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Svetach [21]
1 year ago
13

Will the real atomic model please stand up?,” describe what Dalton’s theory states about a molecule of water.

Biology
1 answer:
Law Incorporation [45]1 year ago
8 0

According to  the Dalton's atomic theory, we can see that there are two atoms that compose the water molecule.

<h3>What is the atomic model?</h3>

The atomic theory states that matter is made up of small particles called atoms. These atoms are capable of independent existence and can take part in chemical reaction.

From the Dalton's atomic theory, we can see that there are two atoms that compose the water molecule. These atoms are hydrogen atom and the oxygen atom.

Learn more about Dalton Model:brainly.com/question/12981565

#SPJ1

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The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A
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Complete question:

The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.

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Use a 0.05 level of significance.

Determine the chi-square test statistic.

Answer:

Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.  

Explanation:

<u>Available data</u>:

  • Crossed genotypes: two
  • Genotypes among the offspring; Three → A, B, and C
  • Expected phenotypic ratio → 1:2:1
  • Total number of individuals, N = 100
  • A = 18 individuals
  • B = 55 individuals
  • C = 27 individuals

So, let us first state the hypothesis:

  • H₀= the population is equilibrium for this locus → F(A) = 25%,  F(B) = 50%, F(C) = 25%  
  • H₁ = the population is not in equilibrium

Now, let us calculate the number of expected individuals, according to their expected ratio.

4 -------------- 100% -------------100 individuals

1 ---------------  25% -------------X = 25 individuals A

2 --------------  50% -------------X = 50 individuals B

1----------------- 25%--------------X = 25 individuals C

<u>                                                   A                             B                           C</u>

  • Observed                         18                            55                         27
  • Expected                         25                           50                         25
  • (Obs-Exp)²/Exp                1.96                        0.5                        0.16

<u>(Obs-Exp)²/Exp</u>

A)  (18 - 25)²/25 = 49/25 = 1.96

B)  (55 - 50)² / 50 = 25/50 = 0.5

C)  (27 - 25)²/25 = 4/25 = 0.16

Chi square = X² = Σ(Obs-Exp)²/Exp  

  • ∑ is the sum of the terms
  • O are the Observed individuals: 2 in chamber B, and 18 in chamber A.  
  • E are the Expected individuals: 10 in each chamber  

X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62

Freedom degrees = 2

Significance level, 5% = 0.05  

Table value/Critical value = 5.99

X² < Critical value

2.62 < 5.99    

<em>These results suggest that there is </em><u><em>not enough evidence to reject</em></u><em> the null hypothesis. We can assume that </em><u><em>the locus under study in this population is in equilibrium H-W.  </em></u>

 

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