When placed in tap water, which is a hypotonic environment, animal cells have a higher solute concentration inside of the cell compared to the tap water. ... This, however, is not the case for plant cells which has a strategy to prevent lysis when put in a hypotonic environment. So the answer to your question is D
<h2>Answer is option " 2 & 4"</h2>
Explanation:
- Two types of posterity can be created subsequent to mating: parental and recombinant. Recombinants are created because of the traverse of non-sister chromatids at the hour of gamete arrangement. Here, mating happens between m+g/mg+ female and mg/mg male so the posterity would be separated as follows:
- m+g/mg : Parental
- mg+/mg : Parental
- m+g+/mg : Recombinant
- mg/mg : Recombinant
- m+g+/mg (wild sort) and mg/mg (smaller than usual wings, garnet eyes) are the recombinants thus they were created because of the hybrid occasion.
- Hence,the right answer is option 2 & 4 "wild type and miniature wings,garnet eyes"
Genotype - RR - 25%, Rr - 50%, rr - 25% (1:2:1)
Phenotype - Round seeds - 75%, Wrinkled seeds - 25% (3:1)
<h3>How explain your answer?</h3>
Let the letter "r" stand for the alleles, where R is round seeds and r is wrinkled seeds. A genotype is an individual's genes represented through alleles. Phenotypes are how the genes express themselves. In other words, genotypes will be written using letters, the alleles, and phenotypes will be the possible outcomes of the alleles.
Both of the parent seeds have the genotypes Rr and the phenotype of round seeds.
If you create punnet square (which had four boxes in total) 1 will have RR, 2 will have Rr, and 1 will have rr. These are the ratios for the genotypes. Each box represents 25%, so the percentages will be 25, 50, and 25. Finally, 3 of these boxes (RR and Rr) will result in round seeds because those are dominant. Only the genotype rr will result in wrinkled phenotype. Therefore, the ratio is 3:1 or 75% to 25%.
Thus, this could be the answer.
To learn more about genotypes and phenotypes click here:
brainly.com/question/20730322
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Answer:
5
4
1
3
Explanation:
In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon 1 when looking at the Fischer projection.
In D-Fructose, there is a ketone functional group, and the carbonyl group is at carbon 2 when looking at the Fischer projection.
Both of your questions say "glucose" so I gave you my answer to the best of my understanding which required me to change the second answer to FRUCTOSE.