Answer:
b. AB
e. aB
Explanation:
Gametes are haploid and are produced through meiotic division in the reproductive cells.. The given individual in the question has genotype AaBB. Since the given individual has heterozygous for Aa and homozygous for BB. Thus, two types of gametes will be formed after meiosis. Therefore, meiosis in AaBB individual will produce AB and aB type of gametes.
Answer and Explanation:
Bovine pancreatic trypsin inhibitor (BPT I) is a small globular protein and it inhibits the proteolytic enzymes like trypsin. BPTI is composed of ∝ helices, β sheets and 3 disulfide bonds. Due to these BPTI is a stable protein in its tertiary structure. It is almost inert to denaturation by urea and exhibits denaturation below 100 degree, only in highly acidic solutions. When all the disulfide bonds in BPTI are reduced, the protein is unfolded at room temperature and can reform three correct S-S pairings in native confirmation. if the 6 cysteine residues are reduced and unfolded in urea, the re-oxidation would yield 3 pairs with probability of first pair with 5, second pair with 3 and the third pair with 1 cysteine residues. Therefore, 5 x 3 x 1 = 15 combinations are possible accounting for 7% of protein refolding.
5 is a happens immediately after <span>meiosis.
</span>6 is b r<span>esults in genetically identical cells.</span>
