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Ostrovityanka [42]
3 years ago
15

Is there DNA in your food? How do you know?

Biology
1 answer:
Pachacha [2.7K]3 years ago
4 0
<span>YES,THERE IS DNA IN OUR FOODS.
All living organisms both plants and animals have DNA in their nucleus. Humans eat both plants and animals materials in their food, thus, the foods eaten by humans have DNA in their cells. DNA is a thread like structure that is made up of nucleic acid and is found in all living cells. It is the basis of inheritance in both plants and animals. During cell division, the DNA replicate in order to pass inheritable traits to the newly formed cells.</span>
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C. to determine the compatibility of antibodies of donor and recipient.
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Which type of organism is E. coli?
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I believe the answer is restriction organism.
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The cause of respiratory distress syndrome in premature infants is
FrozenT [24]

Answer:

The cause of respiratory distress syndrome in premature infants is <u>lack of surfactant in the lungs because the lungs has not fully developed.</u>

Explanation:

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8 0
3 years ago
7. Pyrethrums are chemicals used to kill insects. In bed bugs, the mutant-type allele, r, confers resistance, but only in homozy
natita [175]

Answer:

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

Explanation:

It is given that 99% of the bugs were killed after the spray of  pyrethrum. This suggests that 1% of the bugs that were not killed must be homozygous for the mutant type allele "r"

Thus, the frequency of homozygous "rr" species i.e q^2 = 0.01

From this we can evaluate the frequency of mutant "r" allele.

Thus, q = \sqrt{0.01} \\

q = 0.1

As per Hardy-Weinberg first equilibrium equation, p + q = 1

Substituting the value of q in above equation, we get

p = 1 - q\\p = 1 - 0.1\\p = 0.9

Thus, the frequency of homozygous normal bug is equal to

p^2 = 0.9^2 = 0.81

As per Hardy-Weinberg second equilibrium equation-

p^2 + q^2 + 2pq = 1\\

Substituting all the available values we get -

0.81 + 0.01+ 2pq = 1\\2pq = 0.18

Allele frequency

Normal allele  = 0.9

Mutant r allele = 0.1

Genotype frequency

Homozygous normal bugs = 0.81

Homozygous mutant bug = 0.01

Heterozygous normal bug with one mutant r allele and one normal allele = 0.18

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3 years ago
How do ukaryotic cells differ from each other
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