Answer:Eukaryotes only (A) and (D), bacteria only (E) both (B) and (C).
Explanation:
(A) Only eukaryotes have a nucleus.
(B) Both replicate the DNA before performing cell division, so each cell keeps a copy.
(C) This structural feature is called centromere in eukaryotes and in bacteria this role is played by the origin of replication.
(D) Centromeres are eukaryotes exclusive.
(E) Only bacteria have a replication origin.
Answer:
The correct answer will be option-D
Explanation:
The evolution of vascular tissues in plants from non-vascular plants was an adaptive feature to survive in the terrestrial environment. The group of plants in which vascular tissues evolved and developed is the Pteridophytes which contained stellar organization of the vascular bundles.
The pteridophytes exhibited xylem and phloem which allows the rapid uptake of the water and nutrients from the soil which allowed the growth of the plant and since plants got enough water and nutrients from the soil, they adapted to grow tall so that they can maximize the output of photosynthesis in bright sunshine.
Thus, option-D is the correct answer.
A-G-T-C-G-T or in other words the answer is C
Answer:
D. some components of the cycle are used as building blocks for certain anabolic pathways
Explanation:
Kreb's cycle starts when pyruvate enters the cycle in the form of acetyl CoA. The cycle completely breakdown acetyl CoA into CO2 and H2O. The released energy is stored in the form of NADH and FADH2 which enter oxidative phosphorylation to drive ATP synthesis. Kreb's cycle is an amphibolic pathway as it is involved in both catabolic and anabolic processes.
Some of the intermediates of Kreb's cycle serve as precursors for other anabolic pathways. For example, alpha-Ketoglutarate and oxaloacetate from Kreb's cycle enter the anabolic processes of synthesis of amino acids aspartate and glutamate respectively. Succinyl CoA from Kreb's cycle serves as one of the intermediates for the synthesis of porphyrin rings present in "heme" groups. The heme group is a structural and functional part of hemoglobin and myoglobin proteins.
Answer:
Frequency of B allele is 0.6681
Explanation:
If p represents the frequency of dominant allele and q represents the frequency of recessive allele, according to Hardy-Weinberg equilibrium:
p + q = 1
p² + 2pq + q² = 1
where p² = frequency of homozygous dominant genotype
q² = frequency of homozygous recessive genotype
2pq = frequency of heterozygous genotype
Given that number of recessive chestnut horse = 28
Total horses = 226 + 28 = 254
frequency of b² genotype = 28/254 = 0.1102
frequency of recessive b allele = √0.1102 = 0.3319
So, frequency of B allele =
1 - 0.3319 = 0.6681
Hence frequency of B allele is 0.6681
