Mechanisms of antibody variability
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Mechanisms of antibody variability
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[post_date] => 2024-12-23 18:35:01
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[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
The heavy chain and light chain subunits of antibodies both contain three regions called complementarity determining regions (CDR1, CDR2, and CDR3). The CDRs are responsible for the high specificity of the fragment antigen binding region (Fab region) of antibodies. CDRs are hypervariable among antibodies, allowing for an extensive range of antigens with which different antibodies can interact.
During B cell maturation, the gene for CDR3 of the heavy chain is formed when a random Vₕ segment, a random Dₕ segment, and a random Jₕ segment are joined together, as shown in Figure 1. To further randomize CDR3, DNA breaks are introduced at the junction sites, leading to nucleotide addition or subtraction before the breaks are sealed. A similar process occurs with only Vₗ and Jₗ segments to form CDR3 of the light chain.
Figure 1 Joining of Vₕ, Dₕ and Jₕ gene segments
Three different mechanisms cause random nucleotide alteration at the junction sites – these include P-addition, N-addition, and junctional flexibility:
- P-addition (palindromic-sequence addition) involves the creation of a DNA hairpin at the ends of gene segments; these hairpins are enzymatically cleaved, causing the unpairing of nucleotides to form a single strand. Repair enzymes then add complementary nucleotides, generating a new palindromic sequence.
- N-addition entails the insertion of a random number of nucleotides (“N” number of random nucleotides, up to 15) between two gene fragments.
- Lastly, in the phenomenon known as junctional flexibility, a random number of nucleotides may be lost on the adjoining end of either gene fragment. These random nucleotide alterations come with the risk of non-productive rearrangements, which occur when the altered nucleotide sequence disrupts the reading frame of the heavy chain or light chain gene, resulting in a non-functional antibody. Non-productive rearrangements cause the B cell to initiate cell death.
These processes are summarized in Figure 2.
Figure 2 P-addition, N-addition, and junctional flexibility
In addition to these nucleotide alteration mechanisms, another mechanism that increases Fab variability is somatic hypermutation, which enhances the diversity and specificity of antibodies produced in response to antigens. During somatic hypermutation, the DNA of the B cell's antibody genes undergoes random mutations, introducing point mutations in the variable region of the antibody. In a process known as affinity maturation, B cells with mutated antibodies that exhibit higher affinity for the target antigen receive survival signals and continue to proliferate, while those with lower affinity undergo apoptosis.
[post_title] => Mechanisms of antibody variability
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[question] => If there were 44 Vₕ gene segments, 27 Dₕ gene segments, and 6 Jₕ gene segments, how many potential SEQUENCES (x) can the antibody heavy chain have?
[value] => Array
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[answer] => 4
[description] => Reason for Correct Answer:
Because there are 44 Vₕ gene segments, 27 Dₕ gene segments, and 6 Jₕ gene segments, there are 44 * 27 * 6 different combinations of gene segments that can create an antibody heavy chain.
However, additional mechanisms besides V-D-J recombination can contribute to the sequence of the heavy chain. Think about the effects of N-addition, P-addition, and junctional flexibility.
Because additional changes can be made to the final antibody heavy chain sequence by N-addition, P-addition, and junctional flexibility, the total potential number of antibody heavy chain sequences is much GREATER than 44 * 27 * 6
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[answers] => Array
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[0] => Array
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[each_answer] => A. x = 44 * 27 * 6
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[each_answer] => B. x = 44 * 27 * 6 / 2
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[2] => Array
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[each_answer] => C. x < 44 * 27 * 6
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[each_answer] => D. x > 44 * 27 * 6
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[1] => Array
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[quiz_unique_key] => 1403770772
[question] => Which B-cell process(es) is expected to occur in a germinal center of a lymph node?
I. Affinity maturation
II. B-cell production
III. Somatic hypermutation
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[answer] => 3
[description] => Reason for Correct Answer:
B cell development starts in the bone marrow (BM) and continues in the spleen to final maturation.
The passage states that somatic hypermutation occurs in response to antigens.
B cell activation in response to antigens occurs in the secondary lymphoid organs (SLOs), such as the spleen and lymph nodes.
Somatic hypermutation is a part of B cell activation and provides small differences between B cells of the same lineage. Those that bind the antigen most tightly are selected for in the process of affinity maturation. These processes both occur in the germinal centers of lymph nodes.
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[answers] => Array
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[0] => Array
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[each_answer] => A. I only
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[1] => Array
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[each_answer] => B. II only
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[2] => Array
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[each_answer] => C. I and III only
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[3] => Array
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[each_answer] => D. II and III only
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[2] => Array
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[quiz_unique_key] => 1403770772
[question] => Antibodies against a bacterial strain were isolated from a patient immediately after infection and again four weeks after infection. How will the dissociation constant, Kd of the antibodies isolated after four weeks compare to the Kd of the antibodies isolated after one week?
[value] => Array
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[answer] => 2
[description] => Reason for Correct Answer:
The dissociation constant, Kd is found by the formula [Ab][Ag]/[AbAg], where [Ab] is the equilibrium concentration of unbound antibody, [Ag] is the equilibrium concentration of unbound antigen, and [AbAg] is the equilibrium concentration of antibody-antigen complexes.
According to the formula for Kd, a small Kd represents a high concentration of antibody-antigen complexes and therefore a strong binding affinity of the antibody.
After a B-cell becomes activated, somatic hypermutation alters the binding region of the antibody and affinity maturation selects for B-cells that have a high affinity for antigen.
A few weeks after infection, one would expect that the host’s antibodies would have developed a stronger binding affinity for antigen, and therefore a smaller Kd.
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[answers] => Array
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[0] => Array
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[each_answer] => A. Both Kd’s will be the same.
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[1] => Array
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[each_answer] => B. The Kd at four weeks will be smaller.
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[2] => Array
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[each_answer] => C. The Kd at four weeks will be slightly larger.
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[each_answer] => D. The Kd at four weeks will be much larger.
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[3] => Array
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[quiz_unique_key] => 1403770772
[question] => Which part of the antibody contains the CDR region coded for by VH, DH and JH gene segments?
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[answer] => 1
[description] => Reason for Correct Answer:
The VH, DH and JH gene code for part of the heavy chains, or larger chains, of the antibody.
The CDR regions formed by VH, DH and JH regions should also be part of the variable domains of the antibody, where the antibody actually binds to antigens.
Here is a diagram of the antibody’s heavy chains and light chains and their variable and constant regions.
You can see that the VH, DH and JH regions should be in the variable part of the heavy chain, labeled A in the question image.
)
[answers] => Array
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[0] => Array
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[each_answer] => A. Part A
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[1] => Array
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[each_answer] => B. Part B
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[2] => Array
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[each_answer] => C. Part C
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[3] => Array
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[each_answer] => D. Part D
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[quiz_unique_key] => 1403770772
[question] => What is a possible palindromic sequence created as a result of P-addition?
[value] => Array
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[answer] => 3
[description] => Reason for Correct Answer:
Figure 1 shows how P-addition results in palindromic sequences of DNA. DNA hairpins are created at the ends of gene segments and then cleaved.
Palindromic sequences are sequences of nucleic acids within double-stranded DNA and/or RNA that are the same when read from 5′ to 3′ on one strand and 5′ to 3′ on the complementary strand.
If the strand generated by P-addition is palindromic, then the complementary strand added by repair enzymes will also be palindromic, as they will read the same from 5’ to 3’. (Note that the passage also says that “Repair enzymes then add complementary nucleotides, generating a new palindromic sequence.”) So, you’re looking for a palindromic sequence in the answer choices.
The only palindromic sequence in the answer choices is Choice C. See how it reads the same from 5’ to 3’ on each complementary strand:
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[each_answer] => A. 5’-G-A-T-T-A-G-3’
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[1] => Array
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[each_answer] => B. 5’-C-T-A-A-T-C-3’
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[2] => Array
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[each_answer] => C. 5’-C-T-A-T-A-G-3’
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[each_answer] => D. 5’-G-A-T-T-T-C-3’
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[quiz_unique_key] => 1997864699
[question] => Which of the following enzymes is responsible for the addition of random nucleotides in the N-addition mechanism?
[value] => Array
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[answer] => 2
[description] => Reason for Correct Answer:
In N-addition, random nucleotides are inserted between the two gene fragments. This is achieved by adding random nucleotides to the blunt ends of each gene fragment, before the fragments are rejoined by DNA repair enzymes.
Because the reaction is the addition of nucleotides, this will be catalyzed by a DNA polymerase (therefore, you can eliminate Choices C and D).
A template-dependent DNA polymerase uses a DNA ‘guide’ strand (the template) to direct the synthesis of a complementary DNA strand (for example, during DNA replication or during the repair of the opened hairpin in P-addition).
However, in the case of N-addition, random nucleotides are added; therefore, no template is used. The enzyme that is responsible for DNA synthesis in this case is a template-independent DNA polymerase.
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[answers] => Array
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[0] => Array
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[each_answer] => A. A template-dependent DNA polymerase
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[each_answer] => B. A template-independent DNA polymerase
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[each_answer] => C. A DNA ligase
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[each_answer] => D. A DNA helicase
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[554845|1] => D
[554845|2] => C
[554845|3] => B
[554845|4] => A
[554845|5] => C
[554845|6] => B
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Figure 1
Figure 2
