Answer:
The correct answer is
1. 5' end.
2. guanosine
3. 5′ to 5′ triphosphate linkage.
4. On the 7th position.
5. 2′ hydroxy-groups of the first 2 ribose sugars.
Explanation:
The eukaryotic mRNA is processed or modified with a guanosine cap at 5' end and poly-A tail at 3'end before translation.
The process of adding a cap at 5' end of a growing mRNA is known as capping. The process of capping adds guanosine nucleotide at the 5'-end of mRNAs by 5′ to 5′ triphosphate linkage. This guanosine is modified by methylating at the 7th position of the atom. The mRNA is further modified by methylation of first two ribose sugars at their 2′ hydroxy-groups. This capping is advantageous to the sequence as it protects the end from phosphatases and nucleases.
Protists are unicellular eukaryotes, whereas Eubacteria and Archaebacteria are unicellular prokaryotes.
Eubacteria and Archaebacteria belong to kingdom Monera; whereas Protists belong to kingdom Protista.
All Monerans have prokaryotic cell structure. Protists have eukaryotic cell structure and are unicellular.
Protists either lack cell wall or have cell wall made up of cellulose.
Eukaryotes have cell wall made up of peptidoglycan or murein.
In Archaebacteria cell wall lacks peptidoglycan but contains proteins and non-cellulosic polysaccharides.
Protists have typical sexual reproduction involving fusion of gametes. In Eubacteria and Archaebacteria typical sexual reproduction is absent.
Cell division is mitotic type in Protists and amitotic in Eubacteria and Archaebacteria.
<span>to sue the government in cases of unjust laws to persuade legislators to share the scientists' views on issues to make honest, ethical presentations of data to represent the interests of the business community</span>
During the exercise period (10-15min) the blood lactic acid concentration increases to about 13.2 mmol/dL (same units as on graph) as the individual is having problems keeping up their aerobic respiration. After 15min, they stop exercising and the lactic acid concentration starts to return to normal as their body is able to take in enough oxygen and catches up with the excess lactic acid, metabolizing it into CO2 and H2O. The period between 15-20 min shows the fastest reduction in concentration.
Answer:
Uno de los progenitores es heterocigoto para color naranja (Nn) y el otro parental es homocigoto recesivo para gris (nn). Al haber una cruza entre un homocigoto recesivo y un heterocigoto, la 50% de la progenie expresa color naranja (Nn), mientras que el otro 50% expresa color gris (nn).
Explanation:
<u>Datos disponibles:</u>
- Cruce entre peces naranjas y peces grises
- 50% de la F1 son peces grises
- Naranja dominante sobre gris
Podemos nombrar el alelo dominante para color naranja <em>N</em>, y al alelo recesivo para color gris <em>n</em>.
Para que en un cruce entre dos fenotipos distintos, el 50% de la primer camada exprese uno de estos fenotipos, entonces uno de los parentales debe ser heterocigoto, mientras que el otro parental debe ser homocigoto recesivo. De esta forma 50% de la primera generación expresara uno de los fenotipos, mientras que el otro 50% expresará el otro fenotipo.
Supongamos que uno de los parentales lleva el genotipo <em>Nn</em>, y el otro parental es <em>nn</em>.
Cruce:
Parental) Nn x nn
Gametas) N n n n
Fenotipos) Naranja Gris
Cuadro de Punnett) N n
n Nn nn
n Nn nn
F1) 2/4 = 1/2 = 50% de la progenie tendrá genotipo heterocigoto, Nn
2/4 = 1/2 = 50% de la progenie tendrá genotipo homocigota recesivo,
nn
50% de la progenie será color naranja (Nn)
50% de la progenie será color gris (nn)
