Answer: DF508 mutation. A Genetic, Hereditary, Autosomal and Recessive Mutation.
Explanation:
Cystic fibrosis (CF) is a recessive autosomal lethal disease, it is most common on Caucasoid populations. Its diagnosis is suggested by the clinical features of chronic obstructive pulmonary disease, persistent pulmonary colonization (particularly with mucoid Pseudomonas strains), meconium ileus, pancreatic insufficiency with or familiarity history of the disease. The FC gene is large, with about 250 Kb of genomic DNA, 27 exons representing about 5% of genomic DNA; encodes a 6.5 kb transcribed mRNA. This mRNA is transcribed into a protein of 1480 amino acid called CFTR (Regulator Transmembrane Conductance Cystic Fibrosis). When a three-base pair deletion, adenosine-thymine-thymine (ATT) identified in the CFTR gene, exon 10, it results in the loss of a single amino acid phenylalanine at position 508 of the protein. This mutation is called DF508; “D” stands for deletion and “F” for phenylalanine amino acid.
Cast- solid copy
Mold- hollow form
Carbon film- details left after an organism decays
Trace fossil-record of an organisms activity
Petrified fossils- living things turned into rock
It would decrease as their prime food source would be taken away. This would result in a decline in population as the birds wouldn’t have sufficient food.
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Answer: and Explanation:
A.)The reason for the different products of glycogen breakdown in the two tissues is that glucose 6-phosphotase which is
a known enzyme that brings about hydrolysis of glucose 6-phosphate as a result of the creation of a phosphate group and free glucose is not available in heart and skeletal muscle, therefore,any glucose 6-phosphotase that is produced will just enters the glycolytic pathway and get converted to lactate through pyruvate, in the absence of Oxygen O2.
B) Whenever a situation involving fight or flight arises, the concentration of glycolytic precursors becomes high in order to prepare for muscular activity. Since the membrane is impermeable to any charged species, and at the same time glucose 6-phosphotase enzyme cannot be moved through the glucose transporter, then there cannot be a release of Phosphorylated intermediates from the cell. The blood glucose level must be maintained by the liver by releasing of glucose.
glucose that is later formed from glucose 6-phosphotase then enters the bloodstream.
Yes this is true 2 diploids cells can fuse to form haploid cell
