60 centimeters (2 ft 0 in) to over 260 cm( 8 ft 6 in)
The most likely answer choice should and would be B, given that rabbits are herbivores (meaning that they only eat vegetation and plants, including grass). Buffalo's food source(s) and diet consists of grass as well. If enough rabbits are introduced into a grassland area, they will furthermore consume more grass, in turn limiting the food source of the buffalo.
''Attacking'' prey doesn't necessarily make sense, given that buffalos do not 'prey' on any animal and rabbits themselves are prey.
A to a lesser extent can make sense, although I doubtable. B's most likely the answer.
<span>You would know you were in the troposphere if you see stuff like the clouds because it is earths lowest layer. </span>
Answer:
DNA → TACCATGGAATTACT
RNA → AUGGUACCUUAAUGA
PROTEIN → Methionine-Valine-Proline-Stop codon-Stop codon (AUG GUA CCU UAA UGA)
Explanation:
In nucleic acids (i.e., DNA and RNA), base complementarity refers to the interaction between antiparallel strands. In the double helix DNA molecule, adenine always interacts with thymine (uracil in RNA), while cytosine always interacts with guanine. Moreover, amino acids are encoded by codons, i.e., triplets of nucleotides in the messenger RNA (mRNA). Finally, stop codons are triplets of mRNA nucleotides (e.g., UAG, UAA, UGA) that indicates the end of the protein-coding sequence.
<span>F- allele for freckles
f- </span><span>allele without freckles
1) The man is heterozygote and has freckles, its indicating that the allele for freckles is dominant.
A cross between him and a woman who is also </span><span>heterozygote: Ff x Ff
it would result in the following probabilities:
- 1/4 - homozygote with freckles: FF
- 2/4 - </span><span>heterozygote with freckles: Ff
- 1/4- </span><span>homozygote without freckles:ff
Their son would have a probability of 75% of being born with freckles.
2) The cross resulted in this probabilities:
</span><span><span>- 1/4 - homozygote with freckles: FF
- 2/4 - </span><span>heterozygote with freckles: Ff
- 1/4- </span><span>homozygote without freckles:ff
So, the chance of being born heterozygote for this gene is 2/4, which is the same as half (50%).
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