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:
substitution - a base was changed
Explanation:
The nucleotide sequence CTT was changed to the sequence CAT. The T was substituted with an A. This changed the encoded amino acid from Glu to Val.
An insertion is where an additional base is added (e.g. if the sequence changed from CTT to CATT)
A deletion is when a base is lost (e.g. if the sequence changed from CTT to CT)
Answer:
0.033
Explanation:
Tay–Sachs disease is an autosomal recessive disorder. The possible genotypes and phenotypes are:
- TT = normal
- Tt = carrier
- tt = Tay-Sachs disease
I will use <em>p </em>to call the frequency of the dominant <em>T</em> allele, and <em>q</em> the frequency of the recessive <em>t</em> allele.
If the population is in equilibrium, the frequency of the tt genotype is q².
The frequency of the Tt genotype is 2pq.
The Tay–Sachs carrier frequency will be 0.033
<h2>
Answer: T t</h2>
<h2>
Explanation: T TT Tt</h2><h3> t Tt tt</h3><h3>this is the way your supposed to set it up</h3><h3>hopes this helps</h3>
<span>The ability to choose a mate does improve offspring fitness in fruit flies. The choice made by the female fruit fly is based on a recognition pattern of the male's ability to highlight its potency and "genes".
As a conclusion, the better the male's genes, the better the offspring's fitness.</span>
