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kherson [118]
3 years ago
13

In the gene TATTCATTGTTA—TGATTT—ATTCG, CATTGTTA encodes for pepsin, a digestive enzyme. The rest of the sequence doesn’t code fo

r any protein. Which sequence contains a mutation that will affect the formation of pepsin?
TATTCATTCATTA—TGATTT—ATTCG



TATTCATTGTTA—TGACTTT—ATTCG



TATTCATTGTTA—TGATTT—ATTGGCG



TATTCATTGTTA—TGAT—ATTCG



TGCATTCATTGTTA—TGATTT—ATTCG
Physics
1 answer:
MatroZZZ [7]3 years ago
8 0

Answer:

TATTCATTCATTA—TGATTT—ATTCG

Explanation:

A mutation is a permanent change in the nucleotide sequence of DNA. A mutation occurs during replication or recombination. It may be due to base substitutions, deletions and insertions. As per the question, DNA segment forms encodes for the enzyme pepsin is CATTGTTA.

Option TATTCATTCATTA—TGATTT—ATTCG is the correct answer. In the DNA segment which encodes pepsin, a purine base (G) guanine is replaced by another purine (A) adenine. This type of mutation is called a transition type point mutation.

Due to base substitution, the mutated segment CATTCATTA will nor encode pepsin.

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Answer:

The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.

Explanation:

  • It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:

       C = \frac{\epsilon*A}{d}

  • Where ε, is the dielectric constant of the material that fills the space between plates.
  • When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
  • At the same time, the capacitance of a capacitor, by definition, is as follows:

       C =\frac{Q}{V}

  • If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
  • As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
  • So, the change in the charge of the positive plate is +2.5 nC.
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3 years ago
Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?
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The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:

f = c / λ

where, f = frequency of light

            c = speed of light

            λ = wavelength of light

Given data = f = 1.72×10^{15}Hz

Therefore, λ = 3×10^{8} / 1.72×10^{15}

                  λ = 1.74×10^{-7}m

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

Learn more about light here;

brainly.com/question/15200315

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The work done by a constant force in a rectilinear motion is given by:

W=Fd\cos\theta

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

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