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Fed [463]
1 year ago
8

16 - Albinism: Prom Genotype to Phenotype

Biology
1 answer:
Mrac [35]1 year ago
6 0

A correct flow chart is DNA i.e., gene (transcription) >> mRNA + tRNAs (translation) >> amino acids or protein >> phenotype.  It is the flow of the genetic information.

<h3>What is the flow of genetic information?</h3>

The flow of genetic information refers to the molecular process from DNA to the final phenotype (i.e., protein product).

During gene transcription, a particular fragment of DNA called gene is used as a template to generate an mRNA.

Subsequently, the mRNA serves as a template to generate a linear chain of amino acids, i.e., a protein, by a process called translation.

The final protein can be considered as the phenotype (in this case, the albinism trait) because it is the observable feature.

Learn more about genetics here:

brainly.com/question/1480756

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Example of active immunity
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Example of natural active immunity is fighting off cold and artifical active immunity is building up a resistance to a disease due to immunization and allergic reaction is an response to antigen resulting from active immunity

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Which of the following is an example of the endocrine system maintaining homeostasis?
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Answer:

The correct answer is "C".

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The endocrine system is an organ in the body which is in charge of secreting hormones that regulate numerous functions in the body, including metabolism, growth, sexual function, sleep, and mood among others. It is an essential organ which contributes to the state of homeostasis in the body.

Explanation:

4 0
2 years ago
You discover a population of mustard plants and find it to be at Hardy–Weinberg equilibrium with respect to the R locus. Suppose
natulia [17]

Answer:

The fraction of heterozygous individuals in the population is 32/100 that equals 0.32 which is the genotipic proportion for these endividuals.

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example, the r-6 allelic frequency is 0,2. This means that if r-6=0.2, then the other allele frequency (R) is=0.8, and the sum of both the allelic frequencies equals one. This is:

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0.2 + 0.8 = 1

Then, the genotypic proportion for the homozygous individuals RR is 0.8 ² = 0.64

The genotypic proportion for the homozygous individuals r-6r-6 is 0.2² = 0.04

And the genotypic proportion for heterozygous individuals Rr-6 is 2xRxr-6 = 2 x 0.8 x 0.2 = 0.32

5 0
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