Answer:
Venous blood
Explanation:
Arterial blood is carrying oxygen to tissues.....after delivering the oxygen it is venous blood and has less oxygen.
The correct answer should be D) Qualitative and inferential
<h2>Law of independent assortment </h2>
Explanation:
- There are 8 distinct phenotypes, each one has a 12,5% of appearance: Since M=solid leaves D=normal height P= smooth skin.
- Recessive gene characters can only be taken in homozygous recessive mmddpp.
1. MmDdPp: M=solid leaves D=normal height P= smooth skin.
2. MmDdpp M=solid leaves D=normal height pp= peach skin.
3. MmddPp M=solid leaves dd=dwarf height P= smooth skin.
4. Mmddpp M=solid leaves dd=dwarf height pp= peach skin.
5. mmDdPp mm=mottled leaves D=normal height P= smooth skin.
6. mmDdpp mm=mottled leaves D=normal height pp= peach skin.
7. mmddPp mm=mottled leaves dd=dwarf height P= smooth skin.
8. mmddpp mm=mottled leaves dd=dwarf height pp= peach skin.
Hence,The answer is all eight possible phenotypes from the cross MmDdPp x mmddpp are equally likely, resulting in a phenotypic ratio of 1:1:1:1:1:1:1:1
Purines have 2 rings pyrimidines have 1 ring.
A certain bacterial mRNA is known to represent only one gene and to contain about 800 nucleotides. If you assume that the average amino acid residue contributes 110 to the peptide molecular weight, the largest polypeptide that this mRNA could code for would have a molecular weight of about _______.
A) 800.
B) 5,000.
C) 30,000.
D) 80,000.
E) An upper limit cannot be determined from the data given
Answer:
c. 30,000
Explanation:
The gene has a total of 800 nucleotides and would be transcribed into an mRNA having about 800 nucleotides. The nucleotides of mRNA are read in the form of genetic triplets. This means that the mRNA with 800 nucleotides would have about 800/3= 266.66 or 266 genetic codes. This mRNA would be translated into a polypeptide chain having around 266 amino acids.
One amino acid contributes 110 to the molecular weight. The total molecular weight of the largest possible polypeptide that can be encoded by the said gene would b 266 x 110= 29260 (around 30,000).
