Answer: The two possible explanation for the outcome are as follows:
Explanation:
1. The bacterial DNA in the pellet was not completely isolated properly, therefore, no fluorescence was observed in the pellet under the exposure of UV radiation.
2. After 3 hours of induction with the IPTG and labeling with the G3. The IPGT might not be able to bound to the bacterial DNA that could show response of fluorescence in the ultraviolet light.
Answer:
The correct answer would be:
- mRNA sequence - UCACGGAAG,
- amino acid sequence - Ser-Arg-Lys, and
- body type - dwarf
By central dogma, we know that nucleotide sequence of deoxyribonuceic acid (DNA) form the amino acid sequence of a polypeptide chain.
Nucleotide sequence of DNA is first decoded in the form of nucleotide sequence of mRNA (messenger ribonucleic acid) under the process of transcription. The sequence of RNA is complementary to the nucleotide sequence of template strand of DNA. In addition, uracil is present in RNA in place of thymine.
tRNA (transfer RNA) then deciphers the codon sequence of mRNA into amino acid sequence of polypeptide sequence by the process of translation.
Now, given DNA sequence is AGTGCCTTC.
so, the mRNA sequence would be UCACGGAAG.
Codon sequence is UCA CGG AAG.
So, the amino acid sequence would be Ser-Arg-Lys.
Hence, the trait of showman performer would be dwarfism.
Codon sequence chart is attached for reference.
Hello,
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Hope this helped,
j548831
Answer:
In order to find average speed during each interval, we need to divide the distance during those intervals with the period of time. So, for the first interval (day 0 to day 2) hawksbill started from 0 and reached 10 kilometers by the end of the second day. That means that it crossed 10 kilometers in 2 days, so the average speed is 10/2 which is 5 km/day. Similarly, we can calculate speed for other intervals:
• day 2 - day 3: it went from 10 to 12 km in one day, which means it crossed 2 km in one day, so the average speed is 2/1 = 2 km/day
• day 3 - day 4: at the end of the third day it reached 12 km and at the end of the day 4 it remained at 12 km. That means the hawksbill wasn't moving in that interval so the speed was 0
• day 4 - day 5: it went from 12 km to 18 km, which means it crossed 18-12=6 km in one day, so the average speed is 6/1=6 km/day
• day 5 - day 6: it went from 18 to 24 km, which means it crossed 24-18=6 km in one day, so the speed was 6/1=6 km/day
So, to summarize, during the first interval turtle was moving with average speed of 5 km/day, then 2 km/day, in the third interval it wasn't moving and in the last two intervals, it moved in average speed of 6 km/day.
