2) Cross-fertilization makes for more variation in the traits of plants, giving them immunities and resistances to weather, pests, and disease.
Correct answer: Option D- DNA ligase
Explanation: In option A, thymine is a nucleotide, so it is present throughout the replication process, wherever it is required. It is added to the newly formed DNA. In option B, Helicase enzyme is active during initiation and elongation stage, as it facilitates the opening of the winded DNA strands. Option C is nucleotidase and it has no role in DNA replication. So, the correct answer is DNA ligase, which is option D.
The okazaki fragments formed during DNA replication are sealed at the end. And in this step, DNA ligase is used. It catalyzes the formation of phosphodiester bond between the nucleotides of okazaki fragments. So it is the last active molecule of the process.
Answer:
The cell membrane is semipermeable because allows only certain molecules to pass through.
Explanation:
Semipermeability is a common property of cell membranes, due to their predominantly lipid composition, which gives them a hydrophobic character.
The cytoplasmic membrane is constituted by a double layer of lipids, in addition to structural proteins and glucides, which prevent the entry of polar or charged molecules. This characteristic allows lipids and small molecules without charge to pass freely through it.
Regarding the other options:
<em> A. Cell membranes are associated with lysosomes, structures that contain enzymes. However, this characteristic is not related to semipermeability.</em>
<em> C. They allow the passage of small molecules - devoid of charge - through them.</em>
<em> D. Not all non-polar molecules pass through, as some need a transporter.</em>
Answer + Explanation :
Bacteria :
1) Approximately diameter is 1-5 micrometer
2) They are prokaryotic.
3) They have rigid cell wall containing peptidoglycan.
4) Replicate by binary fission
Fungi :
1) Approximately 3-10 micrometers in diameter.
2) They are eukaryotic.
3) Rigid cell wall containing chitin.
4) Replicate by budding or mitosis.
<u>Viruses :</u>
1) Approximately 0.02-0.2 in diameter.
2) They are eukaryotic.
3) They contain protien capsid and lipoprotien envelope.
4) Donot replicate by binary fission.
<u>Archae :</u>
1) Not typically associated with human disease.
2) Found in extreme environments.
3) Cell wall doesn't contain peptidoglycan.
Answer:
75% would have the dominant traits for coat length, 25% would have recessive trait for coat length.
Explanation:
After completing a punnet square, we could find that our genotypes are 25% LL, 50% Ll, and 25% ll.
If these genotypes were to be physically expressed, LL and Ll would both be expressed as showing the dominant trait.
This means that 75% would have the dominant traits for coat length and 25% would have recessive trait for coat length.
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