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lisabon 2012 [21]
3 years ago
6

PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!

!!!!!!!!!!!!!!!!!!PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!!!!!!!!!!!!!!!!!!!PLSSSSSSS HELPPPPPPP I WILL GIVE BRAINLIESTTTTTTTTTT!!!!!!!!!!!!!!!!!!
DESCRIBE WHAT IS HAPPENING IN EACH DIAGRAM

Biology
1 answer:
salantis [7]3 years ago
6 0

Answer:

In the first diagram the weakened pathogen (weaker version of the disease) is inserted into the bloodstream.

In the second diagram the antibodies kill the weakened pathogen. When the immune systems encounters a new disease it "remembers it".

This is why in the third diagram, when the "real" pathogen enters the bloodstream the immune system can effectively kill the disease. As the immune system "remembers" how to defeat the weakened version it can more effectively kill the "real" version.

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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:

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  • 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)
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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
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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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