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MariettaO [177]
3 years ago
6

If you take a dna sample from a newborn, would it match a dna sample from when the same person is 80?

Biology
1 answer:
vazorg [7]3 years ago
7 0
Yes DNA doesn't change over time DNA is who you are what you look like you get it from your parents
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What type of RNA is a major component of the structure at which protein synthesis occurs
Zolol [24]
Ribosomal RNA (rRNA)
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3 years ago
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A bird that can easily out complete other birds for its food and that can produce many eggs has high what?
love history [14]

Answer:

Fitness or aptitude

Explanation:

Aptitude (or fitness) is the genotype/phenotype that results in the survival, fertility, and capability of having a mate. It is a way of measuring the individual ability to leave fertile offspring.

Aptitude puts together everything that matters in natural selection. Includes the survival capability, finding a mate partner possibility, producing fertile descendants, and leaving the genes to the next generation. The aptitude of a genotype must be significant for natural selection to act in its favor.

3 0
3 years ago
PLZ HELP!!!!!!!!!!!!!!!!
Leni [432]

Answer:

I see this was posted a week ago, do you still need help with this question?

Explanation:

8 0
3 years ago
What would happen to a flower if it had dark petals or no petals at all?
nadya68 [22]

Answer:

it might catch to much sun with black petals or no sun at all with none so in other words it  would probably die.

Explanation:

white reflects heat, black attracts heat

4 0
3 years ago
In a population of cats, the phenotypic frequency of black cats is 91%, and the phenotypic frequency of white cats is 9%. Assumi
Talja [164]

Answer:

1. Allele frequency of b = 0.09 (or 9%)

2. Allele frequency of B = 0.91 (0.91%)

3. Genotype frequency of BB = 0.8281 (or 82.81%)

4. Genotype frequency of Bb = 0.1638 (or 16.38%)

Explanation:

Given that:

p = the frequency of the dominant allele (represented here by B)  = 0.91

q = the frequency of the recessive allele (represented here by b)  = 0.09

For a population in genetic equilibrium:

p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)

(p + q)^2 = 1

Therefore:

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

in which:  

p^2 = frequency of BB (homozygous dominant)

2pq = frequency of Bb (heterozygous)

q^2 = frequency of bb (homozygous recessive)

p^2 = 0.91^2 = 0.8281

2pq = 2(0.91)(0.9) = 0.1638

4 0
3 years ago
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