Completed part of the questionErrors in copying can result in changes in the DNA sequence that could be inherited by future generation.
Answer:
DNA stores genetic information in the sequence of its bases
DNA can be replicated by making complementary copies of each strand
Errors in copying can result in changes in the DNA sequence that could be inherited by future generation.
Explanation:
<u>the fact that DNA is located in the Nucleus is not related to double strands.</u>
<u>the DNA csn change is not relevant ,</u>
<u> therefore the above 3 options are the correct answers.</u>
Acting as levers for movement is a function of: bones.
Movement can be defined as a process through which several skeletal muscles in the body systems act together as a group, in order to cause a change in the motion of a body part.
Basically, bodily movement are produced when skeletal muscles exerts a force on the tendons, which in turn pull on the bones (levers) and other supporting structures such as the skin.
In the body of living organisms, bones act as levers for movement by enhancing limb speed or power of limb movements while joints act as a fixed point of movement (fulcrum).
Read more: brainly.com/question/24279781
The best and most correct answer among the choices provided by the question is the third choice "<span>Epigenetic changes of mother cells are generally maintained in daughter cells; however, they are susceptible to environmental influences." </span>I hope my answer has come to your help. God bless and have a nice day ahead!
Answer:
The genotype of the F1 was wy+/w+y.
Explanation:
One of the given options has a typo: the red eye-brown body offspring count should be 56 instead of 561.
<u>We have two genes with two alleles each:</u>
Red eyes (w+) is dominant over white eyes (w).
Brown body (y+) is dominant over yellow body (y).
The phenotypes of the F2 tesulting from a test cross (F1 x wy/wy) are:
- wy+/ey (white-eye, brown body): 670
- w+y/wy (red-eye, yellow body): 650
- wy/wy (white-eye, yellow body): 38
- w+y+/wy (red-eye, brown body 56
If the genes w and y are linked, two phenotypes in the F2 will be much more abundant than the other two. Recombination during meiosis is a rare event, so the most abundant phenotypes are the parentals (the ones present in the F1 parent).
Every individual in the offpsring has a <em>wy</em> chromosome, as this was the gamete inherited from the test cross individual.
In this case, the most abundant gametes are wy+ and w+y, so the genotype of the F1 was wy+/w+y.
Notice how when recombination occurs in the F1 parent, the recombinant gametes appear: wy and w+y+, which are the less abundant in the F2 progeny.
