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Gre4nikov [31]
3 years ago
13

A cell biologist found that two different proteins with different structures were translated from two different mRNAs. These mRN

As, however, were transcribed from the same gene in the cell nucleus. Which mechanism below could best account for this?
Biology
1 answer:
Ivanshal [37]3 years ago
3 0

Answer:

Alternative splicing

Explanation:

One gene can lead to multiple proteins by the alternative splicing of the mRNA. The alternative splicing is the most common process that contributes to protein diversity at a pot-transcriptional level. This process is carried out by different combinations of including or excluding exons of the mRNA, obtaining proteins that differ in their amino acids sequence, consequently having different biological functions.

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Which one of the following is an example of epithelial tissue?
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B. subcutaneous skin layer

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Match each biodiversity restoration method to its description.
timofeeve [1]

Match each biodiversity restoration method to its description we have:

  • reforestation: using plants to absorb harmful compounds
  • biological augmentation: using plants to control a native plant population
  • bioremediation:using plants to increase biodiversity and food resources

<h3>What are ecological restoration techniques?</h3>

Some examples of induced ecological restoration methodologies are the conduction of natural regeneration, nucleation techniques, enrichment or diversity planting, among others.

In this case, the ecological restoration techniques are:

  • reforestation: using plants to absorb harmful compounds
  • biological augmentation: using plants to control a native plant population
  • bioremediation: using plants to increase biodiversity and food resources

See more about ecological restoration at brainly.com/question/1331136

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6 0
1 year ago
Will Give Brainliest:
Akimi4 [234]

Answer:

C

Explanation:

From the knowledge I've obtained, that is the answer I feel is best but I am not 100% certain.

7 0
3 years ago
In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The gene
NikAS [45]

Answer:

A) 9:3:3:1

B) All bitter fruit, yellow spotted offsprings

C) Phenotypes are bitter yellow spotted (4), bitter no spot (4), sweet yellow spot (4), and sweet no spot (4). 1:1:1:1

Explanation:

This is a typical dihybrid cross involving two genes, one coding for fruit taste and the other for spot color. The allele for bitter taste (B) and yellow spot (S) is dominant over the allele for sweet taste (b) and no spot (s) respectively.

Hence, a heterozygous F1 resulting from a cross between an homozygous dominant (bitter fruit, yellow spot) and homozygous recessive (sweet fruit, no spot) will have a BbSs genotype. The heterozygous F1 offsprings are self-crossed and produce gametes BS, Bs, bS, bs. (See punnet square). The F2 offsprings will have the following phenotypes: Bitter fruit, yellow spot (9)

Bitter fruit, no spot (3)

Sweet fruit, yellow spot (3)

Sweet fruit, no spot (1)

Back cross between a F1 offspring (BbSs) and homozygous dominant parent (BBSS) will produce all bitter fruit, yellow spot offsprings (see attached image). BBSS (4), BBSs (4), BbSS (4), and BbSs (4) are the offsprings' genotypes.

For the back cross between a F1 offspring (BbSs) and a homozygous recessive (bbss) parent, the Phenotypes with their proportions are as follows:

Bitter fruit, yellow spot (BbSs, 4)

Bitter fruit, no spot (Bbss, 4)

Sweet fruit, yellow spot (bbSs, 4)

Sweet fruit, no spot (bbss, 4).

4 0
3 years ago
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