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myrzilka [38]
2 years ago
6

Fossilized stromatolites

Biology
1 answer:
Cerrena [4.2K]2 years ago
6 0

Answer:

(B) resemble structures formed by bacterial communities that are found today in some shallow marine bays.

Explanation:

Extant stromatolites represent real "living fossils" for they are decendents of ancient forms that are associated with one of the first living forms on earth. Particularly, stromatolites are real bacteria communities where the autotrophic organism of the community are represented by cyanobacteria, which live along with heterotrophic bacteria. This clearly indicates that fossilized stromatolites points to bacteria (prokaryotes) as the first living things on earth (dated with not less than 3.5 billion year old)

Nowadays, stromatolites with cyanobacteria allows to reconstruct and understand fossilized forms. These current structures live in shallow marines ambients (e.g. Australia) but also in continental salt flats (e.g. Argentina) where few others bacteria can survive to these extreme conditions (high light exposure and salt concentration).

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You memorize a series of numbers by grouping them into sets of three and thinking of these sets as area codes. This is an exampl
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Answer:

The correct answer is chunking.

Explanation:

Chunking is a term signifying the procedure of taking single pieces of information or chunks and aligning them into bigger units. By aligning each piece into a large whole, one can better the amount of data one can remember.  

Generally, the most common illustration of chunking takes place in phone numbers. By distinguishing dissimilar single elements into bigger blocks, information becomes easier to recall and retain.  

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30. Why can mitochondria be considered the power plants of the aerobic cells?
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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
ratelena [41]

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.

<em><u>Genotype       Observed frequency</u></em>

    A             →        18 individuals

    B             →        55 individuals

    C             →        27 individuals

Do these contradict the genetic model?  

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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Answer:

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