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SashulF [63]
3 years ago
12

C4 plants occur more commonly in desert conditions because _____. a. they can fix carbon at the lower CO2 concentrations that de

velop when the stomata are closed b. they store carbon by incorporating CO2 into organic acids that are later catabolized c. they produce water as a product of their photosynthetic pathways d. they produce carbon dioxide internally via photorespiration the stomata open at night and close in the day
Biology
2 answers:
ahrayia [7]3 years ago
6 0

Answer:

a. they can fix carbon at the lower CO2 concentrations that develop when the stomata are closed

Explanation:

C4 plants are those that have the capacity to fix CO2 even when a tiny concentration of it is available.

Desert condition is characterized by dryness and strong heat with both condition capable of creating water stress in plant as a result of evapotranspiration. In order to reduce evapotranspiration rate, desert plants (most C4 plants) close their stomata during the day.

<em>Stomata closure limit the diffusion of CO2 into desert plants and the small concentration of CO2 that results is utilized by a special enzyme in the plants.</em>

The correct option is a.

Setler79 [48]3 years ago
5 0

Answer:

They can fix carbon at the lower CO2 concentrations that develop when the stomata are closed.

Explanation:

In order to conserve   the substrate Ribulose biphosphate(RuBP)  which  must react with CO2,, and to avoid  wasteful photorespiration reaction  of RuBP  with oxygen. which are  both  catalysed by enzyme Rubisco; C4 plants formed a 4C-compounds -oxaloacatate . While keeping RuBP and enzyme Rubisco  in the bundle sheath away from oxygen.

The C02 is rather absorbed  by mesopyll cell in contact with air and  combines with 3C-Phosphoenopyruvate, a reaction catalysed by Phosphoenocarboxylase, to form 4C-oxaloacetate.

The latter is converted to 4C -Malate, which is shunted from the mesophyll into the bundle shealth where Rubisco and RuBP are concentrated.

Therefore when the stomata is closed,C4 plants can  fixed Carbon as Oxaloacetate and  Malate,  in the bundle sheath,despite the closed Stomata, because of the anatomy which separated the externally contact mesophyll cells from the airtight bundle sheath cells.

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Which model of replication did the actual results of the Meselson-Stahl experiment support? Explain your answer.
AfilCa [17]

Answer:

Semi-conservative replication

Explanation:

After the double-helix discovery of Watson and Crick, there were three possible models about the DNI replication:

  • The Conservative model stated that the two strands of DNI together were the template of another new molecule. The final product was the original double-stranded molecule and the new molecule.
  • The semi-conservative model stated that the original DNI molecule separated into two strands, and each of them served as a template for the synthesis of a new complementary strand. The replication product would be two double-stranded DNA molecules, each carrying an original strand a new one.
  • The Dispersive moles stated that the replication product would be two molecules made by a mixture of segments of the original and the new molecules.

Meselson and Stahl joined to discover which of the models was the correct one. To do it they used E. coli and Nitrogen isotopes.

  • First, they extracted DNI from bacteria grown in a medium with N¹⁴ and got its density band by centrifugation.
  • Then they grew bacteria in a medium with N¹⁵, extracted their DNI molecules, centrifugated them, and got the density band, which was heavier than the firsts ones.
  • The researchers then transferred bacteria grown in medium with N¹⁵  to a medium with N¹⁴, and they allowed only one replication process to occur. DNI was extracted and centrifugated again, and a new band appeared. This band was an intermediate form between bands of DNI-N¹⁵ and DNI-N¹⁴.

This event <em>eliminated the conservative model</em>. If this model were correct, the expected result would be to get two bands: one corresponding to the density DNI-N¹⁵ and the other corresponding to the density DNI-N¹⁴.

  • Bacteria grown in a medium with N¹⁵ and then transferred to a medium with N¹⁴ were finally allowed to replicate twice. Their DNI was extracted and centrifugated. The result was two bands: one of them coincided with the intermediate band, and the other one with the DNI-N¹⁴.

<u>This result was conclusive</u> because if the dispersal model were correct, these two bands should not appear, as all the DNI strands would have part of the original molecule.

With this experiment, Meselson and Stahl proved that the correct replication model was the semi-conservative one.

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You should wash your cut before and after swimming so that bacteria in the water can't infect the wound.

Let me know if its right <3

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Defend the metric system
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A dihybrid cross produces 30 recombinant offspring out of a total of 1,000 offspring. What is the recombination frequency of the
uranmaximum [27]

The recombination frequency of the two gene pairs is 3%.

The number of recombinant offspring(r.o.) / total number of offspring x 100% = recombination frequency(θ):

r.o./total x 100% = θ

30/100 x 100% = 3%

Recombinant offspring are children that have a different allele combination to their parents.

For example, say a mother has a haploid cell with the alleles AB and the father has a haploid cell with the alleles ab. These combine to make a diploid cell with the sequence Aa+Bb.

Formation of Recombinant Offspring :

Recombination can happen in two different ways; independent assortment and crossing over.

  • Independent assortment is when the maternal and parental DNA are mixed during meiosis, creating a new gene sequence.
  • Crossing over happens during the first stage of meiosis when the two homologous chromosomes are paired and a portion breaks off on the same loci then reconnects to a different end. Crossing over can only happen when there isn't a physical linkage of the parental alleles.

Recombination frequency (θ) is the frequency with which a single chromosomal crossover will take place between two genes during meiosis. A centimorgan (cM) is a unit that describes a recombination frequency of 1%. In this way we can measure the genetic distance between two loci, based upon their recombination frequency. This is a good estimate of the real distance. Double crossovers would turn into no recombination. In this case we cannot tell if crossovers took place. If the loci we're analysing are very close (less than 7 cM) a double crossover is very unlikely. When distances become higher, the likelihood of a double crossover increases. As the likelihood of a double crossover increases we systematically underestimate the genetic distance between two loci.

When two genes are close together on the same chromosome, they do not assort independently and are said to be linked. Whereas genes located on different chromosomes assort independently and have a recombination frequency of 50%, linked genes have a recombination frequency that is less than 50%.

To learn more about Recombination frequency :  brainly.com/question/7299933

#SPJ4

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1 year ago
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