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Rama09 [41]
3 years ago
9

Which mating of four-o'clock plants would produce progeny that were either green, variegated, or white

Biology
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

variegated female x green male

Explanation:

The four o'clock (<em>Mirabilis jalapa</em>) plant is a common ornamental species of <em>Mirabilis</em> (clade Angiosperms, flowering plants) whose leaf pigmentation is a well-known case of maternal inheritance. Moreover, chloroplasts are organelles that have their own genome which follows a maternal inheritance pattern in the majority of plant and animal species because they are contained in the cytoplasm (i.e., chloroplast DNA is inherited only from the mother because only the maternal cytoplasm persists after fertilization). Thus, chloroplast DNA has an inheritance pattern that challenges Mendel's Laws of inheritance (i.e., distinct from nuclear DNA). In the case of the four o'clock plant, the genes responsible for leaf color are located in the chloroplast genome and therefore these genes are transmitted from the female parent to the progeny. In this case, the leaf variegation is caused by two different types of chloroplasts that are inherited from the mother: normal green chloroplasts and defective chloroplasts (without chlorophyll pigment).

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The heterozygous cell will form one or more subunits has the mutant structure when the gene is expressed. From the question if there is a mutation in dominant negative allele, it will cause the inactivation of one trimer resulting in 87.5% inactive trimers for a heterozygous individual.

There are 87.5 % percent of the trimers present in the cell which will be inactive for a heterozygous individual, If at least one of the subunits has the mutant structure.

The probability for each individual subunit to be a mutant is 1/2

Gene codes for a protein that forms trimer so the percentage of the trimers present in the heterozygous individual is third power of 1/2 =1/8

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=0.875 will be inactive.

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Therefore 87.5% will be inactive.

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