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3 years ago

How are the energy needs of plant cells similar to those of animal cells? How are they different?

1 answer:

7 0

Well plant cells have both chloroplasts and mitochondria, and what happens is photosynthesis happens in the chloroplast (they produce their own food to eat for energy (sugar/glucose). The outcome of photosynthesis is glucose and oxygen which then goes to the mitochondria (where the process of cellular respiration takes place) which takes in the glucose and oxygen and turns that into carbon dioxide, water, and ATP energy. Where the plant goes through the process of making the food first before going through the process of turning into energy, animals just have the mitochondria, so they consume foods for the sugars needed and the mitochondria do the same work they did in the plant, just minus the photosynthesis.

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Cellular Process Explain How the Process Assists in Maintaining Homeostasis for each word

WINSTONCH [101]

Cellular reproduction is the answer

5 0

2 years ago

Which of the following actions occurs during DNA replication?

Nata [24]

Replication occurs in three major steps: the opening of the double helix and separation of the DNA strands, the priming of the template strand, and the assembly of the new DNA segment. During separation, the two strands of the DNA double helix uncoil at a specific location called the origin.

5 0

2 years ago

In a species of bird, the allele for the striped trait is recessive to the allele for the speckled trait. Suppose that a coastal

mylen [45]

Answer:

The answer is "0.42"

Explanation:

Please find the complete question in the attached file.

Coastal community striped allele intensity $= q = 0.62$

Landlocked community, strip intensity allele $=q'=0.40$

Its eliminated allele frequency for movement $= q"=0.42 \ in \ indoors$ community And now in the sturdy vineyard optimum communities, genotype frequency of stripped characteristic (heterozygous device) in coasts $=q_2=0.39$

Heritable (striped) allele frecency:

$\to q=\sqrt{q_2}$

$=\sqrt{0.39} \\\\ =0.624 \\\\ =0.62$

(round up to the closest cent) is therefore the only frequency of coastal people.

The genome of pulling function in host population $= q'_2=0.16$ Heterozygous feature regularly.

Therefore the inland community rate of recessed (striped), allele rate:

$\to q'= \sqrt{q'_{2}}= \sqrt{0.16}=0.40$

Following migration;

Its percentage of coastal migrants:

$m=10\% \\\\$

$= \frac{10}{100} \\\\= 0.1$

Coastal population Non-immigrant percentage of coastal residents:

$=1- m\\\\=1-0.1\\\\ =0.9$

Stripping coastal community of allele rate after immigration:

$(q")=\text{(stripping of migrant only freq)}\times (m)+\text{(stripping allele baud rate for non-immigrants)} \times ( 1-m)$

$= [q \times m]+[[q' \times(1-m)]\\\\=(0.62 \times 0.1)+(0.40 \times 0.9)\\\\=0.062+ 0.36\\\\=0.422\\\\=0.42$

8 0

2 years ago

If mitotic cell division is the only way a particular species of single-celled organism can reproduce, it is most likely that

leva [86]

If mitotic cell division is the only way a particular species of single-celled organism can reproduce, it is most likely that the rate of evolution in this species is slower than in one that reproduces sexually. This is asexual reproduction, which causes mutations in a population to accumulate and combine more slowly than sexual reproduction in a cell line, as the mutations would have to arise multiple times in one line, rather than a few separate times and then be allowed to combine sexually for the descendants to have those traits.

4 0

2 years ago

"A blood bank for a local hospital has a supply of plasma for which some have been contaminated with a virus causing hepatitis.

djverab [1.8K]

Answer:

The probability that the victim not getting hepatitis virus is 0.4547

Explanation:

We have given that 7 of 48 units of blood is contaminated with virus causing hepatitis.

So probability of getting hepatitis virus is $=\frac{7}{48}=0.1458$

So probability of not getting hepatitis virus = 1 - 0.1458 = 0.8542

It is given that severely injured accident victim given 5 units of plasma

So probability that the victim not getting hepatitis virus = $0.8542^5=0.4547$

So the probability that the victim not getting hepatitis virus is 0.4547

5 0

3 years ago