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Mastering Biology CH 14 homework Questions with correct Answers
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Mastering Biology CH 14 homework
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Mastering Biology CH 14 homeworkAround the mid-1850s, Mendel crossed true-breeding purple-flowered pea plants with
true-breeding white-flowered pea plants. The results of his research provided us with
the basic principles of heredity.
Drag the terms on the left to the appropriate blanks on the right to complete the
sentences. Not all terms will be used. - answer Every gene is a sequence of DNA
nucleotides at a specific position along a chromosome called a locus.
Diploid cells have two sets of chromosomes, one set inherited from each parent, that
form homologous pairs.
The homologs of a chromosome pair contain the same genetic loci. Therefore, each
genetic locus is represented twice in a diploid cell.
Variations in inherited characteristics is due to the presence of alleles, which are
alternative versions of genes.
You know that alleles are alternative versions of a gene. What makes alleles different
from each other? - answerThey have different sequences of DNA nucleotides.
In the video, the paternal parent carries the allele for purple flowers, and the maternal
parent carries the allele for white flowers. What would happen if the maternal parent
carried the purple allele, and the paternal parent carried the white allele? - answerThe
F1 hybrid pea plants would still produce purple flowers.
In pea plants, the purple allele is sufficient for making purple flowers, even if one of the
homologous chromosomes carries the white allele. Which of the following statements
are true in this case?
Select all that apply. - answerThe purple allele is the dominant allele.
The white allele is the recessive allele.
One purple allele results in enough purple pigment for the flowers to be purple.
Mendel studied pea plants dihybrid for seed shape (round versus wrinkled) and seed
color (yellow versus green). Recall that
the round allele (R) is dominant to the wrinkled allele (r) and
, the yellow allele (Y) is dominant to the green allele (y).
The table below shows the F1 progeny that result from selfing four different parent pea
plants.
Use the phenotypes of the F1 progeny to deduce the genotype and phenotype of each
parent plant.
Complete the table by dragging the correct label to the appropriate location. Labels can
be used once, more than once, or not at all. - answerParent phenotype plant 1: green
round
Parent genotype plant 1: Rryy
Parent phenotype plant 2: yellow round
Parent genotype plant 2: RrYy
Parent phenotype plant 3: yellow round
Parent genotype plant 3: RRYy
Parent phenotype plant 4: green wrinkled
Parent genotype plant 4: rryy
A plant grown from a [round, yellow] seed is crossed with a plant grown from a
[wrinkled, yellow] seed. This cross produces four progeny types in the F1:
[round, yellow], [wrinkled, yellow], [round, green], and [wrinkled, green].
Use this information to deduce the genotypes of the parent plants.
Indicate the genotypes by dragging the correct label to the appropriate location. -
answerParent genotype with yellow round: RrYy
Parent genotype with yellow wrinkled: rrYy
For the cross in Part B, predict the frequencies of each of the phenotypes in the F1
progeny, and determine the genotype(s) present in each phenotypic class.
Complete the diagram by dragging the correct label to the appropriate location. Labels
can be used once, more than once, or not at all. - answerExpected frequencies yellow
round: 3/8
Progeny genotypes yellow round: RrYY, RrYy (x2)
Expected frequencies yellow wrinkled: 3/8
Progeny genotypes yellow wrinkled: rrYY, rrYy (x2)
Expected frequencies green round: 1/8
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