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BIO 340 UNIT 3 Questions And Answers With Verified Solutions Graded A+ Latest Update 2025.

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BIO 340 UNIT 3 Questions And Answers With Verified Solutions Graded A+ Latest Update 2025.

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Bio 340
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Institución
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Bio 340

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Subido en
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Escrito en
2024/2025
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BIO 340 UNIT 3 Questions And Answers With
Verified Solutions Graded A+ Latest Update 2025.

A+A+ Genotype - ANSWER fully dominant. normal (dominant wild type functional protein,
dominant wild type functional protein)
A+a Genotype - ANSWER recessive. presents as normal (dominant wild type functional
protein, mutant recessive non functional protein)
aa Genotype - ANSWER reccessive. presents as mutant (mutant recessive nonfunctional
protein, mutant recessive nonfunctional protein)
recessive mutation - ANSWER only seen in homozygotes
A+A Genotype - ANSWER fully dominant. MUTANT (dominant wild type, mutant allele
nonfunctional protein) loss of function=single copy is not enough to produce dominant wild type, present
as mutant
what causes the dominance of one allele over another? - ANSWER determined by the protein
product of the allele: the phenotype is the result of the activities. (DNA-RNA-PROTEIN) dominance
relationships between alleles have a molecular basis. gene expression can be affected by interactions with
other genes
loss of function mutation - ANSWER causes lots of recessive phenotypes
ex: phenylketonuria (PKU)
cause a gene to lose some or all of its normal function. there are two different types of these mutations:
hypomorphic mutations-have lost only some of their function
null mutations-have lost all of their function
haplosufficient - ANSWER only one working copy is necessary/sufficient for normal
expression of the gene's function. a single copy is sufficient to produce the wild-type phenotype in the
heterozygous genotype
allele - ANSWER variant of a gene
complementation tests - ANSWER determine the number of genes causing a mutant phenotype
phenylketonuria (PKU) - ANSWER caused by a recessive mutation in the PAH gene.
PAH encodes phenylalanine hydroxylase.
Phenylalanine hydroxylase is an enzyme that converts the amino acid phenylalanine into tyrosine.
Without the enzyme Phenylalanine hydroxylase people accumulate a lot of phenylalanine in the body,
which is turned into phenylpyruvic acid, and is toxic

,fully dominant mutations - ANSWER wild type allele is haploinsufficient if one mutant allele is
present (A+A=mutant AA=mutant)
haploidinsufficient - ANSWER a single copy of the gene is not suffient to produce the wild type
phenotype in the heterozygote genotype
AA - ANSWER fully dominant, present as mutant (non functional protein, non functional
protein)
gain of function (mutant alleles) - ANSWER the gene product acquires a new function or
express increased wild type activity
loss of function (mutant alleles) - ANSWER there is a significant decrease or complete loss of
functional gene product
codominance - ANSWER both alleles for the same characteristic are simultaneously expressed
in the heterozygote. ex: AB blood type. 1:2:1 ratio
incomplete dominance - ANSWER example: snapdragon (Antirrhinum majus) flower. a
homozygous parent with white flowers (CWCW) and a homozygous parent with red flowers (CRCR) will
produce offspring with pink flowers (CRCW).
alleles per gene, blood type - ANSWER Although individual humans (and all diploid
organisms) can only have two alleles for a given gene, multiple alleles may exist at the population level,
such that many combinations of two alleles are observed.
example: ABO blood system, 3 allele types circulating in the population but each individual gene only
gets 2 of their alleles from the parents. 6 genotypes for 3 alleles (1A1A, 1A1B, 1A1, 1B1A, 1B1B, 1B1,
11A, 11B, 11(O))
alleles per gene, rabbit coat color - ANSWER four alleles for C gene:
CC=expressed as black
c^ch c^ch=chincilla phenotype: expressed as black-tipped white fur
c^h c^h=himalayan phenotype: expressed as black fur on the extremities and white fur elsewhere.
cc=albino, or "colorless" phenotype: expressed as white fur.
multiple alleles - ANSWER (allelic series) dominance hierarchies can exist in multiple alleles
complementation test - ANSWER Occasionally multiple mutations of a single wild-type
phenotype are observed. (h1, h2, etc)
In order to determine if any of the mutations are in a single gene, or if each mutation represents one of the
several genes necessary for a phenotype to be expressed, geneticists use a complementation test.
performing a complementation test - ANSWER two organisms with different homozygous
recessive mutations, that produce the same mutant phenotype are crossed.
the F1 is examined:

, if the F1 expresses the wild type phenotype we conclude that the mutations are in different genes. in this
scenario each strains genome supplies the wild type allele to compliment the mutated allele of the other
strain's genome. Because mutations are recessive the offspring will display the wild-type phenotype


if the F1 does not express the wild type phenotype, but the mutant, we conclude that both mutations occur
in the same gene
Complementary Gene Interaction (9:7 Ratio) - ANSWER harebell color: complementation tests
of two different mutants who both produce white plants show that the mutations are in different genes,
because they produce a blue offspring, the wild type color. the two mutations compliment one another
mutant 1: aaBB (white)
mutant 2: AAbb (white)
offspring: AaBb (blue)
The 9:7 ratio indicates two genes interacting in the same pathway. Mutations in either gene disrupt the
pathway, causing the mutant phenotype.
gene interaction: when two or more genes influence the same pathway, affecting the same phenotype.
if two genes do interact there are less phenotypes than in the offspring of a typical dihybrid cross
Duplicate Gene Interaction (15:1 Ratio) - ANSWER The genes in a redundant system have
duplicate gene action; they encode the same product, or they encode products that have the same effect in
a pathway or compensatory pathways. Whenever a dominant gene is present, the trait is expressed.


EXAMPLE IN BEAN FLOWER COLOR:
in bean flower color, one allele is sufficient to produce the pigment. having P or R will produce a purple
phenotype


Parents: PpRr Purple x PpRr Purple
Offspring:
PPRR: PURPLE 9/16
P_rr: (having the "P" is enough to produce purple) PURPLE 3/16
ppR_:(having the "R" is enough to produce purple) PURPLE 3/16
pprr: (no pigment) WHITE 1/16


Bean flower color is an example of duplicate gene interaction. Having P or R will produce a purple
phenotype.
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