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ruslelena [56]
3 years ago
14

In the experiment determining the parental genotype of plant seedlings from a monohybrid cross, if the dominant color was green

(C) and the recessive color was white (c), what would the parental genotype have been if plate 1 only contained white seedlings instead of all green
Biology
1 answer:
eimsori [14]3 years ago
7 0

Answer:

cc × cc

Explanation:

This question involves a single gene coding for plant seedling. The green allele (C) is dominant over the white allele (c). This means that the green allele will mask the phenotypic expression of the white allele in a heterozygous state (Cc).

In this experiment where plate 1 only contained white seedlings instead of all green, this illustrates that all the offsprings were recessive. This is because the parental genotypes were both recessive for the color trait i.e. cc.

Note that, the recessive trait can only be expressed when the recessive alleles are present in a gene. Therefore, the parental genotype would have been cc × cc, in order to give rise to all offsprings with the recessive trait (white colour).

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Answer:

The concentration of

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Explanation:

Given

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The formula weight of sodium acetate is 82.03 g/mol.

Moles = 11.1g/82.03g/mol

= 0.1353M

Since the volume is 1L, the molarity of sodium acetate will still be 0.1353M

The starting concentration of sodium acetate is 0.1353M

The final concentration of sodium acetate is the starting amount of sodium acetate, minus the amount that was protonated.

Assume that, x = concentration of acetic acid, which is the amount of sodium acetate that was protonated.

From the Henderson-Hasselbalch equation

PH = pKa + log[A-]/[HA]

[A-] = 0.1353 - x

[HA] = x

pKa = 4.75

PH = 5.25

Substitute in these values

5.25 = 4.75 + log(0.1353-x)/x

log(0.1353 - x)/x = 5.25 - 4.75

log(0.1353-x)/x = 0.5

log(0.1353 - x)/x = ½

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(0.1353 - x)/x = 3.1623

0.1353 - x = 3.1623x

0.1353 = x + 3.1623x

0.1353 = 4.1623x

x = 0.1353/4.1623

x = 0.0325M

Therefore [HA] = x = 0.0325M

[A-] = 0.1353 - x = 0.1353 - 0.0325 =

[A-] = 0.1028M

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