Gene Expression

 

Gene Expression [18,787 bytes]

 

 

Section 11-1     Control of Gene Expression

 

1. Cells use ______________________ to build hundreds of different________________each with a unique ____________________________.

2. Are all proteins used by a cell at any one time? If not, how do cells control this?

3. Define gene expression.

4. When are proteins produced?

5. What is the genome?

6. What are the 2 steps of gene expression?

7. What 2 scientists determined how genes are expressed in prokaryotes?

8. What gene & in what organism did Jacob & Monod make their discoveries about gene expression?

9. Name the 3 regulatory elements on the DNA of the E. coli bacterium and tell the function of each.

10. What is an operon & what 3 things is it made up of?

11. What name did Jacob & Monod give their gene & why?

12. If lactose is not present, what attaches to the operator?

13. Define repressor protein and give its function.

14. Define repression.

15. What occurs if lactose is present in the E. coli in lactose metabolism?

16. What is an inducer?

17. What is an inducer for E. coli in lactose metabolism?

18. How does the genome of eukaryotes compare with that of prokaryotes?

19. Are operons found in eukaryotes?

20. Each eukaryotic cell contains a ___________________ set of genes, but only some genes

are ______________________ at a given time.

21. What controls much of the gene expression in eukaryotes?

22. What is euchromatin?

23. Some sections of chromatin always remain coiled preventing what process?

24. Name & define the 2 kinds of segments found behind the promoter in eukaryotes.

25. Where do the processes of transcription & translation take place in prokaryotes?

26. Where do these processes take place in eukaryotes?

27. Are introns and/or exons transcribed?

28. What is pre-mRNA and how is mRNA formed from this?

 

Section 11-2     Gene Expression and Development

 

29. Multicellular, sexually reproducing organisms begin life as a _____________________with all cells containing the same _______________________.

30. Genes may be turned ______________ and _____________as various ___________________ are needed by the cells.

31. What is cell differentiation?

32. Define morphogenesis.

33. What genes determine what anatomical structures an organism will develop during morphogenesis?

34. What is a tumor and what are the 2 main types?

35. Define benign tumor.

36. Are benign tumors dangerous? Explain.

37. What treatment do doctors use with benign tumors?

38. Define malignant tumor.

39. Malignant tumors are commonly known as ____________________________.

40. What is metastasis & what happens to the body when this occurs?

41. How are malignant tumors categorized?

42. Name & describe 4 types of malignant tumors.

43. Lung cancer & breast cancer are what type of tumors?

44. When do normal cells stop dividing? Do cancer cells respond the same way? Explain.

45. What trait of cancer cells facilitates the spread of cancer cells in the body?

46. What is a carcinogen & give 5 examples?

47. What causes most lung cancer?

48. What is the effect of mutagens on cells?

49. What are oncogenes?

50. Certain ____________________ can cause cancer in plants & animals.

BACK

 

Genetic Notes Bi

 

Mendelian Genetics 

 

 

Mendel 1862 Mendel 1868 Mendel 1880
1862 1868 1880

 

Genetic Terminology:

  • Trait – any characteristic that can be passed from parent to offspring
  • Heredity – passing of traits from parent to offspring
  • Genetics – study of heredity
  • Alleles – two forms of a gene (dominant & recessive)
  • Dominant – stronger of two genes expressed in the hybrid; represented by a capital letter (R)
  • Recessive – gene that shows up less often in a cross; represented by a lower case letter (r)
  • Genotype – gene combination for a trait (e.g. RR, Rr, rr)
  • Phenotype – the physical feature resulting from a genotype (e.g. tall, short)
  • Homozygous genotype – gene combination involving 2 dominant or 2 recessive genes (e.g. RR or Rr); also called pure 
  • Heterozygous genotype – gene combination of one dominant & one recessive allele    (e.g. Rr); also called hybrid
  • Monohybrid cross – cross involving a single trait
  • Dihybrid cross – cross involving two traits
  • Punnett Square – used to solve genetics problems

Blending Concept of Inheritance:

  • Accepted before Mendel’s experiments
  • Theory stated that offspring would have traits intermediate between those of its parents such as red & white flowers producing pink
  • The appearance of red or white flowers again was consider instability in genetic material
  • Blending theory was of no help to Charles Darwin’s theory of evolution 
  • Blending theory did not account for variation and could not explain species diversity
  • Particulate theory of Inheritance, proposed by Mendel, accounted for variation in a population generation after generation
  • Mendel’s work was unrecognized until 1900

Gregor Mendel:

  • Austrian monk
  • Studied science & math at the University of Vienna
  • Formulated the laws of heredity in the early 1860’s
  • Did a statistical study of  traits in garden peas over an eight year period

 

drawing of a flower cross-section showing both male and female sexual structures

 

Why peas, Pisum sativum?

  • Can be grown in a small area
  • Produce lots of offspring
  • Produce pure plants when allowed to self-pollinate several generations
  • Can be artificially cross-pollinate

Picture of Pisum sativum
GARDEN PEA

Mendel’s Experiments:

  • Mendel studied simple traits from 22 varieties of  pea plants (seed color & shape, pod color & shape, etc.)
  • Mendel traced the inheritance of individual traits & kept careful records of numbers of offspring
  • He used his math principles of probability to interpret results
  • Mendel studied pea traits, each of which had a dominant & a recessive form (alleles)
  • The dominant (shows up most often) gene or allele is represented with a capital letter, & the recessive gene with a lower case of that same letter (e.g. B, b)
  • Mendel’s traits included:

         a. Seed shape —  Round (R) or Wrinkled (r)
            b. Seed Color —- Yellow (Y) or  Green (y)
            c. Pod Shape — Smooth (S) or wrinkled (s)
            d. Pod Color —  Green (G) or Yellow (g)
            e. Seed Coat Color —  Gray (G) or White (g)
            f. Flower position — Axial (A) or Terminal (a)
            g. Plant Height — Tall (T) or Short (t)
            h. Flower color — Purple (P) or white (p)


  •  Mendel produced pure strains by allowing the plants to self-pollinate for several generations
  • These strains were called the Parental generation or P1 strain
  • Mendel cross-pollinated two strains and tracked each trait through two
    generations (e.g. TT  x  tt )

     

                  Trait – plant height

                  Alleles – T tall, t short

    P1 cross    TT  x  tt

    genotype      —    Tt
    t t phenotype    —    Tall
    T Tt Tt genotypic ratio –all alike
    T Tt Tt phenotypic ratio- all alike

     

 

  • The offspring of this cross were all hybrids showing only the dominant trait & were called the First Filial or F1 generation
  • Mendel then crossed two of his F1 plants and tracked their traits; known as an F1 cross

 

              Trait – plant height

              Alleles – T tall, t short

F1 cross    Tt  x  Tt

genotype      —    TT, Tt, tt
T t phenotype    —    Tall & short
T TT Tt genotypic ratio —1:2:1
t Tt tt phenotypic ratio- 3:1

 

 

  • When 2 hybrids were crossed, 75% (3/4) of the offspring showed the dominant trait & 25% (1/4) showed the recessive trait; always a 3:1 ratio
  • The offspring of this cross were called the F2 generation
  • Mendel then crossed a pure & a hybrid from his F2 generation; known as an F2 or test cross

 

Trait   –  Plant Height
Alleles – T  tall, t  short

F2 cross       TT  x Tt

F2 cross       tt  x Tt

T t T t
T TT Tt t Tt tt
T TT Tt t Tt tt
          genotype – TT, Tt           genotype – tt, Tt
          phenotype  –  Tall           phenotype  –  Tall & short
          genotypic ratio  – 1:1           genotypic ratio  – 1:1
          phenotypic ratio – all alike           phenotypic ratio – 1:1

 

  • 50% (1/2) of the offspring in a test cross showed the same genotype of one parent & the other 50% showed the genotype of the other parent; always a 1:1 ratio

Problems: Work the P1, F1, and both F2 crosses for all of the other pea plant traits & be sure to include genotypes, phenotypes, genotypic & phenotypic ratios.

  • Mendel also crossed plants that differed in two characteristics (Dihybrid Crosses)
    such as seed shape & seed color
  • In the P1 cross, RRYY  x  rryy, all of the F1 offspring showed only the dominant form for both traits; all hybrids, RrYy

 

Traits:      Seed Shape & Seed Color

Alleles:     R round                Y yellow
r wrinkled             y green

 P1 Cross:     RRYY          x     r r yy  

      

ry Genotype:      RrYy
RY RrYy
Phenotype:      Round yellow seed
Genotypic ratio:      All alike
Phenotypic ratio:      All Alike

 

  • When Mendel crossed 2 hybrid plants (F1 cross), he got the following results

 

 

Traits:       Seed Shape & Seed Color

Alleles:     R round                Y yellow
r wrinkled             y green

     F1 Cross:     RrYy           x     RrYy                   
RY Ry rY ry
RY
RRYY

RRYy

RrYY

RrYy
Ry
RRYy

RRyy

RrYy

Rryy
rY
RrYY

RrYy

r rYY

r rYy
ry
RrYy

Rryy

r rYy

r ryy

 

 

 

Genotypes Genotypic Ratios Phenotypes Phenotypic Ratios
RRYY 1 Round yellow seed
9
RRYy 2
RrYY 2
RrYy 4
RRyy 1 Round green seed
3
Rryy 2
r rYY 1 Wrinkled yellow seed
3
r rYy 2
r ryy 1 Wrinkled green seed
1

 

Problems: Choose two other pea plant traits and work the P1 and F1 dihybrid crosses. Be sure to show the trait, alleles, genotypes, phenotypes, and all ratios. 

Results of Mendel’s Experiments:

  • Inheritable factors or genes are responsible for all heritable characteristics
  • Phenotype is based on Genotype
  • Each trait is based on two genes, one from the mother and the other from the father
  • True-breeding individuals are homozygous ( both alleles) are the same
  • Law of Dominance states that when different alleles for a characteristic are inherited (heterozygous), the trait of only one (the dominant one) will be expressed. The recessive trait’s phenotype only appears in true-breeding (homozygous) individuals

 

Trait: Pod Color
Genotypes: Phenotype:
GG Green Pod
Gg Green Pod
gg Yellow Pod

 

  • Law of Segregation states that each genetic trait is produced by a pair of alleles which separate (segregate) during reproduction

 

Rr
R r

 

  • Law of Independent Assortment states that each factor (gene) is distributed (assorted) randomly and independently of one another in the formation of gametes

 

RrYy

RY Ry rY ry

 

 

Other Patterns of Inheritance:

  • Incomplete dominance occurs in the heterozygous or hybrid genotype where the 2 alleles blend to give a different phenotype
  • Flower color in snapdragons shows incomplete dominance whenever a red flower is crossed with a white flower to produce pink flowers

  • In some populations, multiple alleles (3 or more) may determine a trait such as in ABO Blood type
  • Alleles A & B are dominant, while O is recessive

 

Genotype Phenotype
IOIO Type O
IAIO Type A
IAIA Type A
IBIO Type B
IBIB Type B
IAIB Type AB

 

  • Polygenic inheritance occurs whenever many variations in the resulting phenotypes such as in hair, skin, & eye color
  • The expression of a gene is also influenced by environmental factors (example: seasonal change in fur color)

 

Genetics of Drosophila Melanogaster

 

 

Genetics of Drosophila melanogaster

Introduction:
Gregor Mendel revolutionized the study of genetics. By studying genetic inheritance in pea plants, Gregor Mendel established two basic laws of that serve as the cornerstones of modern genetics: Mendel’s Law of Segregation and Law of Independent Assortment. Mendel’s Law of Segregation says that each trait has two alleles, and that each gamete contains one and only one of these alleles. These alleles are a source of genetic variability among offspring. Mendel’s Law of Independent Assortment says that the alleles for one trait separate independently of the alleles for another trait. This also helps ensure genetic variability among offspring.
Mendel’s laws have their limitations. For example, if two genes are on the same chromosome, the assortment of their alleles will not be independent. Also, for genes found on the X chromosome, expression of the trait can be linked to the sex of the offspring. Our knowledge of genetics and the tools we use in its study have advanced a great deal since Mendel’s time, but his basic concepts still stand true.
Drosophila melanogaster, the common fruit fly, has been used for genetic experiments since T.H. Morgan started his experiments in1907. Drosophila make good genetic specimens because they are small, produce many offspring, have easily discernable mutations, have only four pairs of chromosomes, and complete their entire life cycle in about 12 days. They also have very simple food requirements. Chromosomes 1 (the X chromosome), 2, and 3 are very large, and the Y chromosome – number 4 – is extremely small. These four chromosomes have thousands of genes, many of which can be found in most eukaryotes, including humans.
Drosophila embryos develop in the egg membrane. The egg hatches and produces a larva that feeds by burrowing through the medium. The larval period consists of three stages, or instars, the end of each stage marked by a molt. Near the end of the larval period, the third instar will crawl up the side of the vial, attach themselves to a dry surface, and form a pupae. After a while the adults emerge.
Differences in body features help distinguish between male and female flies. Females are slightly larger and have a light-colored, pointed abdomen. The abdomen of males will be dark and blunt. The male flies also have dark bristles, sex combs, on the upper portion of the forelegs.

Hypothesis:
After performing a dihybrid cross between males with normal wings and sepia eyes and females with vestigial wings and red eyes, we expect to see only hybrids with normal wings and red eyes in the first filial generation. Then we expect to observe a 9:3:3:1 ratio of phenotypes in the second filial generation.

Materials and Methods:
The materials used for this lab were:  culture vial of dihybrid cross, isopropyl alcohol 10%, camel’s hair brush, thermo-anesthetizer, petri dish, 2 Drosophila vials and labels, Drosophila medium, fly morgue.

A vial of wild-type Drosophila was thermally immobilized and the flies were placed in a petri dish. Traits were observed. A vial of prepared Drosophila was immobilized and then observed under a dissecting microscope. Males and females were separated and mutations were observed and recorded. The parental generation was placed in the morgue. The vial was placed in an incubator to allow the F1 generation to mature.
The F1 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded. Five mating pairs of the F1 generation were placed into a fresh culture vial, and the vial was placed in an incubator. The remaining F1 flies were placed in the morgue. The F1 flies were left in the vial for about a week to mate and lay eggs. Then the adults were removed and placed in the morgue. The vial was placed back in the incubator to allow the F2 generation to mature. The F2 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded.

Results:  

Table 1 Phenotypes of the Parental Generation

Phenotypes Number of Males Number of Females
Normal wings/red eyes 0 0
Normal wings/sepia eyes 3 0
vestigial wings/red eyes 0 4
vestigial wings/sepia eyes 0 0

Table 2  Phenotypes of the F1 Generation

Phenotype Number of Males Number of Females
Normal wings/red eyes 78 95
Normal wings/sepia eyes 0 0
vestigial wings/red eyes 0 0
vestigial wings/sepia eyes 0 0

Table 3  Phenotypes of the F2 Generation

Phenotypes Number of Males  Number of Females
Normal wings/red eyes 4 7
Normal wings/sepia eyes 4 5
vestigial wings/red eyes 0 1
vestigial wings/sepia eyes 0 0
normal red/mutated body shape 2 0
normal sepia/mutated body shape 1 0

Questions

  1. How are the alleles for genes on different chromosomes distributed to gametes? What genetic principle does this illustrate?
    The alleles on different chromosomes are distributed independently of one another, demonstrating Mendel’s Law of Independent Assortment.
  2. Why was it important to have virgin females for the first cross (yielding the F1 generation), but not the second cross (yielding the F2 generation)?
    It was important to have virgin females for the first cross to ensure that the offspring are the result of the desired cross. It was not necessary to isolate virgin females for the second cross because the only male flies to which they had been exposed were also members of the F1 generation.
  3. What did the chi-square test tell you about the validity of your experiment data? What is the importance of such a test?
    The chi-square test showed that the results of our first cross were valid, but that the results of our F1 cross were not normal. It is important to conduct such a test to determine how much your experimental data deviated from what was expected.

Discussion and Conclusion:
The results of our parental cross turned out just as expected, but our F2 generation was not normal. Some sort of mutation must have occurred that caused the strange body shape seen in several individuals of our F2 generation.

Enzyme Catalysis

 

Enzyme Catalysis

Introduction:
In general, enzymes are proteins produced by living cells, they act as catalysts in biochemical reactions. A catalyst affects the rate of a chemical reaction. One consequence of enzyme activity is that cells can carry out complex chemical activities at relative low temperatures. In an enzyme-catalyzed reaction, the substance to be acted upon ( the substrate = S ) binds reversibly to the active site of the enzyme (E). One result of this temporary union is a reduction in the energy required to activate the reaction of the substrate molecule so that the products (P) of the reaction are formed.

In summary:     E + S —> ES –> E + P

Note that the enzyme is not changed in the reaction and can even be recycled to break down additional substrate molecules. Each enzyme is specific for a particular reaction because its amino acid sequence is unique and causes it top have a unique three-dimensional structure. The active site is the portion of the enzyme that interacts with the substrate, so that any substance that blocks or changes the shape of the active site affects the activity of the enzyme. A description of several ways enzyme action may be affected follows:

1. Salt Concentration. If the salt concentration is close to zero, the charged amino acid side chains of the enzyme molecules will attract to each other. The enzyme will denature and form an inactive precipitate. If, on the other hand, the salt concentration is too high, normal interaction of charged groups will be blocked, new interactions will occur, and again the enzyme will precipitate. An intermediate salt concentration such as that of human blood (0.9% ) or cytoplasm ins the optimum for many enzymes.

2. pH. Amino acid side chains contain groups such as – COOH and NH2 that readily gain or lose H+ ions. As the pH is lowered an enzyme will tend to gain H+ ions, and eventually enough side chains will be affected so the enzyme’s shape is disrupted. Likewise, as the pH is raised, the enzymes will lose H+ ions and eventually lose its active shape. Many of the enzymes function properly in the neutral pH range and are denatured at either an extremely high or low pH. Some enzymes, such as pepsin, which acts in the human stomach where the pH is very low, have a low pH optimum.

3. Temperature. Generally, chemical reactions speed up as the temperature is raised. As the temperature increases, more of the reacting molecules have enough kinetic energy to undergo the reaction. Since enzymes are catalysts for chemical reactions, enzyme reactions also tend to go faster with increase temperature. However, if the temperature of an enzyme-catalyzed reaction is raised still further, a temperature optimum is reached; above this value the kinetic energy of the enzyme and water molecules is so great that the conformation of the enzyme molecules is disrupted. The positive effect of speeding up the reaction is now more than offset by the negative effect of changing the conformation of more and more enzyme molecules. Many proteins are denatured by temperatures around 40-50 degrees C, but some are still active at 70-80 degrees C, and a few even withstand boiling.

4. Activation’s and Inhibitors. Many molecules other than the substrate may interact with an enzyme. If such a molecule increases the rate of the reaction it is an activator, or if it decreases the reaction rate it is an inhibitor. These molecules can regulate how fast the enzymes acts. Any substance that tends to unfold the enzyme, such as an organic solvent or detergent, will act as an inhibitor. Some inhibitors act by reducing the -S-S- bridges that stabilize the enzyme’s structure. Many inhibitors act by reacting with the side chains in or near the active site to change its shape or block it. Many well known poisons such as potassium-cyanide and curare are enzyme inhibitors that interfere with the active site of critical enzymes.

The enzyme used in this lab, catalase, has four polypeptide chains, each composed of more than 500 amino acids. This enzyme is ubiquitous in aerobic organisms. One function of catalase within cells is to prevent the accumulation of toxic levels of hydrogen peroxide formed as a by-product of metabolic processes. Catalase might also take part in some of the many oxidation reactions that occur in the cell.

2H2O ——-> 2 H2O + O2 (gas )

In the absence of catalase, this reaction occurs spontaneously, but very slowly. Catalase speeds up the reaction considerably. In this experiment, a rate for this reaction will be determined. Much can be learned about enzymes by studying the kinetics of enzyme-catalyzed reactions. For example, it is possible to measure the amount of product formed, or the amount of substrate used, from the moment the reactants are brought together until the reaction has stopped. If the amount of product formed is measured at regular intervals and this quantity is plotted on a graph, a curve like the one that follows is obtained.

Figure 2.1 Enzyme Activity

Study the solid line of the graph of this reaction. At time 0 there is no product. As time progresses the production of product increases at a steady rate. After a period of time this rate slows down and at a certain point the reaction rate is very slow.

General Procedure:
In this experiment the disappearance of the substrate, H2O2, is measured as follows:

1. A purified catalase extract is mixed with substrate ( H2O2) in a beaker. The enzyme catalyzes the conversion of H2O to H2O and O2 (gas ).

2. Before all the H2O2 is converted to H2O and O2 , the reaction is stopped by adding sulfuric acid ( H2SO4 ). The sulfuric acid lowers the pH, denatures the enzyme, and thereby stops the enzyme’s catalytic activity.

3. After the reaction is stopped, the amount of substrate (H2O2) remaining in the beaker is measured. To measure this quantity, potassium permanganate is used. Potassium permanganate (KMnO4), in the presence of H2O2 and H2SO4 reacts as follows:

5 H2O2 + 2 KMnO4 + 3 H2SO4 ————–> K2SO4 + 2 MnSO4 + 8 H2O + 5 O2

Note that H2O2 is a reactant for this reaction. Once all the H2O2 has reacted, any more KMnO4 added will be in excess and will not be decomposed. The addition of excess KMnO4 causes the solution to have a permanent pink or brown color. Therefore, the amount of H2O2 remaining is determined by adding KMnO4, until the whole mixture stays a faint pink or brown, permanently. Add no more KMnO4 after this point.

Figure 2.2 The General Procedure for the above exercise and Exercise 2C.
The figure below represents the complete Exercise 2C.

Exercise 2A: Test of Catalase Activity:
1. To observe the reaction to be studied, transfer 10 mL of 1.5% (0.44M) H2O2 into a 50 ml glass beaker and add 1 mL of freshly made catalase solution. The bubbles coming from the reaction mixture are oxygen, which results from the breakdown of H2O2 by catalase. Be sure to keep the freshly made catalase solution on ice at all times.

a. what is the enzyme in this reaction? ____________________________________________________

b. What is the substrate in this reaction? ___________________________________________________

c. What is the product in this reaction? ____________________________________________________

d. How could you show that the gas evolved is oxygen ? _____________________________________

2. To demonstrate the effect of boiling on enzymatic activity, transfer 5 mL of purified catalase extract to a test tube and place it in a boiling water bath for five minutes. Transfer 10 mL of 1.5% H2O2 into a 50 mL glass beaker and add 1 mL of the cooked, boiled catalase solution. How does the reaction compare to the one using the unboiled catalase? Explain the reason for this difference.

_____________________________________________________________________

_____________________________________________________________________

3. To demonstrate the presence of catalase in living tissue, cut 1 cm3 of potato or liver, macerate it, and transfer it into a 50 mL beaker containing 1.5% H2O2 . What do you observe? What do you think would happen if the potato or liver was boiled before being added to the H2O2?

_____________________________________________________________________

_____________________________________________________________________

Exercise 2B: The Baseline Assay:
To determine the amount of H2O2 initially present in a 1.5% solution, one needs to perform all the steps of the procedure without adding catalase to the reaction mixture. This amount is known as the baseline and is an index of the initial concentration of H2O2 un solution. In any series of experiments, a baseline should be established first.

Procedure for Establishing Baseline:
1. Put 10 mL of 1.5% H2O2 into a clean glass beaker.

2. Add 1 mL of H2O ( instead of enzyme solution).

3. Add 10 mL of H2SO4 (1.0 M) Use Extreme care in Handling Acids.

4. Mix well.

5. Remove a 5 mL sample. Place this 5 mL sample in another beaker, and assay for the amount of H2O2 as follows: Place the beaker containing the sample over white paper. Use a burette or 5 mL pipette to add potassium permanganate a drop at a time to the solution until a persistent pink or brown color is obtained. Remember to gently swirl the solution after adding each drop. Record your data below.

Baseline calculations

Final Reading of burette ________ mL

Initial reading of burette ________mL

Baseline (Final -Initial) _________mL KMnO4

Figure 2.4: Proper Reading of a Burette

 

 

Exercise 2C: An Enzyme-Catalyzed Rate of H2O2 Decomposition

Refer to figure 2.2 to complete this section and record the data in Table 2.1 below.

Table 2.1

Potassium Permanganate (ml)

Time (Seconds)

10 30 60 120 180 360
A. Baseline
B. Final Reading
C. Initial Reading
D. Amount of KMnO4 consumed (B-C)
E. Amount of H2O2 used (A-D)

Graph the data for enzyme-catalyzed H2O2 decomposition.

Graph Title: ___________________________________________________________________

 

Graph 2.1

Analysis of Results:
1. Explain the inhibiting effect of sulfuric acid on the function of catalase. Relate this to enzyme structure and chemistry.

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

2. Predict the effect lowering the temperature would have on the rate of enzyme activity. Explain you prediction.

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

3. Design a controlled experiment to test the effect of varying pH, temperature, or enzyme concentration.

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

AP LAB PAGE

 

Enzyme PowerPoint Worksheet

Enzymes
ppt Questions

Enzyme Structure & Function

1. Most enzymes are what type of macromolecule?

2. Most enzymes are ______________ or ______________ structures.

3. Enzymes act as ___________ in reactions.

4. Are enzymes permanently changed in the chemical reactions they are involved in?

5. Will an enzyme work on any substance? Explain.

 

6. Can enzymes be reused?

7. What ending is found on many enzymes?

8. Give 3 examples of enzymes with this ending.

 

9. How does an enzyme work?

 

10. What effect does an enzyme have on activation energy needed to start a reaction?

11. Hydrogen peroxide H2O2 is a common waste product of cells. Enzymes called catalases in cells break this down into harmless ________________.

12. What is meant by the term substrate?

 

13. What is meant by active site?

 

14. Sketch and label the enzyme-substrate complex.

 

 

15. What is meant by induced fit?

 

16. What induces an enzyme to change the shape of its active site?

17. List 4 factors that can affect enzyme activity.

 

18. What is the effect of high temperature on an enzyme (running fever)?

 

19. What temperature do most enzymes do best at?

20. Most enzymes like a pH near ______________.

21. To denature an enzyme means the enzyme becomes _______________ and can no longer work properly.

22. Name 3 inorganic substances (cofactors) that are often needed for enzymes to work properly.

 

23. Give an example of an enzyme & its needed inorganic substance.

 

24. Give one example of an enzyme inhibitor.

25. Explain how competitive inhibitors work.

 

 

26. If a competitive inhibitor blocks the active site, the ____________ can’t fit.

27. Explain noncompetitive inhibitors. 

 

 

28. Do noncompetitive inhibitors bind to the active site? Explain.