Hemoglobin is an oxygen‑transport protein found in the red blood cells of all vertebrates. in mice, hemoglobin is a complex of four subunits, 2 α‑globin chains and 2nd β‑globin chains, with each subunit containing a heme molecule. the mouse β‑globin gene is approximately 1,500 nucleotides long; however, the cytoplasmic β‑globin mrna is only about 750 nucleotides long. what's the best explanations for the difference in length.

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Hemoglobin is an oxygen‑transport protein found in the red blood cells of all vertebrates. In mice, hemoglobin is a complex of four subunits, 2 α‑globin chains and 2nd β‑globin chains, with each subunit containing a heme molecule. The mouse β‑globin gene is approximately 1,500 nucleotides long; however, the cytoplasmic β‑globin mRNA is only about 750 nucleotides long. What's the best explanations for the difference in length.

a. The B globin gene is interrupted by one or more introns.  

b. The beginning of the Bglobin gene is not transcribed  

c. The globin mRNA underwent splicing in the cytoplasm.  

d. The B-globin mRNA is processed after being transcribed.  

e.The B-globin gene encodes both of the B-globin subunits.

a. The B globin gene is interrupted by one or more introns.  

d. The B-globin mRNA is processed after being transcribed.  

Explanation:

Vertebrates are eukaryotes and eukaryotic DNA has introns and exons. The introns are the non-protein coding sequences that are present between the protein-coding exons. However, transcription of eukaryotic genes forms the primary transcripts which in turn undergo post-transcriptional modification to become mature mRNA. It includes the splicing of the introns and ligation of exons.

The mature mRNA serves as a template for protein synthesis. Therefore, the mature mRNA of eukaryotes are shorter than their respective genes. The mouse β globin gene must have one or more introns. However, the removal of introns from the primary transcript after transcription would produce a relatively shorter mRNA. The process of splicing of introns occurs in the nucleus itself.

answer: a) a compound microscope has a greater magnification ability than a simple microscope.

explanation:

answer: glucoamylase presented optimum activity at temperature of 55â°c, and, in the glycosidic bonds from the non-reducing ends of starch, resulting in the production of glucose. source and type of fermentation can affect the enzyme's properties

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