers into the room temperature water bath and the other set in the ice water bath. Elevate the respirometers by setting the pipettes onto masking tape and allow them to equilibrate for five minutes. Next, lower the respirometers into the water baths and take reading at 0, 5, 10, 15, and 20 minutes. Record the results in the table.

 

Results:

 

Table:

 

Beads Alone Germinating Peas

Dry Peas and Beads

Reading at time X Diff. Reading at time X Diff. Corrected Diff.

 

Reading at time X Diff. Corrected Diff.
Initial 13.2 12.7 12.9
0 to 5 11.0 2.2 10.5 2.2 0.0 11.1 1.8 -0.4
5 to 10 10 3.2 9.0 3.7 0.5 10.0 2.8 -0.3
10 to 15 9.2 4.0 8.0 4.7 0.7 9.4 3.5 -0.5
15 to 20 9.1 4.0 7.5 5.1 1.2 9.3 3.6 -0.4
Initial 14.0 13.5 14.0
0 to 5 13.3 0.7 12.1 1.4 0.7 13.6 0.4 -0.3
5 to 10 12.9 1.1 11.0 2.5 1.4 13.2 0.8 -0.3
10 to 15 12.6 1.4 10.0 3.5 2.1 12.9 1.1 -0.3
15 to 20 12.2 01.8 9.0 4.5 2.7 12.5 1.5 -0.3

 

 

Questions

1. In this activity, you are investigating both the effect of germination versus non-germination and warm temperature versus cold temperature on respiration rate. Identify the hypothesis being tested in this activity. Increasing the temperature could increase the oxygen consumption. Germinating peas have a higher respiration rate than non-germinating.

 

2. This activity uses a number of controls. Identify at least three of the controls, and describe the purpose of each control. One control was the respirometer with only beads in it because it didn’t use respiration. Another control was that the water temperatures were constant. The final control was that there was the same amount of KOH in each vial.

 

3. Graph the results from the corrected difference column for the germinating peas and dry peas at both room temperature and at 10 degrees Celsius. On graph paper.

 

4. Describe and explain the relationship between the amount of oxygen gas consumed and time. The oxygen gas consumed increased fairly constantly in respect to time.

 

5. From the slope of the four lines on the graph, determine the rate of oxygen gas consumption of germinating and dry peas during the experiments at room temperature and at 10 degrees Celsius. Recall that rate = _y over _x.

 

Condition Show Calculations Here Rate in mL oxygen gas/minute
Germinating Peas/10°C (1.2-0.7)/5 0.1
Germinating Peas/ Room Temperature (2.7-2.1)/5 0.12
Dry Peas/10°C (-.4-0)/5 -0.08
Dry Peas/Room Temperature (-.3-(-.3))/5 0

 

6. Why is it necessary to correct the readings from the peas with the readings from the beads? The gas changes in the beads were only due to pressure and temperature, and not gas consumption, so the beads act as a control.

 

7. Explain the effect of germination (versus non-germination) on pea seed respiration. Germinating peas consumed more oxygen than non-germinating.

 

8. Below is a sample graph of possible data obtained for oxygen consumption by germinating peas up to about 8 degrees Celsius. Draw in predicted results through 45 degree Celsius. Explain your predictions.

 

 

 

 

 

 

 

 

 

 

 

 

The amount of oxygen consumed will steadily increase until the temperature reaches a point at which the enzymes become denatured.

 

9. What is the purpose of KOH in this experiment? The purpose of the KOH was to remove the effect of carbon dioxide from the readings.

 

10. Why did the vial have to be completely sealed around the stopper? The vial had to be completely sealed so that gases couldn’t escape and water couldn’t leak into the respirometer.

 

11. If you used the same experimental design to compare the rates of respiration of a 25 g. reptile and a 25 g. mammal, at 10 degrees Celsius, what results would you expect? Explain your reasoning. The reptile would use less oxygen because it is cold-blooded and wouldn’t be as active at a colder temperature as the mammal would.

 

12. If respiration in a small mammal were studied at both room temperature (21°C) and 10°C, what results would you predict? Explain your reasoning. The respiration of the small mammal would be higher at 10 degrees Celsius because it would need more energy to keep its normal body temperature.

 

13. Explain why water moved into the respirometers’ pipettes. Water moved into the respirometer’s pipettes because pressure decreased when the amount of oxygen was decreased.

 

14. Design an experiment to examine the rates of cellular respiration in peas that have been germinating for 0, 24, 48, and 72 hours. What results would you expect? Why?

 

Set up five respirometers containing beads, non-germinating peas, peas that have germinated for one day, peas that have germinated for two days, and peas that have germinated for three days. Measure the water readings in intervals of five minutes for twenty minutes. The peas that have been germinating for three days will have the highest rate of respiration and the beads will have the lowest rate of respiration.

 

15. According to your graph, what happens to the rate of oxygen consumed by germinating peas over time? What does this indicate to you? The rate of oxygen consumption is fairly constant.

 

16. How did the KOH affect the water movement in the respirometer? It allows more water into the pipette.

 

17. Which of the two pea types, germinating or non-germinating, consumes the most oxygen? Why? Germinating peas consume more oxygen because they are growing and are more active than non-germinating peas.

 

18. What was the effect of temperature on pea respiration? Warmer temperatures allow for the peas to respire at a faster rate.

 

19. During aerobic respiration, glucose is broken down to form several end products. Which end products contain the carbon atoms from glucose? The hydrogen atoms from glucose? The oxygen atoms from glucose? The energy stored in the glucose molecules? Carbon dioxide contains the carbon, water contains the hydrogen, both carbon dioxide and water contain the oxygen, and ATP contains the energy.

 

20. What is fermentation? What are the two types of fermentation? What organisms use fermentation? Fermentation is a catabolic process that makes a limited amount of ATP from glucose without an electron transport chain and that produces an end-product such as ethyl alcohol or lactic acid. The two types of fermentation are alcoholic and lactic acid fermentation. Plants use alcoholic while animals use lactic acid.

 

21. Draw a Venn diagram showing how respiration and fermentation are similar and how they differ.

 

 

 

 

 

 

 

 

 

 

22. What are the three pathways involved in the complete breakdown of glucose to carbon dioxide and water? What reaction is needed to join two of these pathways? What are the substrates and products of this reaction and where does it take place? The three pathways are glycolysis, the electron transport chain, and the Krebs Cycle. The reaction of the pyruvate joining with CoA enzyme and NAD to produce acetyl CoA, NADH, and carbon dioxide. The acetyl CoA goes to the Krebs Cycle and NADH to the electron transport chain in the mitochondria.

23. Write the letter of the pathway that best fits each of the following processes.

Pathway

a. Glycolysis

b. Krebs Cycle

c. Electron Transport System

Process

1. Carbon dioxide is given off b.

2. Water is formed c.

3. PGAL a.

4. NADH becomes NAD+ c.

5. Oxidative phosphorylation c.

6. Cytochrome carriers c.

7. Pyruvate a.

8. FAD becomes FADH2 b.

24. Calculate the energy yield of glycolysis and cellular respiration per glucose molecule. Distinguish between substrate-level phosphorylation and oxidative phosphorylation. Where does the energy for oxidative phosphorylation come from? 36 ATPs are formed per glucose moelcule. Four of the ATPs are formed from substrate level and 32 from oxidative.

 

   

Substrate

Oxidative Total
Glycolysis 2 4 6
Transition 0 6 6
Krebs 2 22 24
Total 4 32 36

 

 

Error Analysis:

Some of the errors that could have occurred during the experiment were water leaking into the respirometers, gas escaping the respirometers, water temperature in the bath changing, and mathematical mistakes.

Discussion and Conclusion:

This lab showed that germinating peas consumed more oxygen at a faster rate than the non-germinating peas and the beads did. The non-germinating peas and the beads didn’t consume hardly any oxygen at all. It also showed that the respiration rate of germinating peas was faster than the respiration rate of non-germinating peas. Finally, this experiment showed that respiration rates increase as the temperature increases. This shows that temperature and respiration rates are directly proportional to each other.

AP Lecture Guide 16 – The Molecular Basis of Inheritance

AP Biology: CHAPTER 16

THE MOLECULAR BASIS OF INHERITANCE

1. After Morgan and fellow scientists developed the Chromosomal Theory of Inheritance, the

search was on for the chemical mechanism of inheritance. What are the two components of the chromosome?

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2. From initial logic, which component would be the most likely candidate for the genetic

material and why?

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3. What did Griffith, Avery, and others accomplish with bacteria?

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4. Define transformation. _______________________________________________________

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5. What did the experiments done by Alfred Hershey and Martha Chase show?

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6. What are Chargaff’s rules?

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7. If a species has 35% adenine in its DNA, determine the percent of the other three bases.

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8. What was the role of Maurice Wilkins and Rosalind Franklin in determining the structure of

DNA?

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9. Use the diagram to describe the structure of DNA. Include several comments.

 

10. What is the advantage of the double stranded aspect of the DNA? ____________________

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11. Which model of DNA replication is accepted? ____________________________________

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12. What happens at the DNA replication fork?

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13. Make a list of the enzymes involved in replication and their role.

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14. Why does the DNA have to add nucleotides in the 5’ to 3’ direction?

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15. Label the diagram of DNA replication. Include the directions and the terms.

16. Describe the “priming of the DNA” before replication. _______________________________

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17. List some of the steps involved in DNA repair. ____________________________________

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18. What is the problem that occurs at the ends of the chromosome during replication?

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19. What is a telomere and its role in cell division. ____________________________________

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20. Why was there no selection pressure for prokaryotes to evolve a telomere-like solution on

their chromosome? _________________________________________________________

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21. Why is telomerase an active area in cancer research? _____________________________

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AP Lab 3 Sample 4 Mitosis

 

 

Lab 3     Mitosis & Meiosis

 

 

Introduction

 

All cells come from other cells. New cells are formed during cell division which involves both the replication of the cell’s nucleus and division of the cytoplasm. The two kinds of cellular division are mitosis and meiosis. Mitosis usually makes body cells, somatic cells. Mitosis is used in adult cells for asexual reproduction, regeneration, and the maintenance and repair of body parts. The process called meiosis makes gametes, sperm and eggs, and spores in plants. Gamete or spore cells have half the chromosomes that the parent cell has.

Mitosis is the first of the cell divisions studied in this lab. It is easily observed in cells that are growing at a fast paces such as whitefish blastula or onion root tips, which are used in this lab. The onion root tips have the highest percentage of cells going through mitosis. The whitefish blastula is formed directly after the egg is fertilized. This is a period of a fast paced growth and numerous cellular divisions where mitosis can be observed. Just before mitosis the cell is in interphase, a part of the cell cycle where the cell has a distinct nucleus and nucleoli. Next is prophase, where the chromatin thickens into distinct chromosomes and the nuclear envelope breaks open releasing them into the cytoplasm. The first signs of the spindle apparatus begin to appear. Next the cell begins metaphase, where the spindle attaches to the centromere of each chromosome pair and moves them to the middle of the cell. This level position is called the metaphase plate. Then anaphase begins when the chromatids are separated and pulled to the opposite poles. The final stage is telophase where the nuclear envelope is reformed and the chromosomes gradually uncoil. Cytokinesis then may occur forming a cleavage furrow and then the two daughter cells will separate.

Meiosis is more complex and involves two nuclear divisions. The two divisions are called Meiosis I and Meiosis II. These two divisions result in the production of four haploid gametes. This process allows increased genetic variation due to crossing over where genes can be exchanged. The process, like mitosis, depends on interphase to replicate the DNA. Meiosis begins with prophase I. In this stage, homologous chromosomes move together to form a tetrad. This is where crossing over occurs resulting in the recombination of genes. Metaphase I moves the tetrads to the metaphase plate in the middle of the cell, and anaphase I reduces the tetrads to their original two stranded form and moves them to opposite poles. Telophase I then prepares the cell for its second division. Meiosis II is just like mitosis except that the daughter cells are haploid instead of diploid. DNA replication does not occur in interphase II, and prophase II, metaphase II, anaphase II, and telophase II occur as usual. The only change is the number of chromosomes.

 

Hypothesis

 

Mitosis is easily observed in the whitefish blastula and the onion root tip. Meiosis and crossing over occurs in the production of gametes, in animals, and spores, in plants.

 

Materials

 

Lab 3A. 1

The materials used in this lab are as follows: light microscopes, prepared slides of whitefish blastula and onion root tips, pencil, and paper.

Lab 3A. 2

The materials used in this lab are as follows: light microscopes, prepared slides of onion root tips, paper, and pencil.

Lab 3B. 1

The materials used in this section of the lab are as follows: a chromosome simulation kit, pencil and paper.

Lab 3B. 2

The materials used in this section of the lab are as follows: light microscopes, prepared slides of Sordaria fimicola, pencil, and paper.

 

Methods

 

Lab 3A. 1

Observe prepared slides of whitefish blastula and onion root tips under the 10X and 40X objectives. Sketch and identify each section of cell division.

Lab 3A. 2

Observe every cell and determine what stage the cell is in. Count at least 200 cells total, separating them into groups of the same phase. Consider it takes 24 hours for the onion root-tip cells to complete the cell cycle.

Lab 3B. 1

Use the lab book to show how to make the chromosomes. The simulation kit has plenty of beads to use. There are red and yellow beads to be used to show the different chromatids. There is also a piece that resembles half of a centromere which has a magnet to connect to another one.

Lab 3B. 2

Use a light microscope to observe the prepared slide and record all data.

Results

Lab 3A. 1

The sketches below show the phases of mitosis for the onion root-tip.

 

 

The sketches below show the phases of mitosis for the whitefish blastula.

 

Why is it more accurate to call mitosis “nuclear replication” rather than “cellular division”?

 

In mitosis, two new nuclei are formed and the cytokinesis is just part mitosis.

 

Explain why the whitefish blastula and onion root tip are selected for a study of mitosis.

 

The blastula is generally a ball of cells that are rapidly going under mitosis. The onion root-tip is an area where mitosis also occurs very rapidly.

Lab 3A. 2

The table below shows the data collected for this lab.

 

Number of Cells

Field 1 Field 2 Field 3 Total
Interphase 86 38 53 177 78.67 1132.8
Prophase 9 5 7 21 9.33 134.4
Metaphase 6 2 5 13 5.78 83.2
Anaphase 3 1 2 6 2.67 38.48
Telophase 3 1 3 7 3.11 44.8
225

 

 

If your observations had not been restricted to the area of the root tip that is actively dividing, how would your results have been different?

 

There would have been less cell division since most of the plant’s growth occurs at the root-tip.

 

Based on the data in the table above, what can your infer about the relative length of time an onion root-tip cell spends in each stage of cell division?

 

The most time of an onion root-tip cell’s life is in interphase. Prophase is the next most common phase that the cells spend in.

Lab 3B. 1

 

List three major differences between the events of mitosis and meiosis.

 

Mitosis has only one nuclear division while meiosis has two. Mitosis also makes two diploid cells and meiosis makes four haploid cells. Crossing over is also only found in meiosis.

 

Compare mitosis and meiosis with respect to each of the following:

 

 

Mitosis Meiosis
Chromosome number of parent cells 2n 2n
Number of DNA replications 1 1
Number of divisions 1 2
Number of daughter cells produced 2 4
Chromosome number of daughter cells 2n n
Purpose Repair and growth Gamete and spore production

 

 

How are Meiosis I and Meiosis II different?

 

Meiosis I ends in two chromosomes with two chromatids and Meiosis II ends in four chromosomes with only one chromatid.

 

How do oogenesis and spermatogenesis differ?

 

Oogeneis forms the eggs and spermatogenesis forms the sperm.

 

Why is meiosis important for sexual reproduction?

 

Meiosis makes the chromosome number come out in half so that fertilization can come back and restore the diploid number.

Lab 3B. 2

 

Number of 4:4 Number of Asci Showing Crossover Total Asci % Asci Showing Crossover Divided by 2 Gene to Centromere Distance (Map Units)
39 19 58 9.5 9.5 map units

 

 

Draw a pair of chromosomes in MI and MII, and show how you would get a 2:4:2 arrangement of ascospores by crossing over.

 

 

Error Analysis

 

No errors were reported in this lab but there could have been. The cells in Lab 3A. 1 could have been miscounted or not counted.

 

Conclusion

 

From the data collected in this experiment, it can be concluded that the mitotic stages of the whitefish blastula and the onion root-tip can be observed with a light microscope. The time spent in each phase of mitosis can be recorded and it is concluded that the most time spent in a stage is in interphase. It can also be concluded that the least time spent in a stage is in telophase. It is also understood that someone can simulate meiosis using a chromosome simulation kit. On the last part of the lab, Lab 3B. 2, one could conclude that more asci do not cross over than do the number of asci that do cross over.

BACK

 

AP Lecture Guide 17 – From Gene to Protein

AP Biology: CHAPTER 17

FROM GENE TO PROTEIN

1. How did diseases involving metabolic pathways lead to hypotheses about the nature of

genes?

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2. Identify some genetic diseases that occur along metabolic pathways.

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3. What was Beadle and Tatum’s hypothesis regarding enzymes?

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4. How has that hypothesis been modified?

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5. What occurs during transcription?

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6. What occurs during translation?

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7. How does the protein process differ in prokaryotes and eukaryotes?

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8. Briefly explain how Marshall Nirenberg and Heinrich Matthaei “cracked the genetic code?”

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9. What is the genetic code and why is said to be universal?

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10. List several features about the genetic code.

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11. Give an example of what happens if reading frames are altered?

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12. List the highlights of the three stages of transcription.

a. Initiation _______________________________________________________________

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b. Elongation ______________________________________________________________

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c. Termination _____________________________________________________________

13. What happens to the transcript RNA before it leaves the nucleus?

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14. What is the advantage of the 5’ cap and poly A tail?

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15. Distinguish between exons and introns.

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16. Describe the mechanism for splicing RNA.

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17. What does alternative RNA processing do for cells?

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18. Identify the roles of the players of the translation process.

a. Transfer RNA ___________________________________________________________

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b. Aminoacyl-tRNA synthetase _______________________________________________

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c. Ribosomes _____________________________________________________________

19. Identify and briefly describe the steps of translation. Initiation Elongation Termination

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20. What is the advantage of polyribosomes?

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21. Give an example of how a polypeptide gets into the ER for additional processing.

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22. How does protein synthesis differ between prokaryotes and eukaryotes?

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23. Define point mutations. ______________________________________________________

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24. Define mutations that are:

a. Missense ______________________________________________________________

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b. Nonsense ______________________________________________________________

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c. Insertion or deletion ______________________________________________________

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25. Use the diagram to trace the flow of chemical information from the gene to the protein product.