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Veronika [31]
2 years ago
11

If A has a free energy of 38 units, and C has a free energy of 45 units, and the reaction is exergonic, based on the calculation

of free energy, which is the substrate and which is the product__________
Biology
1 answer:
Vesnalui [34]2 years ago
7 0

Answer:

C is the substrate

A is the product

Explanation:

Exergonic reactions are the ones that release energy during the course of the conversion of substrates into the products. The free energy content of the substrates of exergonic reactions is higher than that of their products. This excess energy is released during the reaction. In the given reaction, the free energy content of C is 45 units which is higher than the free energy content of A (38 units). This means that the substrate "C" has undergone an exergonic reaction and is being converted into the product "A".

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Three linked autosomal loci were studied in smurfs.
cupoosta [38]

Answer:

height -------- color --------- mood

           (13.2cM)      (14.5cM)

C=0.421

I = 0.579

Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

Phenotype Number

  • pink, tall, happy            580
  • blue, dwarf, gloomy     601
  • pink, tall, gloomy         113
  • blue, dwarf, happy      107
  • blue, tall, happy              8
  • pink, dwarf, gloomy        6
  • blue, tall, gloomy          98
  • pink, dwarf, happy      101

Total number of individuals = 1614 = N

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the phenotypes of the parental with the ones of the double recombinants. We can recognize the parental in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

  • Pink, tall, happy            580 individuals
  • Blue, dwarf, gloomy      601 individuals

Simple recombinant)

  • Pink Tall Gloomy           113 individuals
  • Blue, Dwarf, Happy       107 individuals
  • Blue Tall Gloomy             98 individuals
  • Pink Dwarf Happy          101 individuals

Double Recombinant)  

  • Blue Tall Happy                 8 individuals
  • Pink  Dwarf Gloomy           6 individuals  

Comparing them we realize that parental and double recombinant individuals differ in the position of the gene codifying for <u>color</u><u>.</u> They only change in the position of Blue and Pink. This suggests that the position of the color gene is in the middle of the other two genes, height and mood, because in a double recombinant only the central gene changes position in the chromatid.  

So, the alphabetic order of the genes is:

---- height ---- color ----- mood ----

Now we will call Region I to the area between Height and Color, and Region II to the area between Color and Mood.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between Height and color genes, and P2 to the recombination frequency between color and mood.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region (the ones that have an intermediate phenotypic frequency), DR is the number of double recombinants in each region, and N is the total number of individuals.  So:

Region I

Tall------ Pink--------happy  (Parental) 580 individuals

Dwarf ---Pink------- Happy (Simple Recombinant) 101 individuals

Dwarf--- Pink-------Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Tall ------Blue------- Gloomy (Simple Recombinant)  98 individuals

Tall ----- Blue------- Happy   (Double Recombinant) 8 individuals  

Region II

Tall------ Pink--------happy (Parental) 580 individuals

Tall-------Pink------- Gloomy (Simple Recombinant) 113 individuals

Dwarf----Pink------- Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Dwarf ----Blue-------Happy (Simple Recombinant) 107 individuals

Tall ----- Blue------- Happy   (Double Recombinant) 8 individuals

In each region, the highlighted traits are the ones that suffered recombination.

  • P1 = (R + DR) / N

P1 = (101+6+98+8)/1614

P1 = 213/1614

P1 = 0.132    

  • P2= = (R + DR) / N

P2 = (113+6+107+8)/1614

P1 = 234/1614

P1 = 0.145

Now, to calculate the recombination frequency between the two extreme genes, height and mood, we can just perform addition or a sum:

  • P1 + P2= Pt

0.132 + 0.145 = Pt

0.277=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).  

The map unit is the distance between the pair of genes for which every 100 meiotic products, one results in a recombinant product.  

Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.132 x 100 = 13.2 MU = 13.2 cM

GD2= P2 x 100 = 0.145 x 100 = 14.5 MU = 14.5 cM

GD3=Pt x 100 = 0.277 x 100 = 27.7 MU = 27.7 cM

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:  

-observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals

-expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

  • CC= ((6 + 8)/1614)/0.132x0.145

        CC=0.008/0.019

        CC=0.421

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.421

I = 0.579

8 0
3 years ago
Explain how the inheritance pattern shown in figure 1 supports a role for chloroplast genes in the production of chlorphyll in t
Aloiza [94]

Answer:

Answer b) The result of the experiment identifies that the color of the egg donor branch (parent) prevailed and assigned the color of the offspring.

-The female parental branches that were pure green or pure white had children of pure green or pure white, respectively.

- The female parental branches that were variegated allow to obtain the three types of offspring.

It is evident that the chloroplast shows maternal inheritance.

The branch that is pure green will produce eggs with green chloroplasts that will give rise to a pure green offspring. Thus, a pure white branch will have offspring with ovules with an exclusive content of white chloroplasts and will give rise to a pure white offspring.

If a branch is variegated, it is combined, some with only functional chloroplasts, some with only non-functional chloroplasts, and some with a mixture of chloroplasts. All three types of cells can give rise to ovules, leading to green offspring, white offspring, and variegated offspring.

8 0
2 years ago
Write any two adaptational characteristics of camel on the basis of habitat.​
Vera_Pavlovna [14]
Long eyelashes that doesnt allow the sand to get into its eyes
And its ability to store water for long periods of time in its hump
5 0
2 years ago
What is “the balance of nature”?
Kamila [148]

Answer:

<h3>The balance of nature is a theory that proposes that ecological systems are usually in a stable equilibrium or homeostasis,which is to say that a small change will be corrected by some negative feedback that will bring the parameter back to its original "point of balance" with the rest of the system.</h3>

Explanation:

<h3>I hope this will help you/helpful for you.</h3>

5 0
2 years ago
In which process are the products of the reaction glucose and oxygen? photosynthesis cellular respiration
Mariana [72]

Answer:

That would be photosynthesis. Glucose and oxygen are rather reactants for cellular respiration

6 0
2 years ago
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