Answer:
48 amino acids
Explanation:
The wild type gene codes for a protein with 100 amino acids. One amino acid is encoded by one triplet code of the gene. This means that the wild type gene has a total 100 triplets or 300 nucleotides to code for a protein of 100 amino acid. Mutation in this protein has introduced the code "UAA" at the 49th codon. The code "UAA" is a stop codon. Therefore, the mRNA transcribed from the mutant allele would code for a protein having 48 amino acids as the protein synthesis will be stopped once the stop codon at the 49th position is read.
Answer:
The correct answer would be - 25%.
Explanation:
It is given that the Blue flower is dominant over the yellow flower which means Blue is represented by allele B here and allele b for yellow and Bb is a heterozygous case with a blue phenotypic character. Similarly, the Tall plant is dominant over short and represented by T and t respectively.
BbTt is a heterozygous condition and a cross with itself will produce :
Gametes: BT Bt bT and bt
BT Bt bT bt
BT BBTT BBTt BbTT BbTt
Bt BBTt BBtt BbTt Bbtt
bT BbTT BbTt bbTT bbTt
bt BbTt Bbtt bbTt bbtt
Here 4 out of 16 offspring are heterozygous for both traits represented by bold alphabets. Therefore the correct answer is 25%
Answer:
The correct answer is "reduced initiation of translation".
Explanation:
The ribosome-binding site (RBS) is the sequence of DNA responsible for the recruitment of a ribosome, which results in the initiation of protein translation. A mutation that specifically occurs in a bacterial RBS may result in a reduced initiation of translation. A mutated RBS would likely fail to recruit the ribosome, which will affect the level of initiation of translation.
When light reaction occurs in the stroma of chloroplast glucose is produced
The man couldnt be the dad because there is no B in the mother or father genotype. the dads genotype is ii.
