B- black allele;
b- brown allele;
The parental cross is true-breeding because both the male and the female are homo zygotes, so: BBx bb
The F1 generation will be all black heterozygote: Bb
The F2 generation will be between mice of the F1 generation and they all have the same genotype:Bb xBb
In this cross it would result in:
1/4 of the plants being black homozygote-BB 25%
2/4 of the plants being black heterozygote- Bb- 50%
1/4 of the plants being brown homozygote- bb- 25%
answer: 25%+25%=50% are homozygotes
It exhibits both cleavage and fracture.
"Hardness is 10. Specific Gravity is 3.5 (above average) Cleavage is perfect in 4 directions forming octahedrons. Fracture is conchoidal."
-galleries.com (diamonds)
Answer: Bacteria lack a mechanism for splicing out introns
Explanation:
Factor VII gene is 186k nucleotides long while the protein is 2332 amino acids long. <u>This lenght discrepancy is due to introns interrupting gene</u>, because the cell first transcribes the entire gene and then cuts introns out of the transcript. At the end, it splices the remaining pieces. Prokaryotes don't perform splicing so it can not edit out introns from the primary mRNA transcript. To produce an eukaryote gene in prokaryotes it is necessary to use a reverse transcriptase to get a cDNA sequence without the introns, and then insert that into a bacterial genome.
The hormone Progesterone increases in the blood to a level that causes lh to be released from the anterior pituitary.]
The correct answer is “B”
I hope this helps
