Answer:
4. B cells become activated by interacting with helper T cells.
1. B cells display antigens in MHC class II receptors on the cell surface.
2. Antibodies released by plasma cells bind to the antigen so they will be destroyed by other cells of the immune system.
3.B cells rearrange their DNA to create a unique B-cell receptor.
5. B cells undergo clonal expansion.
6. B cells digest antigens that bind to the antibodies on their surface.
Explanation:
B-cells get activated by interacting with helper T cells when they bind to the antigen to receptors i.e (MHC class II receptors on the cell surface) on the surface of the cell. Series of activities such as release by plasma cells which cause rearrangement of B cells causes the cell to divide and proliferate. The process through which daughter cells arise from a parent cell called clonal expansion.
From the protein hydrolysis cycle, the missing words are;
- Amino Acid
- Deamination
- Non-essential Amino Acids
- Free fatty acids and monoglycerides
- acetyl-CoA molecules
- FADH2, NADH and water.
<h3>What is protein hydrolysis?</h3>
Protein hydrolysis is a complicated process that comprises a series of enzymatic processes, each with a unique kinetic response for numerous substrates.
Two enzymes that aid the digestion of protein in the body are;
Learn more about protein hydrolysis at;
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Chromosome mutations can result in changes in the number of chromosomes in a cell or changes in the structure of a chromosome. Unlike a gene mutation which alters a single gene or larger segment of DNA on a chromosome, chromosome mutations change and impact the entire chromosome. Key Takeaways: Chromosome Mutations
Answer:
heat 100ml and 300 ml of water and record the temperature when boiling occurs
Answer:
The correct answer would be C. 0.60
It can be calculated with the help of Hardy-Weinberg equation i.e
p² + q² + 2pq = 1
Also, p + q = 1
- where p represents the frequency of dominant allele.
- q represents the frequency of recessive allele.
- p² represents the frequency of homozygous dominant genotype
.
- q² represents the frequency of homozygous recessive genotype
.
- 2pq represents the frequency of heterozygous genotype.
Number of homozygous dominant (YY) = 36
Number of homozygous recessive (yy) = 16
Number of heterozygous dominant (Yy) = 48
Total number of plants = 100
Now, frequency of homozygous dominant (YY) = 36/100 = 0.36
Therefore, p² = 0.36
So, p = √0.36 = 0.60
Hence, frequency of dominant allele (Y) = 0.60
