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3 years ago

What are 3 adaptions that frogs have that make them amphibians?

2 answers:

6 0

1) They are cold blooded
2) They respire both by Skin and lungs.
3) They lay eggs in water ; they have a 3 chambered heart.

melomori [17]3 years ago

4 0

Webbed feet, cold blooded, and they live in moist areas

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HELP PLEASE!!!!!!

allochka39001 [22]

Answer:

They probably use aerobic respiration.

Explanation:

A travel distance of 11.500 kilometers in 9 days covered by flying surely requires a lot of energy.

If cells are fermenting, the ATP (energy) they generate only comes from glycolysis, which produces 2 ATP molecules.

If they are using aerobic respiration, glucose is completely oxidized to CO₂ through glycolysis and the Citric Acid Cycle, and the electrons enter the electron transport chain to finally reduce oxygen into water. In the complete process, up to 36 ATP molecules are produced.

In sum, aerobic respiration is much more efficient to generate energy than fermentation, so it's probably the main metabolism of the flight muscles in bar-tailed godwits.

4 0

3 years ago

Analyze the model of the cell membrane below. Which of the following statements is true about the phospholipid components of the

MArishka [77]

Answer:

This question appears incomplete but The correct option is the last option

Explanation:

Cell membranes are made up of a lipid/phospholipid bilayer in which each layer consists of an hydrophilic head (water loving) and a hydrophobic tail (water fearing). The hydrophobic tail of each layer is oriented inward and the hydrophilic heads are oriented outward so the cell can interact with water molecules inside (the cytoplasm) and outside the cell environment. This orientation forms a critical part of the function of the cell membrane that makes it a selectively permeable barrier that regulates what goes in and out of the cell.

3 0

3 years ago

Alcohol consumption tends to cause more ___________ behavior.

scZoUnD [109]

The answer is most likely B, "Alcohol consumption tends to cause more [Aggressive] behavior."

7 0

3 years ago

Read 2 more answers

Question 3

shepuryov [24]

The statements Molecule B can be used to produce molecule A and Molecules B and C interact so that amino acids can be joined together are TRUE regarding the structures shown in the diagram (Options B and E).

<h3>What are transcription and translation?</h3>

Transcription is the process by which a DNA template (molecule B) can be used to synthesize a complementary RNA (molecule A), while translation is a process in which an sequence is used as templates to generate a protein, which requires tRNAs (molecule C) to trasnport amino acids to the ribosomes.

Therefore, with this data, we can see that DNA is molecule B used to create RNA, which is molecule A, while tRNA is represented by molecule C and it is used during translation.

Learn more about transcription and translation here:

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7 0

1 year ago

Cystic fibrosis (CF) is one of most common recessive disorders among Caucasians it affects 1 in 1,700 newborns. What is the expe

Phantasy [73]

Answer: The expected frequency of carriers is P(Aa)=0.046.

The proportion of childs with CF is P(aa)=0.024.

25% of having a child with CF (aa).

Explanation:

Hardy-Weinberg's principle states that in a large enough population, in which mating occurs randomly and which is not subject to mutation, selection or migration, gene and genotype frequencies remain constant from one generation to the next one, once a state of equilibrium has been reached which in autosomal loci is reached after one generation. So, a population is said to be in balance when the alleles in polymorphic systems maintain their frequency in the population over generations.

Given the gene allele frequencies in the gene pool of a population, it is possible to calculate the expected frequencies of the progeny's genotypes and phenotypes. <u>If P = percentage of the allele A (dominant) and q = percentage of the allele a (recessive)</u>, the checkerboard method can be used to produce all possible random combinations of these gametes.

Note that p + q = 1, that is, the percentages of gametes A and a must equal 100% to include all gametes in the gene pool.

The genotypic frequencies added together should also equal 1 or 100%, and all the equations can be summarized as follows:

$p+q=1\\(p+q)^{2} = p^{2} +2pq+q^{2} = 1\\P(AA)=p^{2} \\P(aa)=q^{2} \\P(Aa)=2pq1$

So, there are 1700 individuals and only one is affected. Since it is a recessive disorder, the genotype of that individual must be aa. So the genotypic frequency of aa is 1/1700=0.000588.

Then, $P(aa)=q^{2}=0.000588$ . And with that we can calculate the value of q,

$P(a)=q=\sqrt{0.000588}=0.024$

And since we know that p+q=1, we can find out the value of p.

$p+0.024=1\\1-0.024=p\\p=0.976$

Next, we find out the genotypic frequency of the genotype AA:

$P(A)=p=0.976\\P(AA)=p^{2} = 0.976^{2}=0.95$

Now, we can find out the genotypic frequency of the genotype Aa:

$P(Aa)=2pq=2 x 0.976 x 0.024 = 0.046$

Notice than:

$p^{2} + 2pq + q^{2} = 1\\x^{2} 0.976^{2} + 2 x 0.976 x 0.024 + 0.024^{2} = 1$

Then, the expected frequency of carriers is P(Aa)=0.046

The proportion of childs with CF is P(aa)=0.024

If two parents are carriers, then their genotypes are Aa.

Gametes produced by them can only have one allele of the gene. So they can either produce A gametes, or a gametes.

In the punnett square, we can see that there genotypic ratio is 2:1:1 and the phenotypic ratio is 3:1. So, there is a probability of 25% of having an unaffected child, with both normal alleles (AA); 50% of having a carrier child (Aa) and 25% (0.25) of having a child with CF (aa).

5 0

3 years ago