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Aleksandr-060686 [28]
3 years ago
7

How is a branching tree diagram used?

Biology
2 answers:
Wittaler [7]3 years ago
8 0

Branching tree diagrams are groups within groups. Starting at the bottom, organisms branch off as characteristics are added. ... It's in this way that branching tree diagrams are useful for understanding evolutionary relationships between different organisms.Answer:

Explanation:

vodka [1.7K]3 years ago
4 0

Answer:

Branching tree diagrams are groups within groups. Starting at the bottom, organisms branch off as characteristics are added. ... It's in this way that branching tree diagrams are useful for understanding evolutionary relationships between different organisms.

Explanation:

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If there were a change in the DNA sequence, what would result?
Natalija [7]

Answer:

mutated dna

Explanation:

there would be either be dna substitutions, deletions or insertions. substitution simply changes the letter. deletion gets rid of a letter. insertion adds another letter.

5 0
3 years ago
What is an ionic bond ?
Volgvan

Answer: ionic bonding is a type of chemical bonding that involves the electrostatic attraction between oppositely charged ions, or between two atoms with sharply different electronegativities

Explanation:

3 0
3 years ago
In a population of plants, the allele for long stems (S) is completely dominant over the allele for short stems (s). If 35% of t
Sergio [31]

Answer:

The correct answer would be 48 percent.

It can be solved by using the Hardy-Weinberg equation.

p² +  2pq+ q²  = 1

p + q = 1

where, p² represents homozygous dominant's frequency that is, SS in this case,

q² represents homozygous recessive's frequency that is, ss in this case,

2pq represents heterozygous dominant's frequency that is, Ss in this case.

p represents dominant allele's frequency and q represents the recessive allele's frequency.

Now, the percentage of homozygous recessive or short stems is 35%. Therefore, the frequency of homozygous recessive would be 0.35.

Thus, the value of q² = 0.35

So, q = \sqrt{0.35} = 0.59 which is approximately equals to 0.6.

As p + q = 1 so, q = 1 - 0.6 = 0.4

The frequency of heterozygous (Ss) = 2pq = 2(0.6)(0.4) = 0.48

Hence, the frequency of heterozygous would be 48 percent.

8 0
3 years ago
Influenza, strep throat, measles, and the common cold are all infectious diseases. Which of the following methods can be used to
KIM [24]

Answer:

D. Washing your hands before eating.

Explanation:

Option a, b and c are examples of not practicing proper hygiene and they increase the chances of you getting diseases. Option D is the only example that practices proper hygiene.

6 0
1 year ago
1. Please describe the signal transmission across a myoneural junction that allows the nervous system to move the muscles of a f
SSSSS [86.1K]
1. Action potential reaches the axon terminal and depolarizes it.

2. Depolarization opens voltage-gated calcium channels, enabling influx of Ca into the neuron.

3. Calcium binds to specialized proteins on vesicles (containing pre-made acetylcholine) and triggers them to fuse with the neuron membrane at the synapse.

4. Exocytosis of acetylcholine into the synaptic cleft occurs.

5. Acetylcholine diffuses across the synapse and binds to nicotinic receptors on the end plate of the myocyte.

6. Activated nicotinic receptors, themselves ion channels, cause cation influx into the myocyte and generate an end plate potential. This eventually gives rise to the full depolarization within the myocyte that enables contraction.
6 0
3 years ago
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