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IgorLugansk [536]
2 years ago
12

Which of the following structures are NOT found in the cytoplasm?

Biology
2 answers:
ElenaW [278]2 years ago
8 0

Answer:

MRCORRECT has answered the question

Explanation:

D)chloroplast

aleksley [76]2 years ago
7 0

Answer:

Nucleus

Explanation:

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Which of these sources add water to lakes? Select three correct answers.
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6 0
2 years ago
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.

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3 years ago
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4 mps increase in 10 seconds = 2 mps increase in 5 seconds = .4meters/2sec
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Read 2 more answers
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