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Natali5045456 [20]
2 years ago
8

Somatic cells of roundworms have four individual chromosomes per cell. How many chromosomes would you expect to find in an ovum

from a roundworm?
Biology
1 answer:
andrezito [222]2 years ago
3 0

Answer:

2 chromosomes

Explanation:

The cell of an eukaryotic organism like roundworm contains the Nucleus, which harbors the genetic material embedded in the chromosome. The number of chromosomes of that organism is contained in each cell.

Somatic cells, also called body cells, are all other cells asides sperm and eggs, that form the tissues and organs of an organism. Somatic cells are usually diploid i.e two sets of chromosomes from each parent. In this question, the roundworm has 4 chromosomes in its somatic cell.

The reproductive cells or sex cells (sperm and eggs) of an organism always result from meiotic division of specialized cells.

Since meiosis is a kind of division that results in cells with their chromosome number reduced by half (haploid), it therefore means that the ovum and sperm cell will be expected to contain 2 chromosomes each.

This way, when fertilization occurs (sperm and egg fusion), the resulting zygote, which will eventually develop into an adult organism, will have 4 chromosomes.

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Modify the existing vector's contents, by erasing the element at index 1 (initially 200), then inserting 100 and 102 in the show
Aleonysh [2.5K]

Answer:

Following are the code to this question:

#include <iostream>//defining header file

#include <vector>//defining header file

using namespace std;

void PrintVectors(vector<int> numsList)//defining a method PrintVectors that accept an array

{

   int j;//defining integer variable

   for (j = 0; j < numsList.size(); ++j)//defining for loop for print array value  

   {

       cout << numsList.at(j) << " ";//print array

   }

   cout << endl;

}

int main()//defining main method  

{

   vector<int> numsList;//defining array numsList

   numsList.push_back(101);//use push_back method to insert value in array

   numsList.push_back(200);//use push_back method to insert value in array

   numsList.push_back(103);//use push_back method to insert value in array

   numsList.erase(numsList.begin()+1);//use erase method to remove value from array

   numsList.insert(numsList.begin(), 100);//use insert method to add value in array

   numsList.insert(numsList.begin()+2, 102);//use insert method to add value in array

   PrintVectors(numsList);//use PrintVectors method print array value

   return 0;

}

Output:

100 101 102 103  

Explanation:

In the above-given code, inside the main method an integer array "numList" is defined, that use insert method to insert value and use the erase method to remove value from the array and at the last "PrintVectors" method is called that accepts a "numList" in its parameter. In the "PrintVectors" method, a for loop is declared, that prints the array values.

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2 years ago
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RUDIKE [14]

Answer:

Basically the translation of CK-B protein is inhibited in the U937D cells, despite the fact that the CK-B protein is bounded to the ribosomes. This is because(<u>mechanisms)</u> the translation is inhibited by the binding of translational repressors to the 3’UTR of the CK-B mRNA rather than the actual CK-B mRNA 3'UTR.

Furthermore, the soluble protein inhibitions is due to the reaction of the U937 cells to the short RNA sequences with the 3’UTR.

<u>The introduction of these sequences(shot segement of RNA) </u>into the U937D cells leads to CK-B synthesis. This makes 3’UTR sequences to bind to the translational repressor proteins, thus preventing them from binding to the CK-B mRNA .

COMPLETED QUE.

A common feature of many eukaryotic mRNAS is the presence of a rather long 3' UTR, which often contains consensus sequences. Creatine kinase B (CK-B) is an enzyme important in cellular metabolism. Certain cells—termed U937D cells—have lots of CK-B mRNA, but no CK- B enzyme is present. In these cells the 5’ end of the CK-B mRNA is bound to ribosomes, the mRNA is apparently not translated. Something inhibits the translation of the CK-B mRNA in these cells. Researchers introduced numerous short segments of RNA containing only 3’UTR sequences into U937D cells. As result, the U937D cells began to synthesize the CK-B enzyme, but the total amount of CK-B mRNA did not increase. The introduction of short segments of other RNA sequences did not stimulate the synthesis of CK-B; only the 3’UTR sequences turned on the translation of the enzyme. Based on these results, <u>purpose a mechanism for how CK-B translation is inhibited in U937D cells. Explain how the introduction of short segments of RNA containing the 3'UTR sequences might remove the inhibition.</u>

<u />

Explanation:

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Then comes the paracrine signals in which signals acts locally on the cells nearby it. The cells close together to the cells producing chemical signals is being affected.

The endocrine signals can be defined as the effect of the hormone on the distant cells. The signals is produced by the cells somewhere else but is carried through the bloodstream to the distant cells.

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