The correct answer is: gaps left at the 5' end of the lagging strand
Telomeres are repetitive regions at the very ends of chromosomes found in eukaryotic organisms. The “problem” with telomeres is that the DNA replication cannot be fully finished in each round which result in a slow, gradual shortening of the chromosome.
Gaps at the end of lagging strand are formed because lagging strand is synthesized via Okazaki fragments (small DNA fragments). So, when replication fork during replication reaches the end of the chromosome, a short stretch of DNA does not get covered by an Okazaki fragment and consequently remains uncopied in each round of replication, leaving a single-stranded overhang.
Telomerase is an enzyme that solves this problem by extending the telomeres of chromosomes.
Answer:
Please see below
Explanation:
The white board displays helpful info with regards to a QR scanner associated with the day's lesson plan. The shelves on the far end of the classroom holds up relevant books and other objects that will prove educational for the students. Some schedules have been put up too for the ease of the students.
Answer: No. You would not withdraw CSF if the needle is in the epidural space.
Explanation: There are 3 spaces that cover the spinal cord, which are epidural, subdural and subarachnoid spaces. The epidural space is the outer most space while the subarachnoid space is the inner most space. CSF flows from the brain where it is produced to the spinal cord. In the spinal cord it flows only in the subarachnoid space. Thus you would not be able to withdraw CSF if the needle is in the epidural space.
Answer:
See the answer below
Explanation:
Let the disorder be represented by the allele a.
Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.
Since the four adults are carriers, their genotypes would be Aa.
Aa x Aa
Progeny: AA 2Aa aa
Probability of being affected = 1/4
Probability of being a carrier = 1/2
Probability of not being affected = 3/4
(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2
(b) The chance that the third child of Sara and James will be free of the condition = 3/4
(c)
(d) If someone has no family history of the disorder, their genotype would be AA.
AA x aa
4 Aa
<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0
(e)
(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0
