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3 years ago

You can't tell me this doesn't look like a shiny piece of bacon lol

Biology

1 answer:

grin007 [14]3 years ago

3 0

I can surely tell you it is super shiny bacon

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3 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).

From this cross the expected ratios are:

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>Chi square test</h3>

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

The observed frequencies were:

34 Red
76 Pink
40 White

Total 150

The expected frequencies for our null hypothesis are:

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

You can't tell me this doesn't look like a shiny piece of bacon lol​

You can't tell me this doesn't look like a shiny piece of bacon lol