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Kruka [31]
3 years ago
11

On a large volcanic island, researchers are studying a population of annual herbaceous plants. Which of the following observatio

ns best supports the prediction that speciation will occur within the existing plant population?(A) Individuals of the species sometimes reproduce asexually by producing runners. (B) Lava has separated the population into two areas: an upland forest and a lowland marsh. (C) Multiple groups of birds depend on the fruit produced by the plants as a source of food. (D) The plants produce more seeds during warm summers than they do during cool summers.
Biology
1 answer:
tangare [24]3 years ago
4 0

Answer:

(B) Lava has separated the population into two areas: an upland forest and a lowland marsh.

Explanation:

In allopatric speciation, individuals of an ancestral species may undergo evolution into two or more progeny species which arises when physical separation causes a geographic boundary, this barrier can be in the form of a mountain range, rockslide, river or lava.

As in this scenario, a lava flow divides the populations of annual herbaceous plants by altering the landscape. This will cause genetic divergence with varying heritable features among its many generations favouring different traits in the separate environment along with genetic drift. The Lava will separates the population into two areas: an upland forest and a lowland marsh.

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Answer:

Normal Strand: alanine - methionine - histidine
Mutated Strand: glutamine - cysteine - no third amino acid.

Explanation:

<h3>mRNA Structure</h3>

Messenger ribonucleic acid (mRNA) is the RNA that is used in cells for protein synthesis. It has a single strand made by the transcription of DNA by RNA polymerase. It contains four nucleotides: Adenine (A), Guanine (G), Cytosine (C), and Uracil (U).

<h3>DNA Replication</h3>

Before transcribing, we need to create the complementary strand of the DNA. We're going to write out the nucleotides of the complementary strand by matching the nucleotides in these pairs: (A & T) and (C & G).

              Normal Strand: GCA ATG CAC
Complementary Strand: CGT TAC GTG

Next, we can transcribe this to find our mRNA. We're going to do the same thing to the complementary DNA strand, but with Uracils instead of Thymines. So our pairs are: (A & U) and (C & G)

Complementary DNA Strand: CGT TAC GTG
                        mRNA Strand: GCA AUG CAC

You'll notice that the mRNA strand is almost exactly like the new mRNA strand, but with Uracil instead of Thymine.

<h3>Reading Codons</h3>

Each set of three nucleotides is known as a codon, which encodes the amino acids that ribosomes make into proteins. To read the codons, you need to have a chart like the one I attached. Start in the middle and work your way to the edge of the circle. Some amino acids have multiple codons. There are also "stop" and "start" codons that signify the beginning and ends of proteins.

mRNA Strand: GCA AUG CAC
Amino Acids:   Ala   Met   His

Our sequence is alanine, methionine, and histidine.

<h3>Frameshift Mutations</h3>

A frameshift mutation occurs when a nucleotide is either added or removed from the DNA. It causes your reading frame to shift and will mess up every codon past where the mutation was. This is different than a point mutation, where a nucleotide is <em>swapped</em> because that will only mess up the one codon that it happened in. Frameshift mutations are usually more detrimental than point mutations because they cause wider spread damage.

<h3>Mutated Strand</h3>

Let's repeat what we did earlier on the mutated strand to see what changed.

              Mutated Strand: CAA TGC AC
Complementary Strand: GTT ACG TG
---
Complementary DNA Strand: GTT ACG TG
                        mRNA Strand: CAA UGC AC
---
mRNA Strand: CAA UGC AC
Amino Acids:   Glu   Cys   X
---
Our amino acid sequence is glutamine, cysteine, and no third amino acid.

As you can see, removing the first nucleotide of the strand caused every codon to change. The last codon is now incomplete and won't be read at all. If this happened in a cell, the protein that was created from this mutated strand would be incorrect and may not function completely or at all.

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