Answer:
recombination and independent segregation of chromosomes during meiosis
Explanation:
Recombination and independent segregation of chromosomes represent the two most important meiotic mechanisms by which sisters from the same parents can inherit different gene variants and therefore look very different from each other:
1- Independent assortment (segregation) of chromosomes: during meiosis, homologous chromosomes are randomly distributed in daughter cells (which will give rise to the gametes), and therefore separate independently of each other. It is for that reason that gametes have unique combinations of chromosomes, which increases genetic variation.
2- Recombination, also known as crossing over, refers to the exchange of chromosome segments between non-sister chromatids during meiosis. This mechanism is well-known to produce new gene variants (alleles) in the daughter cells. In consequence, recombination also increases the genetic variation of the resulting gametes that will produce offspring (in this case, different sisters).
Answer: The percentage of each phenotype to be expressed is *50%* each
Explanation: In codominance, two alleles for a particular trait are equally expressed in an individual carrying it, neither an allele is dominant or recessive over the other. In this case the individual shows the two traits of the allels.
Furthermore, when 2 Christmas parrot are crossed with red and green color in an homozygous codominant trait the heterozygous F1 generation will either be red with traces of green or green showing traces of red.
If parrot 1 (Red) is represented as PR PR and parrot 2 (Green) is PG PG the cross between them will give PR PG therefore, the individual shows both traits.
Answer:
Polypeptides are composed of amino acids, and we know amino acids are differently charged, have different R groups, and also have different isoelectric points. Depending on different isoelectric points and charged groups, the polypeptides can be separated and because a protein has its lowest solubility on its isoelectric point.
So in this question,
(a) (Lys-Ala)3 ; this is highly positively charged (polar) at pH 7 than (Gly)20 which is uncharged except for the amino and carboxyl terminal.
(b) (Glu)20 ; it is highly negatively charged at pH 7 whereas (Phe-Met)3 is much less polar and hence less soluble.
(c) (Asn-Ser-His)5 ; at pH 3, because in (Ala-Asp-Gly)5 the carboxylate groups of Asp residues are partially protonated and neutral, whereas in (Asn-Ser-His)5 , the imidazole groups of His residues are fully protonated and positively charged.
(d) (Asn-Ser-His)5; at pH 6.0; both polymers have polar Ser sidechains, but (Asn-Ser-His)5 also has the polar Asn side chains and partially protonated His side chains.
Explanation:
Answer: hi I 4 my broter wa in thi ap
Explanation:
Answer: pH = 9.4
Explanation: in water, -log(molar concentration OH-) + -log(molar concentration H+) = 14
-log(H+) is pH
so pH = 14 - -log(OH-)
= 14 - -log(2.4*10^-5)
= 14 - 4.6
= 9.4
