Ribosomal RNA<span> (rRNA) associates </span>with<span> a set of </span>proteins<span> to form ribosomes. These complex structures, which physically move along an mRNA </span>molecule<span>, catalyze the assembly of amino acids into </span>protein<span> chains. They also bind tRNAs and various accessory </span>molecules<span> necessary for </span>protein<span> synthesis.</span>
Answer:
All of the above answer choices are correct.
Explanation:
Test cross is done to find out the genotype of an individual displaying dominant phenotype as it can be homozygous or heterozygous. To find this the individual is crossed with a recessive phenotype individual. For example: a dominant trait tall height can be homozygous TT or heterozygous Tt. If it is TT all the offspring of test cross with tt will be tall. If it is Tt half of the offspring will be tall and half of the offspring will be short.
Multiple offspring are required to come to the final result because offspring production happens in random order and it might take a few tries before another type of phenotype is produced. For example: If a test cross produces an individual with dominant phenotype we can still not surely say if the test individual is homozygous or heterozygous because both can produce dominant phenotype in test cross. We need more offspring to check if the recessive phenotype is produced or not and accordingly decide the genotype of test individual.
Hence all of the above answer choices are correct.
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.
Answer: That act helps endangered species from being extinct
Explanation:
