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MIMG 185 > Structure & Properties of Antibodies > Flashcards

Structure & Properties of Antibodies Flashcards

1
Q

What are the 2 types of adaptive immunity?

A

Cell-mediated: T cells, effector cells

Humora immunity: B cells, “ciruculating molecules”

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2
Q

What did Pfeiffer notice when he injected guinea pigs with vibrio cholera?

What was this protection mediated by?

A

He noticed specific protection: This guinea pig was protected against subsequent infections, but not against reltaed organisms.

The protection was mediated by humoral, circulating factors - Antibodies.

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3
Q

What are the hallmarks of Adaptive immunity?

A

Specificity.

Ability to recognize broad spectrum of pathogens.

(Both cell-mediated and humoral.)

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4
Q

What happens when bacteria are injected into organisms?

A

When bacteria are injected, specific substances are found in the blood serum that exhibit different biologic properties. Allows for discrimination beween similar but different organisms.

  1. agglutinins: specifically clump only the bacteria used for immunization.
  2. opsonins: facilitate engulfment (opsonize = phagocytosis; Antibodies facilitate ability to phagocytose.)
  3. antitoxins: neutralize toxins associated with the immunizing bacteria.
  4. cytolysis: lysis of bacteria, often through complement.
  5. precipitins: If organim makes protein that is secreted out, material present in immune serum will precipitate it.

All of these represent different manifestations of Antibody function.

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5
Q

What are Agglutinins?

A

It is one of the manifestations of Antibody function.

Agglutinins specifically clump only the bacteria used for immunization.

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6
Q

What are Opsonins?

A

It is one of the manifestations of Antibody function.

Opsonins facilitate engulfment.

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7
Q

What are Antitoxins?

A

It is one of the manifestations of Antibody function.

Antitoxins neutralize toxins associated with the immunizing bacteria.

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8
Q

What is Cytolysis?

A

It is one of the manifestations of Antibody function.

Cytolysis is lysis of bacteria.

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9
Q

What are Precipitins?

A

It is one of the manifestations of Antibody function.

Precipitins form flocculate precipitates from cell-free supernatants of the bacteria.

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10
Q

What is serum?

A

Serum is blood w/o RBCs.

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11
Q

What is antiserum?

A

Antiserum: circulating material in which the blood of an immune animal and other cells have been removed, leaving soluble material that contains Antibodies.

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12
Q

What is an Antibody?

A

An antibody is an immunoglobulin.

It is found as both cell-associated B-cell receptor (BCR) and as circulating effector molecules.

It is a molecule produced in response to foreign substances (such as Antigen) and can binds the eliciting agent.

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13
Q

What is an Antigen?

A

Any molecule capable of being bound by the combining site of an Antibody or the T cell Receptor (TCR).

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14
Q

What is an Immunogen?

A

A substance capable of eliciting an immune response; all immunogens are Antigens but not all Antigens are immunogens (Not all Antigens can elicit an immune response.)

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15
Q

What 7 experiments allowed the determination of the structure of an Antibody?

A
  1. Electrophoresis: Antibodies migrated as gamma globulins (proteins).
  2. Ultra centrifugation: Ran with a 7S sedimentation coefficient, which indicated a molecular weight of 150kDa.
  3. Precipitation: Each Antibody molecule has a valence of 2 - it can bind 2 molecules of Antigen.
  4. Papain: 2 Fab (“fragment Antigen binding”), and 1 Fc (“fragment crystallizable”). The molecular weight of Fab and Fc are 50kDa each.
  5. Pepsin: (Fab’)2, with a molecular weight of 100kDa. (Fab’)2 binds TWO molecule of Antigen and can precipitate. If the disulfide bonds of (Fab’)2 are broken, it gives 2 molecules of Fab.
  6. Beta-mercaptoethanol breaks disulfide bonds: 2 molecule of H chain (50kDa each), 2 molecules of Light chain (25kDa).
  7. Make Antiserum:
    a. Fab & Fc contain Heavy chain.
    b. Only Fab has the Light chain, Fc does not have Light chain.
    c. Fab has both Heavy and Light chains.
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16
Q

Describe the 1st experiment and its conclusion.

A
  1. Electrophoresis: Separated molecule by charge & size. He immunized rabbits with the protein ovalbumin (Antigen/Immunogen). The electrophoretic pattern of the serum of both hyperimmune animals and animals that they added Ovalbumin to showed precipitation of Antibodies.

CONCLUSION: The Antibodies that had been removed and present in hyperimmune animals were present in the gamma globulin peak. _So Antibodies were gamma globulins (proteins). _

17
Q

Describe the 2nd experiment and its conclusion.

A
  1. Ultra centrifugation: Spin at rapid rate to separate by size. The immune precipitate was taken from the immune serum, dissolved to produce purified Antibodies, and then analyzed in an ultracentrifuge.

CONCLUSION: It migrated with a peak. Ran with a 7S sedimentation coefficient, which indicated a molecular weight of 150kDa.

18
Q

Describe the 3rd experiment and its conclusion.

A
  1. Precipitation: 2 molecules of Antigen were precipitated by each Antibody.

CONCLUSION: So each Antibody molecule has a valence of 2 - it can bind 2 molecules of Antigen.

19
Q

Describe the 4th experiment and its conclusion.

A
  1. Papain: Treated the Antibodies with a protease, Papain. Papain can heal jellyfish bite by chewing up jellyfish toxin.

CONCLUSION: Papain separated the Antibody molecule into 3 fragments; 2 Fab (“fragment Antigen binding”), and 1 Fc (“fragment crystallizable”). Always has this ratio.

Fab binds Antigen but it does not precipitate.

The molecular weight of Fab and Fc are 50kDa each.

20
Q

Describe the 5th experiment and its conclusion.

A
  1. Pepsin: Treated the Antibody with a protease, Pepsin.

CONCLUSION: Digesting the Ab resulted in a molecule, (Fab’)2, with a molecular weight of 100kDa.

(Fab’)2 binds TWO molecules of Antigen and can precipitate (Fab cannot precipitate).

If the disulfide bonds of (Fab’)2 are broken, it gives 2 molecules of Fab.

21
Q

Describe the 6th experiment and its conclusion.

A
  1. Beta-mercaptoethanol breaks disulfide bonds:

CONCLUSION: If separated by size, there are two products - 2 molecule of H chain (50kDa each), 2 molecules of Light chain (25kDa).

22
Q

Describe the 7th experiment and its conclusion.

A
  1. Make Antiserum:

a. Inject a purified Heavy chain to make an antiserum/Antibody to Heavy chain. Anti-H chain will react with Fab & Fc.
CONCLUSION: Fab & Fc contain Heavy chain.

b. Make an Antibody to Light chain: Anti-L chain will react with Fab (but not Fc).
CONCLUSION: Only Fab has the Light chain, Fc does not have Light chain.

c. Make an Antibody to Fab: Anti-Fab will react with Heavy and Light chains.
CONCLUSION: Fab has both Heavy and Light chains.

23
Q

What is the structure of an Antibody?

A

An Antibody is a 4-chain protein. It is made of 2 Heavy chains and 2 Light chains that are held together by disulfide bonds.

This is consistent with the overall molecular weight of 150kDa, because Heavy chain is 50kDa and Light chain is 25kDa. (EXP 2 & 6)

Fab can bind Antigen but not precipitate. (EXP 4)

(Fab’)2 can bind 2 Antigen and precipitate. (EXP 5)

Consistent with Valence of 2. (EXP 3)

Consistent with Heavy chains and Light chains in their respective locations of Fab & Fc. (EXP 7, antiserum)

24
Q

What is a polymeric Immunoglobulin?

A

It is made up of more than 1 copy of the fundamental building block.

Polymeric Immunoglobulins involve IgA & IgM.

25
Q

In an electropheresis pattern, why is the peak of the homogeneous protein albumin narrow compared to the broad peak of the Immunoglobulin?

A

Because Antibodies are very heterogeneous - they can bind many different Antigens. When Antibodies are present in antiserum, they are heterogeneous.

The heterogeneity of Antibodies poses a problem for further understanding of the structure of an Antibody.

26
Q

What is Multiple Myeloma?

A

A tumor of plasma cells that generally secretes a single species of Immunoglobulin, otherwise known as homogeneous Antibodies or Myeloma Proteins.

The electrophoresis pattern of normal human serum shows polyclonal/heterogeneous Immunoglobulins, which represent the product of many different plasma cells (low, broad peak).

The electrophoresis pattern of Multiple Myeloma serum shows proliferation of a homogeneous Immunoglobulin or a single clone of cells (high, sharp peak).

It is a significant finding because it allowed for further understanding of the structure of Antibodies.

27
Q

What is Myeloma protein?

A

Homogeneus Antibodies that are produced by malignant plasma cells; plasma cells are Antibody-producing factories.

28
Q

What are Bence-Jones proteins?

A

Monoclonal/homogenous L chains in patients with Multiple Myeloma.

People with Multiple Myeloma make more L chains than H chains.

L chains are small, so they are filtered out by the kidneys. The urine from people with Multiple Myeloma can be filtered to isolate large quantities of homogeneous L chains for further analysis and understanding of the structure of Antibodies.

Bence-Jones proteins have an unusualy property - If heated until a certain temperature, they precipitate. But if continue heating, the proteins willl go back into solution (normal proteins never go back into solution).

29
Q

How do you explain the ability of Antibodies to bind many different Antigens? (Explain diversity of Antibodies)

A
  1. 2 categories of Light chains: kappa & lambda
30
Q

What are the categories of Light chains?

A

There are 2 categories of Light chains: kappa and lambda.

31
Q

How did they discover the two different types of Light chains?

A

Investigators used Bence-Jones proteins (homogeneous Light chains) from 2 different people with Multiple Myeloma (and therefore from different plasma cells) and produced proteolytic fragments from these different Light chains.

They analyzed the fragments by 2D peptide mapping, in which the fragments are separated by chromatography (solubility) in 1 dimension and electrophoresis (charge) in the other.

RESULT: They found that 1/2 of the cleavage products are identical between the 2 chains, while the other half differ. (Shaded spots indicate peptides shared by the 2 kappa or lambda chains.)

CONCLUSION: Light chains have a Variable region and a Constant region. All kappa Light chains had the same Constant region, while all lambda had the same Constant region (although kappa and lambda Constant region differ from each other.)

32
Q

How did they discover the location of the Variable and Constant region?

A

EXP: They used protein sequencing, which required large quantities of homogeneous protein, and sequenced both Light and Heavy chains.

CONCLUSION: The Variable region of both Light and Heavy chains were at the amino terminus (front) while the Constant region was at the carboxy terminus (tail).

33
Q

What did they discover about the Variable region?

A

The Variable regions had relatively conserved regions and hypervariable regions.

34
Q

How were the hypervariability regions determined?

A

An investigator proposed a Metric of Variability, in which the number of different amino acids in a particular position is divided by the frequency of the most common amino acid.

If there is only 1 amino acid present at the particular position, then the variability is 1 (100%, relatlively constant).

If all amino acids are the same frequency, then the variability is 400 (20 amino acids/5% = 400, very high variability = hypervariable region).

CONCLUSION: He was able to identify hypervariable regions and conserved regions of the Variable region of Light chains.

35
Q

What did he hypothesize about the Hypervariable regions?

A

He hypothesized that the Hypervariable regions were responsible for the different binding specificities.

These Hypervariable regions were called “Complementarity Determining Regions” (CDRs), and predicted that this part of the Antibody could contact the Antigen and determine the binding specificity.

36
Q
A

EXP: X-Ray Crystallography

The crystal structure of a Light chain shows that CDRs of the Variable region are present on loops at the end of proteins.

The conserved regions of Light chains are called “framework” amino acids because they keep the hypervariable loops/CDRs in the proper orientation.

MIMG 185 (6 decks)

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