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charle [14.2K]
3 years ago
14

The genes for alpha-goblin and beta-globin are found:\

Physics
2 answers:
Luba_88 [7]3 years ago
5 0
Actually the third one is introns, trust me I did the test 
BabaBlast [244]3 years ago
3 0

<em><u>Globin is the protein part of hemoglobin. </u></em>

<u>The alpha globin family </u><u>is located on chromosome 16.</u>

<u>The beta globin family</u> <em><u>is located on chromosome 11.</u></em>

<em><u>The answer is</u></em>: <u>d.) On different chromosomes. </u>

-------------

<u>The vast majority of genes encoding a protein encoded</u> <em><u>within the nucleus of metazoan cells</u></em> <u>contain both introns and exons. </u>

<em><u>The answer is</u></em>: <u>b.) both exons and introns. </u>

-----------------

<u><em>In the DNA sequence the coding content of a gene is not continuously distributed along said gene</em></u>, <u>but has discontinuities, called introns, whose sequence does not encode proteins</u>.

<em><u>The answer is</u></em>: <u>a.) Intron. </u>

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Light from the stars, because the orbits make it difficult to see them. 
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Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra
vredina [299]

Answer:

At low pressure- \mu_{k}=0.02315

At high pressure- \mu_{k}=0.00445

Explanation:

Initial speed, V_{i}=3.3 m/s

Final speed, V_{f}=3.3/2= 1.65 m/s

Net horizontal force due to rolling friction F_{net}=\mu_{k} mg where m is mass, g is acceleration due to gravity, \mu_{k} is coefficient of rolling friction

From kinematic relation, V_{f}^{2}- V_{i}^{2}=2ad

For each tire,

V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd

Making \mu_{k} the subject

\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}

Under low pressure of 40 Psi, d=18 m

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315

Therefore, \mu_{k}=0.02315

At a pressure of 105 Psi, d=93.7

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445

Therefore, \mu_{k}=0.00445

4 0
3 years ago
A spring has been compressed inside a toy what kind of energy is this
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Answer:

Elastic Potential Energy

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4 0
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Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

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q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

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back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
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A.

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