Answer:photosynthesis
Explanation:
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Answer:Producers make food for the rest of the ecosystem through the process of photosynthesis, where the energy of the sun is used to convert carbon dioxide and water into glucose
Life is a match. Where we are born in the world to be good and righteous. If death is when our lives are powerless. After life we all must die and after death there is Eternal Life, which is the hereafter. If our deeds are right, we will definitely go to heaven. If our actions are evil, it will be hell.
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that liver contains lots of germs
the liver contains an enzyme that breaks down hydrogen peroxide
that hydrogen peroxide is a powerful enzyme
When a piece of liver is dropped into hydrogen peroxide, the peroxide bubbles vigorously as the hydrogen peroxide decomposes into water and oxygen. This suggests the liver contains an enzyme that breaks down hydrogen peroxide. Thus, the correct option is C.
Catalase is a common enzyme which is found in nearly all living organisms which are exposed to oxygen gas. Catalase is a key enzyme which uses hydrogen peroxide, a nonradical Reactive Oxygen Species, as its substrate. Catalase enzyme is responsible for the neutralization through decomposition of hydrogen peroxide, thereby maintaining an optimum level of the molecule in the cell system which is also essential for the cellular signaling processes.
Catalase is an enzyme which causes or speeds up a reaction without being affected during the reaction. The enzyme catalase speeds up the breakdown of hydrogen peroxide into water and oxygen gas.
Therefore, the correct option is C.
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Answer: C
Explanation:
Catalyst is one enzyme from liver that breaks down harmful hydrogen peroxide into oxygen gas and water
s u Tu 70
S u Tu 21
s u tu 4
S U tu 82
s U tu 21
s U Tu 13
S u tu 17__
Total 230
a. Determine the order of these genes on the chromosome.
b. Calculate the map distances between the genes.
c. Determine the coefficient of coincidence and the interference among these genes.
d. Draw the chromosomes of the parents used in the testcross.
Answer and Explanation:
We have the number of descendants of each phenotype product of the tri-hybrid cross.
The total number, N, of individuals is 230.
In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the thrid not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.
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 genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:
Parental)
Double recombinant)
Comparing them we will realize that between
s u TU (parental)
s u tu (double recombinant)
and
S U tu (Parental)
S U TU (double recombinant)
They only change in the position of the alleles TU/tu. This suggests that the position of the gene TU is in the middle of the other two genes, S and U, because in a double recombinant only the central gene changes position in the chromatid.
So, the order of the genes is:
---- S ---- TU -----U ----
In a scheme it would be like:
Chromosome 1:
---s---TU---u--- (Parental chromatid)
---s---tu---u--- (Double Recombinant chromatid)
Chromosome 2
---S---tu---U--- (Parental chromatid)
---S---TU---U--- (Double Recombinant chromatid)
Now we will call Region I to the area between S and TU and Region II to the area between TU and U.
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 S and TU genes, and P2 to the recombination frequency between TU and U.
P1 = (R + DR) / N
P2 = (R + DR)/ N
Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals. So:
P1 = (21+17+4+2)/230
P1 = 44/230
P1 = 0.191
P2 = (21+13+4+2)/230
P1 = 40/230
P1 = 0.174
Now, to calculate the recombination frequency between the two extreme genes, S and U, we can just perform addition or a sum:
P1 + P2= Pt
0.191 + 0.174 = Pt
0.365=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 one of every 100 meiotic products 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 = 0.191 x 100 = 19.1 MU
GD2= P2 x 100 = 0.174 x 100 = 17.4 MU
GD3=Pt x 100 = 0.365 x 100 = 36.5 MU
To calculate the coefficient of coincidence, CC, we must use the next formula:
CC= observed double recombinant frequency/expected double recombinant frequency
Note:
CC= ((2 + 4)/230)/0.174x0.191
CC=(6/230)/0.0332
CC=0.7857
The coefficient of interference, I, is complementary with CC.
I = 1 - CC
I = 1 - 0.7857
I = 0.2143
Answer:
The exchange of oxygen and carbon dioxide between the alveoli and bloodstream (via passive diffusion)
Explanation:
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Answer: The exchange of oxygen and carbon dioxide between the alveoli and bloodstream
Explanation:
I hope this helps
2. respiration by marine animals
3. photosynthesis
4. eating vegetables
5. forest fire
6. fossil fuel formation
Atmosphere Hydrosphere Biosphere Lithosphere
The carbon cycle operates in nature that maintains the carbon composition in nature. The carbon is released into the atmosphere in the form of oxides, hydroxides and then recycle back into the atmosphere.
The carbon can be released into the atmosphere by the respiration of the animals in the form of carbon dioxide. The burning of fossil fuels and wood combustion also releases carbon into the atmosphere.
The decomposition by microbial organisms also releases carbon dioxide in the nature. The main carbon reservoir present in the atmosphere is lime stones in the oceans and in the biomass of the organisms. The carbon can be used up by the plants during the process of photosynthesis.
Therefore, The carbon cycle operates in nature that maintains the carbon composition in nature. The carbon is released into the atmosphere in the form of oxides, hydroxides and then recycle back into the atmosphere.
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Answer:
1 2 4
Explanation: