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SSSSS [86.1K]
4 years ago
5

In which sealed container would the organisms be able to continuously cycle O2 and CO2 gases?

Biology
2 answers:
MAXImum [283]4 years ago
7 0

Answer:

The correct answer is C. An aquarium with water plants and a snail

Explanation:

In an aquarium with water plants and snail the flow of oxygen and carbon dioxide will be continuous because plants are able to do photosynthesis which release the oxygen as the byproduct and this oxygen will be used by snail and snail during respiration release CO2 which will be used by plants in the process of photosynthesis to make food for itself.

Therefore there will be symbiotic relationship between plants and snail because snail provides carbon dioxide to plant and plant provide oxygen to the snail and support their survival. Therefore in this container organism would be able to continuously cycle O2 and CO2. So the right answer is C.

Levart [38]4 years ago
3 0
The correct answer is C.
In the aquarium, the water plants would provide the O2 molecules necessary for the snail's metabolism and the snail would release the CO2 molecules that are a byproduct of his metabolism into the container. The CO2 would later be used by plants to make sugar in the process of photosynthesis, releasing more O2 and the cycle of two gases would be completed.
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In Drosophila, the genes crossveinless and Stubble are linked, about 7 map units apart on chromosome 3. cv is a recessive mutant
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Answer:

0.035

Explanation:

<u>cv+ is the wild-type dominant allele over cv, therefore:</u>

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<u>Sb is a dominant mutant allele over wild-type Sb+, therefore:</u>

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<h3><u>Test cross</u></h3>

It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.

\frac{cv\   Sb^+}{cv^+\ Sb}  X \frac{cv \ Sb+}{cv \ Sb+}

-The male produces only 1 type of gamete: cv Sb+

-The female produces 4 types of gametes:

  • cv Sb+   ] Parental
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  • cv+ Sb+ ] Recombinant

The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.

The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.

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Explanation:

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