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matrenka [14]
3 years ago
12

Merlin’s paper discusses two types of data: artifactual and paleoethnobotanical. Give specific examples of each and describe the

sources of data and the type of information you can gain.
Biology
1 answer:
Neko [114]3 years ago
8 0

Answer:

Merlin's paper discusses two types of data: artifactual and paleoethnobotanical. Give specific examples of each and describe the sources of data and the ...

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What is used to measure the density of the earths layers/8476472/8516cadb?utm_source=registration
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Seismographs can be used for this.
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Rhizobium A. fix nitrogen inside nodules on the roots of legumes.B. resemble fungi.C. produce antibiotics.D. produce a gall in p
Marina CMI [18]

.Answer:

Option A, fix nitrogen inside nodules on the roots of legumes

Explanation:

Rizhobium bacteria live in symbiotic relationship with roots of leguminous plants such as pulses including alfalfa, soybean etc.

Rhizobium or Bradyrhizobium  is a gram negative bacteria that satisfies the nitrogen starvation of host plant.  

These bacteria get habitation and nutrition from these leguminous plants and in return they fix the nitrogen into the nodules of the roots. This is a kind of facultative association between soil bacteria rizhobium and legume plants.  

Hence, option A is correct

Thus, option A is correct

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Define "Diffusion"..
taurus [48]

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the spreading of something more widely.

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2 years ago
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When pink sweet peas were self-pollinated and the seeds were collected and sown, the following flower colors were obtained: Red
Pavlova-9 [17]

Answer:

c. 1:2:1

The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

Explanation:

If flower color were determined by a gene showing incomplete dominance, the possible genotypes and phenotypes are as follows:

  • RR- red
  • ww - white
  • Rw - pink

If pink sweet peas are self-pollinated, then a cross between two heterozygous individuals is done (Rw x Rw).

<u>From this cross the expected ratios are:</u>

  • 1/4 RR (red)
  • 2/4 Rw (pink)
  • 1/4 ww (white)

So the null hypothesis is that the observed results exhibit a 1:2:1 ratio.

<h3><u>Chi square test</u></h3>

X^{2} = \sum \frac{(Observed - Expected)^2}{Expected}

<u>The observed frequencies were:</u>

  • 34 Red
  • 76 Pink
  • 40 White

Total 150

<u>The expected frequencies for our null hypothesis are:</u>

  • 1/4 x 150 = 37.5 Red
  • 2/4 x 150 = 75 Pink
  • 1/4 x 150 = 37.5 white

X^{2} = \frac{(34- 37.5)^2}{37.5} + \frac{(76- 75)^2}{75} + \frac{(40- 37.5)^2}{37.5}

X^2=0.5067

The degrees of freedom (DF) are calculated as number of phenotypes - 1; in this case DF = 3-1 = 2.

If we look at the Chi square table, for 2 DF and a probability of p0.05, the critical value is 5.991

Our X^2 value of 0.5067 is less than the critical value, so we do not reject the null hypothesis. The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.

8 0
3 years ago
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