Although in many regions of the brain—particularly those that give rise to nuclear cell groups—neurons migrate without the benefit of glial guides, migration along radial glial fibers is always seen in regions where cells are organized into layers, like the cerebral cortex, hippocampus, and cerebellum.
Answer:
The correct answer is c) They are used to form ATP by chemiosmosis
Explanation:
In cellular respiration, glucose is oxidized into carbon dioxide and water and converts into NADH a FADH₂ during glycolysis and Kreb cycle. Then these reduced molecules are used to provide the source of electrons during chemiosmosis.
This transfer of electron to electron transport chain brings the conformation change in proteins that helps in transport of H⁺ions across the inner membrane of mitochondria. This transport of ions increase the proton gradient in the membrane.
So to equalize the proton gradient H⁺ moves towards the matrix through ATPase which allows the formation of ATP from ADP. Therefore the right answer is c) They are used to form ATP by chemiosmosis.
i think it is inner core.
Answer:
Without DNA, cells could not reproduce, which would mean the extinction of the species. Normally, the nucleus makes copies of chromosomal DNA, then segments of DNA recombine, and next the chromosomes divide twice, forming four haploid egg or sperm cells.
Explanation:
a) Evolutionary fitness can be characterized as a person's commitment to the cutting edge's quality pool,which is chiefly founded on the aggregate or the genotype of the individual.This principally identifies with the conceptive achievement of an organic entity.
It is fundamentally determined by the offspring that can endure and pass on the characteristics/traits.
b) Graph is already uploaded in the below attached document.
c) The population 3 will have the most elevated frequency of heterozygotes,45% of the complete population.
d) Since population 2 is as of now having a low number of heterozygotes, the occasion would totally eliminate them from the population or decreases their number
Explanation:
Frequency of animals in population 1 having both alleles 1 and 3
frequency of allele 1 =0.60
frequency of allele 3= 0.25
Frequency of both allele 1 and 3 would be allele 1 + allele 3 frequency x 100
= 0.60+0.25 X 100= 85%
