"The differences in pre-mRNA splicing that results in an altered pattern of exon inclusion" is most likely to have contributed to this phenomenon.
<u>Option: C</u>
<u>Explanation:</u>
The expression of the eukaryotic gene requires several stages and can be regulated by several of them. Different genes are controlled at different locations and it is not unusual for a gene to be controlled at multiple steps, especially a significant or powerful one.
- In accessibility of Chromatin the chromatin structure includes DNA and can be regulated by its assembling proteins. More free or 'relaxed' chromatin allows a gene more transcriptible.
- For many genes transcription is a key regulatory point. Its factor protein sets bind to unique DNA sequences within or near to a gene and encourage or suppress its transcription into an RNA.
- It is possible to control the splicing, capping, and attaching a poly-A tail to an RNA molecule, and thus exit the nucleus. Specific mRNAs might be produced by alternative splicing from the same pre-mRNA.
Although tube worms cannot produce their own food, they often thrive around these underwater vents. They are able to survive because inside the tube is the trophosome, an organ that contains millions of bacteria in symbiotic relationship with the worm, which are responsible for feeding these creatures that do not have any type of digestive system or visual organs, as they live in the most absolute darkness.
Answer: Black allele has frequency of 0.14.
Red allele has frequency of 0.86.
The frequency of heterozygotes is 0.241
Explanation: <em>Hardy-Weinberg model </em>states the frequency of alleles in a population will stay in equilibrium as long as there are external influences. It is also used to determine alleles frequencies using the following equations:
p represents frequency of dominant allele;
q represents frequency of recessive allele;
2pq is frequency of heterozygotes;
For the frequency of red allele, which is dominant, we can use:
p = 0.86
For the black allele:
q = 0.14
Frequency of heterozygotes is
2pq = 2(0.86)(0.14)
2pq = 0.241
<u>Frequencies for a population of Muscovy ducks are </u><u>0.86 for red allele, 0.14 for black allele and 0.241 for heterozygote.</u>
Answer:
Explanation:
The question step provides the question to be studied, while the hypothesis step provides a proposed answer to the question.
