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guapka [62]
3 years ago
7

2. Name a microbe that lacks a cell wall and would find hypotonic environments detrimental.

Biology
2 answers:
Alina [70]3 years ago
7 0

Answer:

I so I am also a bit confused, but I hope this could answer your question!

Explanation:

The bacteria possess a cell wall composed of peptidoglycan molecule formed by the amino acids called murine.  The bacterial cell wall helps in the survival of the bacteria in some extreme condition like when placed in a hypotonic solution.

The hypotonic solution is formed when it has low solute concentration and high water amount when compared. When the bacterial cell will be placed in hypotonic solution then the water will move inside the cell down the concentration gradient as a result of which the bacterial cell will swell but will not burst.

This happens as the cell wall contains expansin protein which allows the cell wall to expand and provides rigidity to the cell.

Drupady [299]3 years ago
4 0

Answer:

1. Photoautotroph

2.Protozoa

3. Passive transport uses no energy and molecules move from high to low concentrations.  Active transport uses energy to move from low to high concentrations.

4. Synergism

Explanation:

Good luck on your assignment

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the forces between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them

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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
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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.

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