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g100num [7]
2 years ago
9

In a laboratory investigation, students measured the rate of fermentation in yeast cells. Data was collected by student groups s

howing how the rate of fermentation is influenced by the type of sugar used as the food supply by the yeast. The students compared glucose (monosaccharide) to fructose (mono) and sucrose (disaccharide). Amount and type of yeast as well as water temperature remained constant in all trials. The results of the experiment is modeled in the bar graph. After collecting the data, the teacher provided a sample of maltose, a disaccharide. Predict how the bar graph would be amended once the maltose trial is run. A) The maltose bars would fall somewhere between sucrose and glucose. B) The maltose bars would be longer than all of the other three sugars. C) The two bars for the maltose would be similar to the bars for sucrose. D) The two maltose bars would be similar in length to bars for glucose and fructose. Eliminate

Biology
2 answers:
sladkih [1.3K]2 years ago
8 0
The answer would be C makes the most sense
 
tigry1 [53]2 years ago
7 0

"C) The two bars for the maltose would be similar to the bars for sucrose." This is because maltose falls under the disaccharide type of carbohydrateWe appreciate your questions. Please, never hesitate to ask more in Brainly your queries. 
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Pedigrees are used to help geneticists understand how traits are inherited between generations. The Romanovs, a Russian royal fa
inysia [295]

Answer:

Waldemar carried the recessive allele.

Explanation:

The carrier is the individual that has the affected allele or mutation but does not express the trait, or might express it in different levels. Although, as the person carries the mutation, she or he might transmit the genetic mutation associated with a disease to the progeny.  In general, these diseases are inherited as recessive traits.

So, in the exposed example we know that:

  • hemophilia is a sex-linked disorder
  • hemophilia is determined by a recessive allele on the X chromosome.
  • Irene is a carrier.
  • Her husband is not a carrier.
  • Her children Waldemar and Henry have hemophilia.

If Irene is a carrier, this means that she is heterozygous and that her genotype is X⁺X⁻ (Being the symbol + the dominant allele, and - the recessive one for that expresses the trait)

The fact that Irene´s husband is not a carrier means that his genotype is X⁺Y

Their boys Waldemar and Henry have hemophilia, so both their genotypes are X⁻Y

The best evidence to prove that Irene was heterozygous for hemophilia is that Alice carried the recessive allele.

  • Alice is Irene´s Mother, and she is a carrier as well. Irene´s father, Louis, is not a carrier, so she could have inherited a dominant allele from her father and a recessive allele from her mother, X⁺X⁻, or she could have inherited two dominant alleles from both her parents X⁺X⁺. This is not proof enough of Irene being heterozygous.
  • The fact that Alexandra, Irene´s sister, was also a carrier does not say anything about Irene´s genotype, because they could both share the same genotype or not. This is not proof of Irene being heterozygous.
  • Frederick (her brother) was hemophilic.  He received a recessive allele from Alice, but this does not say anything about Irene´s genotype.  
  • The fact that Waldemar (her son) was hemophilic, is the best evidence to prove that Irene was heterozygous for hemophilia. Walderman received the Y chromosome from his father and an X chromosome from his mother. The X chromosome that he received from his mother carried the recessive allele for the trait, and this is why he had hemophilia. This means that there is no best evidence for Irene´s genotype than her son´s genotype.

 

6 0
2 years ago
A cell with a predominance of free ribosomes is most likely _____.
Lapatulllka [165]

Answer: primarily producing proteins in the cytosol

Explanation:

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2 years ago
A water molecule has to pass through various organelles of a plant cell before reaching the nucleus. List in order the path that
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Answer:

cell wall, cell membrane, cytoplasm, vacuole, ER and nucleous

Explanation:

6 0
2 years ago
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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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An endospore may survive a drought because it is protected by a
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Answer:

An endospore may survive a drought because it is protected by a THICK WALL

Explanation:

An endospore is a structure that is produced by some bacteria. This structure possesses characteristics such as:

a. Endospores have thick walls

b. They are very resistant and resilient to changes in temperature.

c. They are resistant as well to the action or activity of some chemicals.

d. Endospores serves as a means of protection for microorganisms such as Bacteria.

e. Endospores do not reproduce.

f. Endospores can be dormant or Inactive for a very long period of time.

Endospores may survive drought, activity of chemicals, extreme weather conditions, extreme temperatures due to the presence of thick walls in their structure.

7 0
2 years ago
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