Answer:
<u>1. type IV hypersensitivity: delayed-type hypersensitivity. </u>
<u>2. type I hypersensitivity: IgE </u>
<u>3. type II hypersensitivity: IgG </u>
<u>4. type III hypersensitivity: immune complexes </u>
Explanation:
Type IV hypersensitivity is cell mediated hypersensitivity rather than antibodies as in all other types of hypersensitivities. It is also known as delayed type hypersensitivity because it usually respond in 2-3 days via T-Lymphocytes rather than antibodies like IgE or IgG. So we can say that in option 5 is mismatched as it relates type IV to IgG antibodies.
Answer:
A) The pollen grains will be pure red.
Explanation:
Plants have alternation of generations, this means that there are two different stages in their life-cycle: a sexual haploid (n) phase and an asexual diploid phase (2n). These phases occur in different individuals, so there is an haploid plant called gametophyte that carries gametes and after fecundation, it will rise a diploid sporophyte (asexual).
In seed plants, the sporophyte is the plant that we normally see, and the gametophyte is reduced into an organ of the sporophyte. The male gametophyte is the pollen that is produced in the sporangium in anthers (parts of sporophyte). When a pollen grain fecundes a female gametophyte (egg), it will produce a diploid embryo or new sporophyte.
Therefore, if the researcher exposes pollen to both stains, these grain will stain red, because red stain identifies gametophyte tissue.
Uh the answer is eating food.
The answers would be:
Genotype Phenotype
Tt Tall stemmed
tt Short stemmed
Genotypic ratio : 2:2 or 1:1
Phenotypic ratio: 2:2 or 1:1
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<u>You can read on to see how this was done:</u>
Tall stems (T) are dominant to short stems (t).
First figure out the genotypes of the parents. We have a short-stemmed plant and a heterozygous long-stemmed plant cross.
For short stem to occur, you need 2 pairs of short alleles. So the first parent would have a genotype of tt.
Heterozygous long-stemmed means that the parent has one of each allele. So the genotype of the second parent would be, Tt.
Now we can make our Punnett Square.
tt x Tt
<u> t t </u>
<u>T | Tt | Tt</u>
<u>t | tt | tt</u>
Let's list down the genotypes and phenotypic results.
Genotype no. Phenotype
Tt 2 Tall stemmed
tt 2 Short stemmed
So from that we can answer the other questions:
Genotypic ratio : 2:2 or 1:1
Phenotypic ratio: 2:2 or 1:1
