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Maru [420]
3 years ago
6

In an experiment, you measure the concentration of a polar molecule inside and outside a cell. You find that the concentration i

s high and gradually increasing inside the cell. You also measure the ATP concentration inside the cell and find that it is dropping. What is your best hypothesis for the process you are observing?
a) facilitated diffusion
b) passive transport
c) simple diffusion
d) active transport
e) endocytosis
Biology
1 answer:
zepelin [54]3 years ago
7 0

Answer:

e) endocytosis

Explanation:

Because the ATP concentration inside the cell is dropping, we know that the cell is using energy to transport the polar molecule into the cell. This eliminates a), b) and c) because they all don't require energy, and we know that polar substances have trouble diffusing across membranes because the tails of the phospholipids making up the membrane are hydrophobic and non-polar. We also know that the polar substance concentration is increasing in the cell. Honestly, both d) and e) would work because both require ATP; however, endocytosis is more specific as a hypothesis. Thus, e). :)

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Offspring of certain fruit flies may have yellow or ebony bodies and normal wings or short wings. Genetic theory predicts that t
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Answer:

Yes, the observed results are consistent with the theoretical distribution predicted by the genetic model. The population is in equilibrium.

Explanation:

<u>Available data:</u>

  • Offspring of fruit flies may have yellow or ebony bodies and normal wings or short wings.
  • Expected phenotypic ratio 9:3:3:1
  • 9 yellow, normal: 3 yellow, short: 3 ebony, normal: 1 ebony, short
  • Sample size , N = 100
  • Phenotypic distribution 59:20:11:10

To know if these results are consistent with the expected ones, we need to develop a chi-square analysis. To do it we need the following information

  • Chi square= ∑ ((Obs-Exp)²/Exp)

- ∑ is the sum of the terms

- Obs are the Observed individuals

- Exp are the Expected individuals

  • Freedom degrees = K – 1

- K =genotypes number = 4  

  • Significance level, 5% = 0.05
  • Table value = Critical value  

First, define the hypothesis:

Hypothesis: The allele of this population will assort independently. The population is in equilibrium

H₀= Individuals will be equally distributed.  

H₁ = Individuals will not be equally distributed.  

Second, we need to get the expected number of individuals with each phenotype. To do it, we will use the expected phenotypic ratio and the total number of individuals in the sample. We can just perform a three simple rule, as follows:

  16 ------------------------------------ 100 individuals in the sample ------- 100%

  9 yellow, normal ---------------- X = 56.25 individuals -------------------56.25%

  3 yellow, short ------------------- X = 18.75 individuals --------------------18.75%

  3 ebony, normal -----------------X = 18.75 individuals ---------------------18.75%

  1 ebony, short ---------------------X = 6.25 individuals ----------------------6.25%

Now that we know the expected numbers of individuals with each genotype, we can compare them with the observed ones.

                  <u>yellow, normal      yellow, short     ebony, normal    ebony, short</u>

Expected            <em>56.25                    18.75                   18.75                    6.25</em>

Observed  <em>          59                           20                      11                          10</em>

The chi-square value = Σ(Obs-Exp)²/Exp.

So now we need to calculate (Obs-Exp)²/Exp

  • <u>yellow, normal</u>

(Obs-Exp)²/Exp = (59-56.25)²/56.25 = 0.134

  • <u> yellow, short </u>

(Obs-Exp)²/Exp = (20 - 18.75)²/18.75 = 0.083

  • <u>ebony, normal</u>

(Obs-Exp)²/Exp = (11 - 18.75 )²/18.75 = 3.203

  • <u> ebony, short</u>

(Obs-Exp)²/Exp = (10 - 6.25)²/6.25 = 2.25

X² = Σ(Obs-Exp)²/Exp =  0.134 + 0.083 + 3.203 + 2.25 = 5.67

  • X² = 5.67
  • Significance level = 0.05
  • Degrees of freedom = genotypes number - one = 4 - 1 = 3
  • Critical value or table value = 9.348

P₀.₀₅ > X2

9.348 > 5.67

There is not enough evidence to reject the null hypothesis. The genotypes might be in equilibrium, and there might be an independent assortment.

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