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4 years ago

How can a chromosome regulate transcription and increase it

2 answers:

6 0

As in bacteria, translation in eukaryotic cells is controlled by proteins that dilemma to particular administrative arrangements and regulate the action of RNA polymerase. At the point when a DNA-bowing protein ties to the enhancer, the state of the DNA changes, which permits communications between the activators and translation elements to happen.

svp [43]4 years ago

5 0

In bacteria and eukaryotes, transcription is regulated by proteins which bind to the specific sites and modulate the activities. These sites are located on the chromosome, and they are a cluster of gene sequences known as operons. Operons are under the control of a single promoter, and can be induced/repressed or may work consistently all the time. Main regulation of transcription occurs by rearrangement of histone proteins, which are located on the chromosome itself. These proteins may leave the desired gene sequence inaccessible for further operations.

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May we please check it for me which one is the correct answer.

-Dominant- [34]

Answer: 37. fimbriae. 38. Ductus deferens. 39. Both Cardiac and skeletal muscles. brainliest??

Explanation:

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3 years ago

PLEASE HELP ME I WILL RATE BRAINLIEST ANSWER

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Solid.
I have to write all this extra stuff to answer ur question on Brainly but that is the answer :)

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3 years ago

Read 2 more answers

In humans there is a dominant allele (A) for the absence of moles; while the recessive allele (a) results in the presence of mol

Art [367]

Answer:

a. $(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. i. 0.75

ii. 0.000061

iii. 0.012

iv. 0.17

c. 0.67

Explanation:

a. The expansion of the binomial (p + q)7 would be such that:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

b. Both couples are heterozygous:

Aa x Aa

AA Aa Aa aa

Since A is dominant over a,

probability of having mole (aa) = 1/4

probability of not having moles = 3/4

Therefore, the probability of the first child not having moles = 3/4 or 0.75

ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:

$(p + q)^7$ = $p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7$

Probability that all of the children will have moles = $p^0q^7$

since p = 3/4 and q = 1/4

$p^0q^7$ = $(3/4)^0(1/4)^7$ = 0.000061

iii. Probability that the first two children will have no moles and the last five will have moles = $21p^2q5$

= $21(3/4)^2(1/4)^5$

= 0.012

iv. Probability that 4 will have no moles and 3 will have moles out of the 7 children = $35p^4q^3$

= $35(3/4)^4(1/4)^3$

= 0.17

c. Probability that the child born without moles is a carrier of the a-allele = probability of heterozygous.

From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:

= 2/3 or 0.67

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3 years ago

The prevention of illness through vaccination occurs due to the formation of

posledela

Memory cell lymphocytes!

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3 years ago

What factors causesd the collapse of bluefin tuna population

natta225 [31]

Answer: Overfishing in the main location that bluefin tuna are caught, the Mediterranean. They are being caught before they can reproduce (very low reproduction rate)

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3 years ago