<span>If the energy in glucose was released at once, most of the energy would be lost as light and heat. The light and heat could harm or even destroy the cell. The gradual process of cellular respiration allows the cell to control the release of energy into packages of ATP that can be used efficiently for cell activities.</span>
Answer:it described how carbon is recycled between earth's biosphere, hydrosphere, geosphere and atmosphere. Carbon is an essential element for all life forms so understanding how it is recycled will help us to understand factors and all the biological processes that influence them.
Explanation:
A. are required for the expression of specific protein-encoding genes.
B. bind to other proteins or to a sequence element within the promoter called the TATA box.
C. inhibit RNA polymerase binding to the promoter and begin transcribing.
D. usually lead to a high level of transcription even without additional specific transcription factors.
E. bind to sequences just after the start site of transcription.
Answer: B
Answer:
Cross each to a rabbit who you know is homozygous - i.e. a white rabbit. If there are any white offspring you know this is the heterozygote.
Explanation:
Heterozygous individuals are those with 2 different alleles of a gene. Homozygous individuals have have 2 alleles that are the same.
Imagine the allele for coat color is B black, or b white. If you cross 2 individuals who are homozygous for the black gene (BB), they would only produce black rabbits in the F1 (BB). If you cross the F1, they could only produce BB rabbits in the F2.
However, if you cross a heterozygous rabbit with a homozygous black rabbit (Bb x BB), you would get either BB or Bb rabbits in the F1. However, intercrossing them could produce BB, Bb, or bb rabbits. Therefore, white rabbits can be produced.
You know that white rabbits are bb. So if you are unsure about the genotype of the black rabbit, you can cross it with a white rabbit (either BB x bb or Bb x bb). If any white rabbits appear in the F1, you know there must be a b allele in the black rabbit genotype, so that rabbit must be heterozygous.
The answer is D) 1/2.
If each allele confers a unit of color darkness, then only genotypes S2S3 and S1S4 will have 5 units of color darkness because in genotype S2S3 S2 will have 2 units, and S3 will have 3 units, which is 5 units in total. The similar math is for <span>genotypes S1S4.
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Now, let's cross S1S3 and S2S4:
Parents: S1S3 x S2S4
Offspring: S1S2 S1S4 S2S3 S3S4
The offspring with genotype S1S2 will have 3 units of color darkness. The offspring with genotype S1S4 will have 5 units, as well as the offspring with genotype S2S3. The <span>offspring with genotype S3S4 will have 7 units.
It can be assumed that only 2 genotypes out of 4 will have 5 units:
2/4 = 1/2.</span>
