The Specific Immune Response

Vibrio cholera and Salmonella are two types of fatal bacteria which cause serious intestinal infections followed by death if untreated. If they are submitted to high temperatures, the bacteria die but their antigens are preserved.

Observe well the below experiments.

Image (23)

Note that:

– The variable used here is the type of the antigen. (Different bacteria).

– Ten Days duration was enough for the chicken to develop an effective immune response.

Interpret them, what do you deduce?

In the first experiment, the injection of KILLED Salmonella into a chicken followed by the injection of the same fatal LIVE bacteria after 10 days, leads to the survival of the animal. Likewise, in the second experiment, the injection of KILLED Cholera into another chicken followed by the injection of the same fatal bacteria but LIVE into the same chicken (after the same period of time as in the first experiment 10 days) leads to the survival of the animal. This signifies that in both cases, the first exposure to the  killed bacteria  immunized the animal against the fatal microbes.

In the third experiment, the injection of killed Salmonella into a chicken followed by the injection of a DIFFERENT type of living bacteria (Cholera) into the same chicken 10 days later could not prevent the death of the animal. Likewise, in the fourth experiment, the injection of killed Cholera into another chicken followed by the injection of a DIFFERENT  living bacteria (Salmonella) into the same chicken 10 days later could not prevent the death of the animal too. This signifies that the first exposure to the killed bacteria could immunize the animal only against the same type of bacteria and not against a different type of bacteria.

We deduce that the immune system is effectively specific against the antigen to which it was already exposed to, in the inactive form.

Be specific Smile and live happily ever after Smile

Yusser CHIDIAC

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Role and Components of the Immune Response

The questions below will help you assess your knowledge about the basics of the Immune System… ENJOY finding the answers!!

Note: Two of the following items are not questions, they are hints to be kept in mind  Smile

1- Describe the characteristics of the genes, alleles and molecule types of the MHC or HLA molecules .

2- Define: Autograft, Isograft, Allograft and Xenograft.

3- Explain how the body decides to accept a graft (vascularization appears around it) or to reject it (redness-edema-black-dry).

4- State the types and location of Agglutinins  and the Agglutinogens found in each blood group of the human ABO system.

48%20ABObloodsystem

5- Define agglutination and specify its importance in blood grouping.

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6- Note that for a blood transfusion to be successful or compatible, the recipient’s antibodies should not agglutinate with the donor’s antigens.

230px-Blood_Compatibility_svg

7- Compare the ABO system and the Rhesus System.

blood1

8- Is it possible to determine the blood group of a person by identifying his serum agglutinins? Justify.

9- Define antigen and list its different types.

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10-Define pathogenic agents and give some examples.

11-Realize that the immune system can not differentiate between harmful and harmless substances; it only recognizes and fights the non self in healthy individuals.

12-Describe the formation of a modified self which is then attacked by the body’s immune system.

13-Differentiate between the different leukocytes populations and state their characteristics and role.

Blood-cells

14-Name the primary lymphoid organs and the secondary lymphoid organs, then state the role of each.

immu

15-Describe the steps of the maturation of (B and T) lymphocytes, and then state its importance.

16-State the structural and functional characteristics of Immunoglobulins(antibodies).

09%20Antibody%20Structure

17-What is the significance of the constant and variable regions of an antibody?

18-Define antigen determinant or epitope, cross reaction, Immune complex and agglutination.

19- Explain how a B lymphocyte, TH lymphocyte, and TC lymphocyte recognize antigens or antigenic peptide.

Microbiology cartoon2

Best Healthy Regards,

Yusser CHIDIAC Smile

The 3 Simple Rules to Determine the Amino-acid Chain from a DNA Piece Without Errors!

dna_unraveled

RULE ONE:Observe carefully the double stranded DNA given piece. Specify the transcribed strand ( template) and the non-transcribed strand. Write them  on your copybook specifying the name of each. Try to separate every 3 letters by a dash ( because every 3 nucleotides represent one codon and will code for ONE amino acid only).

Transcribed/template DNA strand:

TAC – TGC- CTA- GTC- GGC- GTT- CGC- CTT- AAC – CGC TGT-ATT.

The Non- transcribed DNA Strand:

ATG–ACG–GAT- CAG–CCG–CAA–GCG–GAA–TTG–GCG–ACA-TAA.

RULE TWO: Write the m-RNA in a complementary way to the transcribed DNA strand. (A of the m-RNA  binding with T of the DNA ; U of m-RNA binding with A  of  DNA).

the m-RNA :

AUG–ACG–GAU– CAG–CCG–CAA–GCG–GAA–UUG–GCG–ACA-UAA.

Note: to verify your m-RNA correct sequence, just compare it to the DNA non-transcribed strand, it should have the same sequence but instead of T & m-RNA will have a U.

The non-transcribed DNA strand:

ATG-ACG- GAT– CAG- CCG- CAA- GCG–GAA–TTG–GCG–ACA- TAA.

RULE THREE: Use the genetic code to determine the correct amino-acid sequence in the chain coded by the m-RNA .

amino-acid-table-singlet-code

This genetic code is like a dictionary translating two languages : that of  m-RNA to that of amino-acids. So what you see on your left is the code of the m-RNA ( three letters) and on your right is the symbolic name of the amino-acid. All you have to do is to follow the positions of the letters ( codons of RNA) on the table and find its match of the amino-acid.

AUG–ACG–GAU- CAG–CCG–CAA–GCG–GAA–UUG–GCG–ACA-UAA.

Met – Thr – Asp- Gln –  Pro-   Gln – Ala – Glu – Leu – Ala – Thr – .

Note: Usually the synthesized amino-acid chain starts with Methionine (Met) and the last codon in the m-RNA is a stop codon that does not code to any amino-acids. (The number of amino acids is usually less than the number of the codons of the m-RNA by one )

And here you are, the amino-acid sequence completed!!

Best Regards,

Yusser CHIDIAC Smile

Independent Genes with a LETHAL Allele!

Alleles that cause an organism to die only when present in homozygous condition are called lethal alleles.

lethal alleles

The following exercise will help you know how to identify a lethal allele in Genetics’ problems.

Exercise:

The cross between two hybrids of F1 of genotypes AaBb gives the following phenotypes:

36 [AB], 12[Ab], 19[aB] and 6[ab].

1- Calculate the proportions of the descendants.

2- Knowing that the genes are independent, how can you explain such a result?

3- Make the factorial analysis verifying the above phenotypic proportions.

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Answers:

1- Proportion Calculation:

  • Divide by the smallest number:

[AB] 36/6=6

[Ab] 12/6= 2

[aB] 18/6= 3

[ab] 6/6= 1

  • Find out the total: Total = 6+2+3+1=12
  • Write the proportions:

6/12 [AB]

2/12 [Ab]

3/12 [aB]

1/12 [ab]

2- Explanation of the Results:

The genes are independent, and this is a selfing between individuals of F1, this means that the resulting theoretical proportions of the phenotypes of F2 should be 9/16[AB]; 3/16[Ab]; 3/16 [aB]; 1/16[ab], the total of the genotypes will be 16.

        Theoretical Results Real Results

9/16[AB]

3/16[Ab]

3/16 [aB]

1/16[ab]

6/12 [AB]

2/12 [Ab]

3/12 [aB]

1/12 [ab]

Comparing the Theoretical to the  Real results, we got a change (less) proportions for the first 2 phenotypes, this means that a lethal allele caused the death of 4 out of the theoretical 16 genotypes (16-12= 4) in the first 2 phenotypes: [AB] and [Ab] phenotypes.

3- Factorial Analysis:

Genotypes of F1*F1: AaBb * AaBb

Gametes: ¼ AB, ¼ Ab, ¼ aB, ¼ ab for each parent

Table of cross:

¼AB ¼Ab ¼aB ¼ab
¼AB 1/16AABB 1/16AABb 1/16AaBB 1/16AaBb
¼Ab 1/16AABb 1/16AAbb 1/16AaBb 1/16Aabb
¼ aB 1/16AaBB 1/16AaBb 1/16aaBB 1/16aaBb
¼ab 1/16AaBb 1/16Aabb 1/16aaBb 1/16aabb

-The genotypes that will enable us to find out the lethal allele are:

We look at the phenotypes that have a change in the real proportions when compared to the theoretical ones:

  • 1/16AAbb, 2/16Aabb: all have [Ab] and in the real phenotypes there are 3-2=1 missing (dead) so the missing should be AAbb because if the dead was of Aabb, we would have been missing 2 not 1!
  • 1/16AABB, 2/16AaBB, 4/16 AaBb, 2/16 AABb: all have [AB] and in the real phenotypes there are 9-6=3 missing (dead) so with comparison to the above results, the missing genotypes should carry AA (1AABB, 2AABb).

So if we consider that the lethal allele is A in the homozygous state, the given results will be verified.

Best Regards,

Yusser CHIDIAC Smile

Golden Paragraph and Statements in Genetics

Golden Paragraph

In genetics’ problems, you will often be given the results of a backcross and sometimes the results of a selfing F1*F1 to find out information about the localization of the considered genes.the following golden paragraph will help you in this mission!

The variety and the proportions of the results of the test cross reflect the gametic variety and proportions of the female F1 since the male contribute with 100% recessive alleles.

  • The given results do not show 2 different phenotypes this means that the studied genes are not absolutely linkedon the same chromosome.
  • The given results do not show 4 different genotypes of equal proportions which mean that there is no independent disjunction of the studied genes; the genes are not found on independent chromosomes.
  • The given results of the test cross show 4 different phenotypes with 2 high proportions and 2 low proportions which means that there is partial linkage or (a linkage followed by a crossing over) which is produced during the formation of gametes of the hybrid female F1.
  • And since we have 2 dominant alleles are found together in the phenotypes of high proportions and the 2 recessive alleles are also together in the high proportions this means the genes are found in the Cis position.
  • And since we have 1 dominant allele and one recessive allele found together in the phenotypes of high proportions this means the genes are found in the Trans position.”

parentss

Golden Statements

The following statements are very helpful in justifying a specific fact in genetics’ problems.

Statement 1: “The recessive trait is not expressed unless in the homozygous state.”

Statement 2: “ The normal couple gave birth to a diseased child, this means that the allele of the disease is hidden in the genotypes of parents , masked by the normal allele. Thus, the allele of the disease is recessive and the normal allele is dominant.”

Statement 3: “This is a case of intermediate inheritance since two pure flowers , one red and one white gave rise to a pink flower, a new phenotype that is an intermediate color between red and white colors, this means that both alleles responsible for red and white colors are dominant and expressed incompletely.”

Statement 4: “ Both parents are of blood group A gave birth to a child of blood group O. The child of blood group O having for genotype oo ( since allele o is recessive with respect to allele “A”) so he should have taken 1 allele “o” from his mother and 1 allele “o” from his father, so the parents should have allele “o” in their genotypes. And since they are of blood group A, each will also have an allele “A” in his/her genotype. Thus, the parents are hybrid for the blood group genotype Ao .”

to be continued……

Best Regards,

Yusser CHIDIAC Smile

Types of Assortments / Recombination in Chromosomes

There are two types of Assortments of chromosomes that take place in Meiosis during the formation of gametes: Interchromosomic Assortment and Intrachromosomic Assortment.

The Interchromosomic Assortment :

This type of assortment takes place during the anaphase I of the reductional Meiosis.

Becker_JPEG_ch20_Builder

As you can see in the above document, there is an independent segregation of homologous chromosomes. Each chromosome was free to go in any daughter cell. The above document shows only one possibility, while another possibility can take place where the 2 red chromosomes go to one daughter cell while the 2 blue chromosomes go to the other daughter cell.

The Intrachromosomic Assortment:

In this type of assortment, the homologous chromosomes exchange segments of their sister chromatids. It takes place in prophase I of the reductional meiosis.

crossingover rec

After this exchange, the sister chromatids of the same chromosome may carry different alleles. After the completion of meiosis and when the gametes are formed,  we will have 2 types of gametes:

  • Parental gametes that will have the unchanged chromatids ( in this case, a gamete having AB and  another having ab).
  • Recombinant gametes which will have the recombinant , changed chromatids ( in this case, a gamete having Ab and another having aB).

Note that the probability of a crossing over to take place is usually low, so the percentages of the recombinant gametes will be much lower than those of the parental gametes.

Best Regards,

Yusser CHIDIAC Smile

The Results of Backcross and Selfing F1*F1

 

    Type of Cross

    Independent Genes

    Back cross

    25%[AB]                  25% [Ab]

    25% [aB]                     25% [ab]

    Selfing
    F1 * F1

    9/16 [AB]                3/16 [Ab]

    3/16 [aB]                  1/16 [ab]

Type of Cross

Partially Linked Genes

Partially Linked Genes

Back Cross

Trans position

[AB] Low proportions

[ab] Low proportions

[aB] High proportions

[Ab] High proportions

Cis position

[AB] High proportions

[ab] High proportions

[Ab] Low proportions

[aB] Low proportions

Selfing F1*F1

Trans position

50% [AB]

25% [aB]

25%[Ab]

Cis position

[AB] High proportions

[ab] medium proportions

[Ab] Low proportions

[aB] Low proportions

Type of Cross

Absolutely Linked Genes

Absolutely Linked Genes

Back Cross

Trans position

50% [Ab]

50% [aB]

Cis position

50%[AB]

50% [ab]

Selfing F1*F1

Trans position

50% [AB]

25% [aB]

25% [Ab]

Cis position

75%[AB]

25% [ab]