Author: Biology Junction Team

Handase Lab

 

 

ENZYME RATE OF REACTION FOR CATALASE

 

Introduction:

Chemical reactions make life possible. Hundred of chemical reactions are involved in the process of digesting a candy bar. If these reactions proceeded too slowly, not only would the candy bar remain in the stomach for long time, but the ordinary activities of life would come to a halt as well. Since this is not the case, something in the body must be responsible for speeding up the process. Four things that can speed up chemical reactions are: (1) heat; (2) increasing the concentration of reactants; (3) decreasing the concentration of products; and (4) enzymes, which speed up reaction without themselves being used up.

Enzymes are important in regulating chemical pathways, synthesizing materials needed by cells, releasing energy, and transferring information. Enzymes are involved in digestion, respiration, reproduction, vision, movement, thought, and even in the production of other enzymes. With few exceptions, enzymes are proteins. Simple cells may have as many as 2000 different enzymes, each one catalyzing a different reaction. An enzyme may accelerate a reaction by a factor of 1010 making it happen 10,000,000,000 times faster. Thus, a reaction that might take place as long as 1500 years without an enzyme can be completed in just 5 seconds with an enzyme.

In this lab, your hands are the enzyme Catalase. This enzyme will split H2O2 (a poisonous waste product made by cells) into H20 and O2. You will split the molecule by rippling the paper model down the middle.

Pre Lab Questions:

1. What is an enzyme? What is its functions in living things?

 

2. What are things that can affect the function of an enzyme?

 

3 Write the chemical equation for the breakdown of hydrogen peroxide by the enzyme catalase.

 

4 An enzyme’s efficiency increases with greater substrate concentration, but only up to a point. Why?

 

 

Procedure:

1 Cut out 100 hydrogen peroxide molecules from the paper template.

2. Place 100 paper hydrogen peroxide molecules into a paper bag.

3. One member of your group will do the following:

a) When told to, you will grab one hydrogen peroxide molecule and rip it down the middle. Only rip one hydrogen peroxide molecule at a time.

b) Place the pieces back into the paper bag and grab another hydrogen peroxide molecule.

e) Repeat steps a and b, as fast as you can for 10 seconds. A member of group will be timing you for 10 seconds.

d) Empty your container and count the number of ripped hydrogen peroxide molecules.

e) Record the data in the following table.

 

Time in seconds Ripped Hydrogen Peroxide Molecules Rate of Reaction
0-10 A A
10-30 A A
30-60 A A
60-120 A A
120-180 A A

 

4. Repeat step a – step e for 30, 60, 120, 180 seconds.

5. Graph the results.

6. Determine the rate of reaction for the following times.

The rate of reaction can be calculated by using the following equation:

 

Rate = M2-M1
t2– t1

 

a. 0-10 seconds

 

b. 10-30 seconds

 

c. 30-60seconds

 

d. 60-120 seconds

 

e. 120-180 seconds

 

7. Record the above rates in a data table.

8. Graph the results.

Graph Title: ______________________________________________

 

Post Lab Questions:

1. If you were allowed to continue this lab and rip hydrogen peroxide molecules for 240 and 300 seconds. What would happen to the rate of reaction and why will this happen?

 

 

2. What can you say about the length of time and the rate of the reaction?

 

 

3. What would happen to the rate of reaction if you remove the H2O and O2 molecules as soon as they are produced?

 

 

 

Paper Molecules:.

 

Hardy-Weinberg Problems

 

POPULATION GENETICS AND THE HARDY-WEINBERG LAW

 

The Hardy-Weinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions must be met:

  1. No mutations must occur so that new alleles do not enter the population.
  2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population).
  3. Random mating must occur (i.e. individuals must pair by chance)
  4. The population must be large so that no genetic drift (random chance) can cause the allele frequencies to change.
  5. No selection can occur so that certain alleles are not selected for, or against.

Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some or all of these types of forces all act on living populations at various times and evolution at some level occurs in all living organisms. The Hardy-Weinberg formulas allow us to detect some allele frequencies that change from generation to generation, thus allowing a simplified method of determining that evolution is occurring. There are two formulas that must be memorized:

 

p2 + 2pq + q2 = 1 and p + q = 1

 

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

Individuals that have aptitude for math find that working with the above formulas is ridiculously easy. However, for individuals who are unfamiliar with algebra, it takes some practice working problems before you get the hang of it. Below I have provided a series of practice problems that you may wish to try out. Note that I have rounded off some of the numbers in some problems to the second decimal place.

PROBLEM #1    You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:

  1. The frequency of the “aa” genotype.
  2. The frequency of the “a” allele.
  3. The frequency of the “A” allele.
  4. The frequencies of the genotypes “AA” and “Aa.”
  5. The frequencies of the two possible phenotypes if “A” is completely dominant over “a.”

PROBLEM #2.    Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these “partially defective” red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

PROBLEM #3.    There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following:

  1. The frequency of the recessive allele.
  2. The frequency of the dominant allele.
  3. The frequency of heterozygous individuals.

PROBLEM #4.    Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:

  1. The percentage of butterflies in the population that are heterozygous.
  2. The frequency of homozygous dominant individuals.

PROBLEM #5.     A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:

  1. The allele frequencies of each allele.
  2. The expected genotype frequencies.
  3. The number of heterozygous individuals that you would predict to be in this population.
  4. The expected phenotype frequencies.
  5. Conditions happen to be really good this year for breeding and next year there are 1,245 young “potential” Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided?

PROBLEM #6.    A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, “aa”. Calculate allelic and genotypic frequencies for this population.

PROBLEM #7.    After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island?

PROBLEM #8.    You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:

 

BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY
M MM 490 0.49
MN MN 420 0.42
N NN 90 0.09

 

Using the data provide above, calculate the following:

  1. The frequency of each allele in the population.
  2. Supposing the matings are random, the frequencies of the matings.
  3. The probability of each genotype resulting from each potential cross.

PROBLEM #9.    Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following:

  1. The frequency of the recessive allele in the population.
  2. The frequency of the dominant allele in the population.
  3. The percentage of heterozygous individuals (carriers) in the population.

PROBLEM #10.    In a given population, only the “A” and “B” alleles are present in the ABO system; there are no individuals with type “O” blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the allelic frequencies of this population (i.e., what are p and q)?

PROBLEM #11.    The ability to taste PTC is due to a single dominate allele “T”. You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies.

ANSWERS

Pig Heart Dissection

 

Heart Dissection

Introduction
Mammals have four-chambered hearts and double circulation. The heart of a bird or mammal has two atria and two completely separated ventricles. The double-loop circulation is similar to amphibians and reptiles, but the oxygen-rich blood is completely separated from oxygen-poor blood. The left side of the heart handles only oxygenated blood, and the right side receives and pumps only deoxygenated blood. With no mixing of the two kinds of blood, and with a double circulation that restores pressure after blood has passed through the lung capillaries, delivery of oxygen to all parts of the body for cellular respiration is enhanced. As endotherms, which use heat released from metabolism to warm the body, mammals require more oxygen per gram of body weight than other vertebrates of equal size. Birds and mammals descended from different reptilian ancestors, and their four-chambered hearts evolved independently – an example of convergent evolution.

Objective
Using a pig heart, students will observe the major chambers, valves, and vessels of the heart and be able to describe the circulation of blood through the heart to the lungs and back and out to the rest of the body. (The pig heart is used because it is very similar to the human heart in structure, size, & function.)

Materials
Dissecting pan, dissecting kit, safety glasses, lab apron, pig heart, & gloves

Procedure – External Structure

  1. Place a heart in a dissecting pan & rinse off the excess preservative with tap water. Pat the heart dry.
  2. Examine the heart and locate the thin membrane or pericardium that still covers the heart. The pericardium or pericardial sac, is a double-layered closed sac that surrounds the heart and anchors it.  The pericardium consists of two tissues layers – the visceral pericardium that covers the surface of the heart & the parietal pericardium covering the inner surface of the parietal sac. These two tissue layers are continuous with each other where the vessels enter or leave the heart. The slender gap between the parietal & visceral surfaces is the pericardial cavity & is filled with fluid to reduce friction between the layers as the heart pumps.
  3. After examining the pericardium, carefully remove this tissue. Located below the pericardium is the muscle of your heart called the myocardium. Most of the myocardium is located in the lower two chambers of the heart called ventricles.
  4. Locate the tip of the heart or the apex. Only the left ventricle extends all the way to the apex.
  5. Place the heart in the dissecting pan so that the front or ventral side is towards you ( the major blood vessels are on the top and the apex is down). The front of the heart is recognized by a groove that extends from the right side of the broad end of the heart diagonally to a point above & to your left of the apex.


Front or Ventral Side of the Heart

  1. The heart is now in the pan in the position it would be in a body as you face the body. Locate the following chambers of the heart from this surface:
    • Left atria – upper chamber to your right
    • Left ventricle – lower chamber to your right
    • Right atria – upper chamber to your left
    • Right ventricle – lower chamber to your left

pig heart dissection

  1. While the heart is still in this position in the dissecting pan, locate these blood vessels at the broad end of the heart:
  • Coronary artery – this blood vessel lies in the groove on the front of the heart & it branches over the front & the back side of the heart to supply fresh blood with oxygen & nutrients to the heart muscle itself.
  • Pulmonary artery – this blood vessel branches & carries blood to the lungs to receive oxygen & can be found curving out of the right ventricle (upper chamber to your left)
  • Aorta – major vessel located near the right atria & just behind the pulmonary arteries to the lungs. Locate the curved part of this vessel known as the aortic arch. Branching from the aortic arch is a large artery that supplies blood to the upper body.
  • Pulmonary veins – these vessels return oxygenated blood from the right & left lungs to the left atrium (upper chamber on your right)
  • Inferior & Superior Vena Cava – these two blood vessels are located on your left of the heart and connect to the right atrium (upper chamber on your left).  Deoxygenated blood enters the body through these vessels into the right receiving chamber.  Use your probe to feel down into the right atrium. These vessels do not contain valves to control blood flow.

 

Procedure – Internal Anatomy:

  1. Use scissors to cut through the side of the pulmonary artery and continue cutting down into the wall of the right ventricle. Be careful to just cut deep enough to go through the wall of the heart chamber. (Your cutting line should be above & parallel to the groove of the coronary artery.)
  2. With your fingers, push open the heart at the cut to examine the internal structure. If there is dried blood inside the chambers, rinse out the heart.
  3. Locate the right atrium. Notice the thinner muscular wall of this receiving chamber.
  4.  Find where the inferior & superior vena cava enter this chamber & notice the lack of valves.
  5. Locate the valve that between the right atrium and right ventricle. This is called the tricuspid valve. The valve consists of three leaflets & has long fibers of connective tissue called chordae tendinae that attach it to papillary muscles of the heart. This valve allows blood flow from the right atrium into the right ventricle during diastole (period when the heart is relaxed). When the heart begins to contract (systole phase), ventricular pressure increases until it is greater than the pressure in the atrium causing the tricuspid to snap closed.


Tricuspid Valve

  1. Use your fingers to feel the thickness of the right ventricle and its smooth lining. Also note the network of irregular muscular cords on the inner wall of this chamber.
  2. Find the septum on the right side of the right ventricle. This thick muscular wall separates the right & left pumping ventricles from each other.
  3. Inside the right ventricle, locate the pulmonary artery that carries blood away from this chamber. Find the one-way valve called the pulmonary valve that controls blood flow away from the right ventricle at the entrance to this blood vessel.
  4. Using your scissors, continue to cut open the heart.  Start a cut on the outside of the left atrium downward into the left ventricle cutting toward the apex to the septum at the center groove. Push open the heart at this cut with your fingers & rinse out any dried blood with water.
  5. Examine the left atrium. Find the openings of the pulmonary veins form the lungs. Observe the one-way, semi-lunar valves at the entrance to these veins.
  6. Inside this chamber, look for the valve that controls blood flow between the upper left atrium and lower left ventricle. This valve is called the bicuspid or mitral valve. This valve consists of two leaflets & blood flows from the left atrium into the left ventricle during diastole.

 

The mitral valve
Bicuspid or Mitral Valve

  1. Examine the left ventricle. Notice the thickness of the ventricular wall. This heart chamber is responsible for pumping blood throughout the body.
  2. Using your scissors, cut across the left ventricle toward the aorta & continue cutting to expose the valve.
  3. Count the three flaps or leaflets on this valve leading from the left ventricle into the aorta and note their half-moon shape. This is called the aortic valve.
  4. Using scissors, cut through the aorta and examine the inside. Find the hole or coronary sinus in the wall of this major artery. This leads into the coronary artery which carries blood to and nourishes the heart muscle itself.
  5. Answer the questions on your lab report.

Click here for questions

When you have finished dissecting the heart, dispose of the heart as your teacher advises and clean, dry, and return all dissecting equipment to the lab cart. Wash your hands thoroughly with soap.

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Heart Dissection Questions

Name___________________________ Date__________ Period___________

Heart Dissection

1. Why are pig hearts used to study the anatomy of the human heart?

2. How can you tell which side of the heart is the ventral surface?

3. How many chambers are found in the mammalian heart? What other group of organisms would have this same number of chambers?

4. What is the advantage in having this number of chambers compared to organisms with fewer number of chambers?

5. Which chambers are the pumping chambers of the heart?

6. Which chambers are the receiving chambers of the heart?

7. How do the walls of the atria compare with the walls of the ventricles and why are they different?

8. What is the purpose of heart valves?

9. Name & compare the heart valves found between the upper & lower chambers of the right and left sides of the heart.

10. Vessels that carry blood away from the heart are called __________, while __________ carry blood toward the heart.

11. Which artery is the largest and why?

12. What is the purpose of the coronary artery and what results if there is blockage in this vessel?

13. Use the diagram of the heart below to trace blood flow through the heart:

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