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3 years ago

THE FIRST PERSON TO ANSWER ASAP

Biology

2 answers:

skad [1K]3 years ago

6 0

Integral should be the correct answer

Alex73 [517]3 years ago

5 0

Answer: Integral membrane proteins.

question answered by

(jacemorris04)

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Nitrogen fixation is a process by which molecular nitrogen in the air is converted into ammonia or related nitrogenous compounds in soil. Atmospheric nitrogen is molecular dinitrogen, a relatively nonreactive molecule that is metabolically useless to all but a few microorganisms.

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Which of the following cellular environments is conducive to the formation of disulfide bonds within or between proteins? Choose

STatiana [176]

Answer:

Rough endoplasmic reticulum and Mitochondria.

Explanation:

Disulfide bonds are known as covalent bonds. They are formed by the oxidation of 2 cysteines and these bonds can provide stability to proteins. These bonds mainly formed in intermembrane space of mitochondria and cellular compartments outside the cytoplasm endoplasmic reticulum. Both of these organelles present in an oxidation state providing an atmosphere for disulfide bond formation.

Cytoplasm and Nuclei mostly exit in reducing state because of the existence of disulfide reductase which is reducing the disulfide bonds between the cysteine residue to thiolate state. So, the disulfide bond formation will not happen.

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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).

<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