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Solnce55 [7]
3 years ago
9

Select the procedure that will not help saturate a thin-layer chromatography (tlc) developing chamber with developing solvent va

pors.
Biology
1 answer:
ELEN [110]3 years ago
3 0

The correct answer is adding ninhydrin to developing solvent.

Ninhydrinis used to detect (color) amino-acids, since amino acids are colourless compounds. So, after development, the TLC plate is sprayed with ninhydrin reagent and dried in an oven, at 105°C for about 5 minutes. It reacts with α- amino acids that results in purple coloured spots.  


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You decide to conduct a genetic analysis of these mutant lines by crossing each with a pure-breeding wild-type line. The numbers
maxonik [38]

Complete question:

You will find the complete question in the attached files

Answer:

  • For the twist trait: The <em>mutant allele is dominant</em> to its corresponding wild-type allele
  • For the forked trait: the <em>mutant allele is dominant</em> to its corresponding wild-type allele
  • For the pale trait: The <em>mutant allele is neither dominant nor completely recessive</em> to its corresponding wild-type allele

Explanation:

  • Cross 1:  twisted x wild-type ----> Pure lines

Parentals)    TT   x      tt

<em>F1) twisted leaves, Tt</em>. ---> Heterozygous

Parentals) Tt    x    Tt

Punnett square)    T     t

                       T    TT   Tt

                        t    Tt    tt

<em>F2) 53 twisted, 18 wild-type </em>

Total number of individuals in the F2 = 53 + 18 = 71

71 plants -------- 100% of the F2

53 twisted------X = 75% TT + Tt

18 wild-type----X = 25% tt

<em>Phenotypic ratio 3:1</em>

The phenotype of the F1 and F2 progeny tells us that the twist trait is dominant over the wild type. The fact that the whole F1 generation was twisted is enough information to assume that the wild type is recessive and the twisted is dominant. Also, the phenotypic ratio of the F2 corroborates this assumption.

  • Cross 2: forked x wild-type ---> Pure Lines

Parentals)  FF     x     ff

F1) 100% forked, Ff----> Heterozygous

Parentals)  Ff     x     Ff

Punnett square)    F      f

                    F       FF    Ff

                    f        Ff     ff

F2) 49 forked and 16 wild-type plants

Total number of individuals in the F2 = 49 + 16 = 65

65 plants -------- 100% of the F2

49 forked------X = 75% FF + Ff

16 wild-type----X = 25% ff

<em>Phenotypic ratio 3:1</em>

The phenotype of the F1 and F2 progeny tells us that the twist trait is dominant over the wild type. The fact that the whole F1 generation was forked is enough information to assume that the wild type is recessive and the forked is dominant. Also, the phenotypic ratio of the F2 corroborates this assumption.

  • Cross 3: pale x wild-type ---> Pure lines

Parentals) PP    x    pp

F1) 100% Pp, intermediate color.

Parentals) Pp   x   Pp

Punnett square)   P       p

                      P    PP    Pp

                       p    Pp    pp

F2) 34 intermediate, 17 wild-types, and 16 pale.

Total number of individuals in the F2 = 34 + 17  + 16 = 67

67 plants -------------- 100% of the F2

34 intermediate ------X = 51% Pp

17 wild-type-------------X = 25% pp

16 pale -------------------X = 24% PP

<em>Phenotypic ratio 1:2:1</em>

The phenotype of the F1 and F2 progeny tells us that the pale trait is not dominant neither recessive to the wild type. The fact that the whole F1 generation was intermediate is enough information to assume that none of the traits dominates over the other. This is a case of incomplete dominance.  The heterozygous individual express an intermediate phenotype between both the parentals´ one. Also, the phenotypic ratio of the F2 corroborates this assumption.  

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4 0
3 years ago
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