10.5: Post-lab Questions - Biology

Standard Plate Count

  1. Using the formula V1D1=V2D2, find the final dilution of the following (remember that D1=1):
    1. 10.0ml of a sample is transferred to 90.0ml of diluent
    2. 4.0ml of a sample is transferred to 7.0ml of diluent
  2. Using the formula determine the original cell density in the sample:[egin{align*} ext {Number of CFU:} ext {OCD }&= ext { Amount plated } imes ext { dilution factor} end{align*} onumber ]
    1. You counted 72 colonies on a plate in your dilution series. The plate was inoculated with 1.0ml from a 10-4 test tube, what is the OCD?
    2. You counted 235 colonies on a plate in your dilution series. The plate was inoculated with 0.1ml from a 10-7 test tube, what is the OCD?
    1. You must make a 10-1 dilution using 5.0ml of your original sample. Show how you would do this. Hint: What volume of diluent do you need to add the 5.0ml to in order to perform and 10-1 dilution?
    2. You must make a 10-2 dilution using 5.0ml of your original sample. Show how you would do this.
    3. Show how you would make a 10-1 dilution using 4.0ml of your original sample.
    4. Show how you would make a 10-2 dilution using 6.0ml of your original sample.
    1. You are instructed to add 2.0ml of your original sample to 2.0ml of diluent. What is your dilution?
    2. You are instructed to add 2.0ml of your original sample to 4.0 ml of diluent. What is your dilution?
    3. In your own words, explain the difference between the above dilutions:
  3. 1.0ml of sample added to 99.0ml of diluent is a 10-2 dilution. What would happen if you added 0.1ml instead of 1.0ml? What would the final dilution be?
  4. The colony count of a 10-8 plate in a dilution series is 162.
    1. What is the bacterial count in the original sample?
    2. What would the colony count be on the 10-7 dilution?
    3. What would the colony count be on the 10-9 dilution?
  5. Set up a dilution scheme, using the following materials, with a final plating of four plates with dilutions of 10-5 through 10-8. You must use exactly all 4 water blanks and plate four plates. Diagram the dilution scheme showing all tubes, plates, and dilutions.

Original sample

3 sterile water blanks (99mL)

1 sterile water blanks (9ml)

4 Petri plates

The Turbidimetric Method

Indirect Determination of Growth by Optical Density


  • 2 Groups per Table-
  • Your Dilution Series
  • Original E. coli
  • DI water (for blank)
  • TSB (for blank)
  • 8 cuvettes and a cuvette rack (Dispose cuvettes with liquid in Biohazard trash can)
  • Kimwipes
  • 5mL pipettes


You will measure the absorbance (OD) of each one of your dilutions, and the original E. coli broth, individually in a spectrophotometer. The spectrophotometer is a machine that shines a light through a liquid sample. A perfectly clear sample will transmit all the light. A sample that is cloudy (turbid) or colored will absorb some of the light and impede the transmission of the light. Therefore there is a direct relationship between turbidity and absorbance and an indirect relationship between turbidity and transmission. Do not write directly on cuvette tubes or cuvette rack. You may label the rack with tape.

  1. Follow the directions on the Spectrophotometer for calibrating it. Ask your instructor if you do not know how to use the instrument, do not attempt to guess. It is important to blank the spectrophotometer with the sterile un-inoculated water to zero absorbance.
  2. Follow the diagram below to set up the cuvettes.
    1. Using a 5ml pipette, dispense 2ml of sterile water into 1 cuvette as your blank (for the dilution tubes).
    2. Using a 5ml pipette, dispense 2ml of TSB into 1 cuvette as your blank (for the E.c. culture).
    3. Using a 5ml pipette, dispense 2ml of the original E. coli culture into a cuvette.
    4. Using a separate 5ml pipette for each dilution tube, dispense 2ml of each dilution into a corresponding cuvette.
    5. Use the blank cuvette (water only) to blank the spectrophotometer. Place the cuvettes sequentially into the spectrophotometer and read the optical density of the original culture and each of your dilutions. Record the O.D values into the table, plot them on a graph, and answer the questions.


  1. Fill out the following table:




1. Original broth







  1. Graph the results on the attached graph. And answer the questions below. Be careful to set the axes correctly and space the dilutions correctly.

Turbidimetric Method Questions

  1. Why is it important to have a plate count along with the absorbance measurement?
  2. What is the relationship between the O.D. values graphed and the cell number from the SPC for your culture?

Impact of pre-lab learning activities, a post-lab written report, and content reduction on evolution-based learning in an undergraduate plant biodiversity lab

Commonplace biodiversity labs in introductory undergraduate biology typically emphasize declarative knowledge. We contend that shifting these labs to emphasize evolution, higher-order cognition, and science reasoning would benefit student learning. Four factors that likely make evolution-based higher-order learning goals difficult to achieve in these labs are: the novelty and quantity of required declarative knowledge, the number of integrated concepts, the theoretical nature of evolution, and limitations on working memory. Thus, we propose that a model to shift learning from lower-order declarative knowledge to evolution-based higher-order integration in these labs would reduce overall lower-order content, increase time efficiency through hands-on pre-lab activities, and increase evidence-based reasoning through written post-labs that emphasize evolution-based higher-order integration. We tested this contention by comparing exam performances of students who did and did not participate in the redesigned lab.


A new plant biodiversity lab design was implemented in an introductory undergraduate biology lab class. The lab class was a separate class from the complementary lecture class, but the content-oriented learning goals were similar between the lecture and lab. We compared achievement of students in lecture + lab to those in lecture only with a pre-assessment and a mid-semester exam which contained questions that were both related and unrelated to the plant biodiversity lab learning goals.


Students in ‘lecture + lab’ relative to ‘lecture only’ did not perform significantly different on the pre-assessment lower or higher-order questions. On the post-assessment, students in lab + lecture performed significantly better on knowledge questions that were unrelated to lab with an improvement of 5.9%. Moreover, students in lab + lecture also performed significantly better on lab-related knowledge questions and lab-related evolution-based integrative reasoning questions with a range of 6.3 to 11% improvement, compared to students in the lecture only group.


The proposed framework was successful in improving student learning for both lower-order declarative knowledge questions and evolution-based questions involving higher-order integration of concepts. In addition, because students in lecture + lab outperformed students in lecture only on questions unrelated to lab content, our proposed model highlights the importance of multiple inquiry-oriented lab experiences in higher education.

Lab 5 Ap Sample 2 Cell Resp

Cellular respiration is the release of energy from organic compounds by metabolic chemical oxidation in the mitochondria in each cell. Cellular respiration involves a number of enzyme mediated reactions. The equation for the oxidation glucose is C6H12O6 + O2 à CO2 + H2O + 686 kilocalories per mole of glucose oxidized. There are three ways cellular respiration could be measured. The consumption of O2 (how many moles of O2 are consumed in cellular respiration). Production of CO2 (how many moles of CO2 are produced in cellular respiration?) and the release of energy during cellular respiration. In this lab, the volume of O2 consumed by germinating and non-germinating peas at two different temperatures will be measured.

PV=nRT is the inert gas law. 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. T is the temperature of the gas in degrees K. This law tells us several important things 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. 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 or volume or both will change in direct proportion to the temperature.

In this lab, CO2 , made during cellular respiration will be removed by potassium hydroxide (KOH) and will make potassium carbonate (K2CO3). Carbon dioxide is removed so the change in the volume of gas in the respirometer will be directly proportional to the amount of oxygen that is consumed. In the experiment water will move toward the region of lower pressure. During respiration, oxygen will be consumed and its volume will be reduced to a solid. The result is a decrease in gas volume within the tube, and a related decrease in pressure in the tube. The respirometer with just the glass beads will allow changes in volume due to changes in atmospheric pressure or temperature changes.

The respirometer with only germinating peas will have a larger consumption of oxygen and will have a larger amount of CO2 that is converted into K2CO3 than the respirometer with beads and dry peas and the respirometer with beads alone.

The materials used in the lab are as follows: a thermometer, 2 water baths, tap water, masking tape, germinating peas, non-germinating (dry) peas, 100 mL graduated n cylinder, 6 vials, 6 rubber stoppers, absorbent and non absorbent cotton, KOH, 5 mL syringe, 6 pipettes, ice, and 6 washers.

First, set up both a room temperature 25oC and a 10oC water bath. Make sure you allow time to adjust the temperature in each bath. To obtain a temperature of 10oC add ice to of the baths until the temperature in the bath is 10oC. Next, obtain a 100 mL graduated cylinder and fill it with 50 mL of water. Drop in 25 germinating peas and determine the amount of water that is displaced. Record the volume of the 25 germinating peas. Then remove these peas and place them on a paper towel. They will be used in respirometer 1. Next, refill the graduated cylinder with 50 mL of water and drop 25 non-germinating peas into it. Then drop glass beads into the respirometer until the volume is equivalent to that of the expanded germinating peas. Remove the beads and peas. They will be used in respirometer 2. Next, refill the graduated cylinder with 50 mL of water. Determine how many glass beads would be required to attain a volume that is equivalent to that of the germinating peas. Remove the beads. They will be used in respirometer 3. Then repeat the procedures used above to prepare a second set of germinating peas, dry peas + beads, and beads to be used in respirometers 4,5,and 6.

Assemble the six respirometers by obtaining 6 vials, each with an attached stopper and pipette. Then put a small wad of absorbent cotton in the bottom of each vial and, using the syringe, saturate the cotton with 15 % KOH making sure not to get the KOH on the sides of the respirometer. Then place a small wad of dry cotton on top of the KOH-soaked absorbent cotton. Repeat these steps to make the other five respirometers. Make sure to use about the same amount of cotton in each vial.

Next, place the first set of germinating peas, dry peas + beads and beads in vials 1,2, and 3. Place the second set of germinating peas, dry peas + beads, and beads in vials 4,5, and 6. Insert the stoppers in each vial with the proper pipette. Place a washer on each of the pipettes to be used as a weight.

Make a sling using the masking tape and attach it to each side of the water baths to hold the pipettes out of the water during the equilibration period of 10 minutes. Vials 1,2, and 3, should be in the bath containing water of 25o C. Vials 4, 5, and 6 should be in the bath containing water that is 10oC. After the equilibration period completely immerse all six respirometers in the water completely. Water will enter the pipette for a short distance and stop. If it does not stop, there is a leak. Make sure the pipettes are facing so you can read them. The vials should not be shifted during the experiment and your hands should not be placed in the water during the experiment.

Allow the respirometers to equilibrate for three more minutes and then record the initial water in each pipette time 0. Check the temperature in both baths and record in table 5.1. Every five minutes for 20 minutes, take readings of the water’s position in each pipette, and record the data in table 5.1.

Table 5.1: the Measurement of Oxygen Consumption by Soaked and Dry Pea Seeds at Room Temperature 25o C and 10oC Using Volumetric methods.

Temp o C Time (min) Reading at time X Diff. Reading at time X Diff. Corrected Diff. Reading at time X Diff. Corrected Diff.
25 Initial- 0 14.4 13.9 14.2
25 0 to 5 14.1 .3 13 .9 .6 14.1 .1 -.2
25 5 to 10 14.0 .4 11.1 2.8 2.4 13.9 .3 -.1
25 10 to 15 13.9 .5 10.3 3.6 3.1 13.7 .5 0
24 15 to 20 13.9 .5 8.8 5.1 4.6 13.5 .7 .2
10 Initial – 0 14.2 14.2 14.7
10 0 to 5 14.8 -.6 14.0 .2 .8 15.2 -.5 .1
10 5 to 10 14.6 -.4 13.5 .7 1.1 15 -.7 -.3
10 10 to 15 14.8 -.6 13.2 .9 1.5 15 -.7 -.1
10 15 to 20 14.9 -.7 12.6 1.6 2.3 15 -.7 0

Graph: Consumption of Oxygen for Germinating Peas and Dry Peas at 10oC and 25o C.

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.
The hypothesis being tested in this activity is that the germinating peas in a water bath of 25 o C will have a higher respiration rate than the other vials.

2. This activity uses a number of controls. Identify at least three of the controls, and describe the purpose of each control.

One control is each vial had the same volume. This showed that the volume of the vial did not effect respiration rate. Another control was the vial with beads alone. The beads carried out no respiration. The final control was the 10 minute equilibration period. This allowed the contents of the vials to carry out respiration for a short period of time before they were completely immersed in the water.

3.Graph the results from the corrected difference column for the germinating and dry peas at both room temperature and at 10oC.

4. Explain the relationship between the amount of oxygen consumed and time.

As time increased oxygen consumption increased.

5. From the slope of the four lines on the graph, determine the rate of oxygen consumption of germinating and dry peas during the experiments at room temperature and at 10o C.

Condition Show Calculations Rate in mL O / minute
Germinating peas at 10oC 2.3-1.5=.8/5 .16mL O2 /minute
Germinating peas at room temperature 4.6-3.1/5 .3mL O2 /minute
Dry peas at 10oC (.1)/5= .02 mL O2 /minute
Dry peas at room temperature (.2-0 )/5= .04 mL O 2 /minute

6. Why is it necessary to correct the readings from the peas with the readings from the beads?

The beads carried out no cellular respiration. The peas did. Changes in atmospheric pressure could have caused changes in respiration rate and correcting the readings provided the most accurate results under the given conditions.

7. Explain the effect of germination versus non-germination on pea seed respiration.

Germination causes a higher rate of respiration than the non-germinating peas.

8. Graph the predicted results through 45o C. Explain your prediction.

As the temperature increased cellular respiration increased, but after a certain temperature the respiration rate will start to go down. The peak is the optimal temperature.

9. What is the purpose of KOH in this experiment?

KOH removes carbon dioxide formed during cellular respiration.

10. Why did the vial have to be completely sealed around the stopper.

The stopper was completely sealed to prevent water from entering the respirometer.

11. If you used the same experimental design to compare the rates of respiration of a 25g. reptile and a 25 g. mammal at 10oC what results would you expect? Explain your reasoning.

The mammal would carry out a higher rate of cellular respiration. This is because the mammal maintains a constant temperature that is higher than the temperature of the cold blooded reptiles that will have a temperature of 10 C.

12. If respiration in a small mammal were studied at both room temperature 21 o C and 10oC what results would you predict? Explain your reasoning.

The rate of cellular respiration would be higher at 21 degrees C because the 10 degrees C temperature could cause the overall body of the mammal temperature to drop the most.

13. Explain why water moved into the respirometers’ pipettes.

Water moved into the pipettes because oxygen was being consumed and allowed water to move only partially into the pipette.

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?

I would use the same format using respirometers to measure the cellular respiration rate of the peas. The peas that had been germinating for 72 hours would have a higher respiration rate because they have a higher energy demand.

Error Analysis:
Several factors could have caused inaccurate results in this experiment. First, not maintaining a constant temperature in the water bath could have caused inaccurate results. Also moving the vials in the water after the experiment began could have caused inaccurate results. Putting your hands in the water bath while the vials were in the water could have caused inaccurate results. Allowing the peas to come into contact with the KOH could have also caused inaccurate results. Finally not having the same amount of cotton in each vial could have caused an error in the results.

In this experiment the vial with just germinating peas had the greatest consumption of oxygen. This is because germinating peas carried out a more rapid process of cellular respiration than the non-germinating peas. The beads carried out no cellular respiration. The non-germinating peas require less energy than the germinating peas so the dry peas carry out a slower process of cellular respiration. This in turn caused less oxygen to be consumed in the vials with non-germinating peas than the vials with germinating peas. The higher temperature caused cellular respiration to occur at a higher rate which in turn caused a greater consumption of oxygen.

10.5: Post-lab Questions - Biology

Answers for Laboratory Calculation Problem Set #1

1. You need to make a 1:5 dilution of a solution. You need 10 ml of the diluted solution. How much initial sample and diluent should you use?

Answer : 1:5 dilution = 1/5 dilution = 1 part sample and 4 parts diluent in a total of 5 parts. If you need 10 ml, final volume, then you need 1/5 of 10 ml = 2 ml sample. To bring this 2 ml sample up to a total volume of 10 ml, you must add 10 ml - 2 ml = 8 ml diluent.

2. How would you prepare 500 ml of a 10% NaCl solution?

Answer : In this problem, the % solution is the number of grams solute in 100 ml solvent, so a 10% solution of NaCl is 10 grams NaCl in 100 ml water. But you need 500 ml, final volume, so 10 g x 5 = 50 g NaCl.

3. If you have DNA with a concentration of 2 µg/µl, how much DNA (in µl) must be added to make a 20 µl solution with a DNA concentration of 1 µg/µl?

Answer : Since you know the initial concentration (2 µg/µl), the final concentration (1 µg/µl), and the final volume (20 µl), the following formula can be used to calculate the amount of DNA needed (initial volume)

  • (initial concentration)(initial volume) = (final concentration)(final volume)
  • (2 µg/µl)(X µl) = (1 µg/µl)(20 µl)
  • X µl = (1 µg/µl)(20 µl) / 2 µg/µl
  • X µl = 10 µl DNA

4. You have a 10x TBE buffer. To run a gel, you need 500 ml of a 2x solution of TBE. How do you make a 500 ml solution of 2x TBE buffer from the 10x buffer?

Answer : Since you know the initial concentration (10x), the final concentration (2x), and the final volume (500 ml), you can use the formula:

  • (initial concentration)(initial volume) = (final concentration)(final volume)
  • (10x)(X ml) = (2x)(500 ml)
  • X ml = (2x)(500 ml) / 10x
  • X ml = 100 ml of 10x TBE

Then, to calculate the amount of water needed, use the following formula:

  • final volume - initial volume = volume of diluent
  • 500 ml total - 100 ml of 10x TBE = 400 ml water

5. You want to make a 0.5% agarose gel. How much agarose (in grams) do you need to make up a 50 ml gel solution?

Answer : There are at least two methods for solving this question (as with many dilution problems): logically and mathmatically.

  • Logically :
  • 0.5% means 0.5 grams in 100 ml, so if you only need 50 ml, you need 0.5 g / 2 = 0.25 g agarose for a 50 ml gel solution.
  • Mathematically :
  • 0.5 g/100 ml = X g/50 ml
  • (0.5 g) (50 ml)/100 ml = X g
  • 0.25 g = X g

6. What is the DNA concentration of a 50 µl solution which contains 10 µl of DNA at a concentration of 4 µg/µl?

Answer : There are two ways to solve this problem:

  • Calculate the total amount of DNA in the solution, then divide by the total volume:
    • 10 µl x 4 µg/µl = 40 µg of DNA
    • 40 µg DNA/ 50 µl = 0.8 µg/µl
    • (initial concentration)(initial volume) = (final concentration)(final volume)
    • (4 µg/µl)(10 µl) = (X µg/µl)(50 µl)
    • X µg/µl = (4 µg/µl)(10 µl) / 50 µl
    • X µg/µl = 0.8 µg/µl

    7. How would you make a 3x TBE buffer from a 12x TBE buffer for a total volume of 200 ml?

    Answer : Since you know the initial concentration (12x), the final concentration (3x), and the final volume (200 ml), you can use the formula:

    • (initial concentration)(initial volume) = (final concentration)(final volume)
    • (12x)(X ml) = (3x)(200 ml)
    • X ml = (3x)(200 ml) / 12x
    • X ml = 50 ml of 12x TBE

    Then, to calculate the amount of water needed

    Thanks are given to Tim Allen and the MMG graduate students for providing these problems.

    Proteins (Exam Questions and Answers) | Microbiology

    Ans: A protein is a macromolecule composed of one or more polypeptide chains, each with a characteristic sequence of amino acids linked by peptide bonds. The cells of organisms contain thousands of different proteins each with a different function or biological activity. These functions include enzymatic activity, molecular transport, nutrition, cell or organismal motility, structural roles, defense regulation etc.

    The protein consists of very long polypeptide chains with 100 to 2000 amino acid residues joined by peptide linkages. However, all proteins are made from the same set of 20 amino acids. The genetic information of a cell is ultimately expressed as protein. There is a segment of DNA (a gene) for each protein which encodes information specifying its sequence in amino acids.

    Q.2. How do fireflies produce light in darkness? Also give some other examples of bioluminescence.

    Ans: The production of light by a firefly is as a result of light producing reaction involving luciferin a complex carboxylic acid and ATP which is catalyzed by enzyme luciferase. Some fungi that include some mushrooms, marine dinoflagellates, jellyfish and crustaceans are also able to generate bioluminescence which needs a considerable amount of energy.

    In the firefly ATP is used in a set of reactions which converts chemical energy into light energy. McElroy and his colleagues at Johns Hopkins University isolated the major biochemical components involved luciferin, a complex carboxylic acid and luciferase an enzyme.

    The regeneration of light flashes or bioluminescence needs activation of luciferin by an enzymatic reaction with ATP in which a pyrophosphate cleavage of ATP takes place to form luciferyl adenylate. This compound is then acted upon by molecular oxygen and luciferase to bring about the oxidative decarboxylation of the luciferin to produce oxyluciferin. This reaction which has some intermediate steps is accompanied by emission of light as shown in Fig. 39.1.

    Q. 3. How can proteins be classified according to their biological roles?

    Ans: The proteins according to their biological functions can be classified into the following types:

    (i) Enzymes or the proteins with catalytic activity.

    (ii) Transport proteins e.g., hemoglobin of erythrocytes.

    (iii) Nutrient and storage proteins, e.g., proteins of seeds, ovalbumin the major protein of egg white and casein (the milk protein).

    (iv) Contractile or motile proteins, e.g., actin and myosin in skeletal muscle and protein dynein in flagella and cilia.

    (v) Structural proteins, e.g., collagen (protein in leather), keratin (in hair, nails and feathers), fibroin (silk fibres and spider webs), and resilin (hinges of some insects).

    (vi) Defense proteins, e.g., immunoglobulin’s or antibodies fibrinogen and thrombin (blood clotting proteins) and snake venoms, bacterial toxins and plant proteins like ricin.

    (vii) Regulatory proteins, e.g., insulin which regulates sugar metabolism and growth hormones of pituatory.

    (viii) Monellin, a protein of an African plant which has intensely sweet taste.

    (ix) Antifreeze proteins, found in Antarctic fish.

    Q.4. What are conjugate proteins? How are they classified?

    Ans: Some proteins also have chemical components in addition to amino acids. These are called conjugated proteins. The non-amino acid part of the conjugated protein is usually called its prosthetic group. The conjugate proteins are classified on the basis of the chemical nature of their prosthetic groups.

    The major classes of proteins with their prosthetic group and examples are given below:

    1. Lipoproteins, with lipids as prosthetic groups e.g., P, – lipoprotein of blood.

    2. Glycoproteins, with carbohydrates (sugars) as prosthetic group, e.g., immunoglobulin G

    3. Phosphoproteins, with phosphate group, e.g., casein of milk.

    4. Haemoproteins, with haeme or iron porphyrin, prosthetic group, e.g., haemoglobin.

    5. Flavoproteins, with flavin nucleotides as prosthetic group, e.g, succinate dehydrogenase.

    6. Metalloproteins, with iron as prosthetic group, is ferritin, with zinc as alcohol dehydrogenase, with calcium is calmodulin, with molybdenum is dinitrogenase and with copper is plastocyanin.

    Q.5. Give the techniques to study structure and function of proteins.

    Ans: Study of individual proteins can be done by using following techniques:

    (i) Separation and purification of proteins

    (ii) Proteins can be characterized by electrophoresis using detergent sodium dodecyl sulphate (SDS).

    (iv) The antibody-antigen interaction is also used to quantify and localize proteins, e.g. ELISA and RIA.

    Q.6. What do you mean by structure of proteins? Elaborate.

    Ans: There are four levels of structure or architecture in proteins, which are known as primary structure, secondary structure, tertiary structure and quaternary structure.The primary structure covers all the covalent bonds between amino acids and is characterized by the sequence of amino acids in the peptide. It also tells about locations of disulphide bonds. The secondary structure results from hydrogen bonding between side chains of amino acids to form α- helices and β- sheets.

    The tertiary structure is because of folding of proteins as a result of interaction between the amino acid side chains usually controlled by chaperone proteins. The quaternary structures results when more than one polypeptide makes up the functional protein.

    Q.7. Why do woolen clothes shrink when washed in hot water or dried in an electric dryer and the silk on the other hand does not shrink under similar conditions? Explain.

    Ans: The major components of silk fibres and spider webs are fibroin, a type of structural protein and a member of fibrous proteins called |3- keratins. |3- conformation makes fibroin of silk soft and flexible in comparison to a- keratin of hair, feather and nails (which has a helix cross-linked by disulphide bonds) due to which a- keratin is tough, insoluble protective structure of varying hardness and flexibility.

    Heating and steaming of wool increases the spacing of structural units. It is because of soft and flexible nature of β- conformation of β- keratins of silk. The Indian masses have been obtaining silk by boiling the cocoons in hot water since ancient times.