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ohaa [14]
3 years ago
11

_____ plants grow in moist environments so they can absorb water and nutrients by osmosis and diffusion.

Biology
2 answers:
BARSIC [14]3 years ago
8 0
Nonvascular plants are the plants that grow in moist places or environment so they can absorb water and nutrients by osmosis and diffusion. Example of this plant is the water lily. Water lily are seen in the places that has waters and grows healthier.
Shkiper50 [21]3 years ago
5 0

Non-vascular plants grow in moist environments so they can absorb water and nutrients by osmosis and diffusion.

Non-vascular plants are plants that do not have specialized tissues (such as xylem and phloem) that conduct or circulate water throughout the plant. Due to the absence of a vascular tissue system, non-vascular plants usually grow in moist environments so they can absorb water and nutrients directly from the soil by osmosis and diffusion.  Non-vascular plants do not grow very tall and they lack true leaves, stems, and roots.

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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
Rashid [163]

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.

8 0
3 years ago
Which pair of structures forms the hip joint? 
Hatshy [7]

Answer: The answer is A

Explanation: The acetabulum is a concave area of the pelvis that forms a socket into which the femoral head fits. Together they form the hip joint.

5 0
3 years ago
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____ is the condition of the troposphere at a particular time and place
san4es73 [151]
The answer to this question is whether

8 0
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What factor would result in an increased mortality in a population due to increased competition for water?
inn [45]
Mortality, or death rate is a measure of the number of deaths in a particular population, scaled to the size of that population, per unit of the time. In this case a factor that would result to an increased mortality rate in a population due to increased competition of water will be the density-dependent factor. 
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How do new cells form in plants and animals?
frozen [14]
In both animals and plants, cells produce new cells by mitosis - but they split differently. A cleavage farrow forms in the animal cell and it splits. For the plant cell, a cell plate forms and then the cell splits.
7 0
3 years ago
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