Quiz 2: WK 3 - Extensions and Modifications of Basic Principles Flashcards
A change in phenotype produced by environmental factors that mimic the phenotype of a mutant organism is called a:
a) genocopy
b) phenocopy
c) Polymorphism
d) Pleiotropy
b
the Siamese allele is a temperature-sensitive allele of the Tyrosinase gene and produces a protein that is inactive at higher temperatures resulting in cat coat pigment variations. This is an example of :
a) Co-dominance
b) incomplete dominace
c) incomplete penetrance
d) variable expressivity
d
Phenylketonuria (PKU) is a disease caused by a single loss of function mutation in the gene responsible for the conversion of phenylalanine to tyrosine. This loss of function results in mental retardation and reduced hair and skin pigmentation. This is an example of:
a) dominance
b) pleiotropy
c) polymorphism
d) an allele
b
The gene associated with the recessive disorder Cystic Fibrosis (CFTR) produces a transmembrane cellular chloride channel. Multiple alleles for this gene exist, some of which contain mutations that render this channel non-functional. An individual with two mutated alleles will have Cystic Fibrosis. However, an individual that is heterozygous for the wildtype and mutated alleles will not have symptoms as a healthy copy of this gene will produce enough of the chloride channel to maintain appropriate function. This is an example of:
a) A gain-of-function mutation
b) haploinsufficiency
c) incomplete penetrance
d) haplosufficiency
d
Inbred lines of self-fertilizing plants are typically homozygous for many genes and are often less vigorous than hybrid lines, this is referred to as:
a) inbreeding depression
b) hybrid vigor
c) allelism
d) heterosis
a
For the ABO blood group gene I, the I^A and I^B alleles are examples of:
a) incompletely dominant alleles
b) co-dominant alleles
c) Epistatic alleles
d) null alleles
b
A woman with Type O blood and a man with Type B have a child. What is the probability that is the child will have type A Blood?
a) 25%
b) 0%
c) 50%
d) 75%
b
If a rare recessive single-gene genetic disorder occurs at a rate of 1/14600 in a given population, use the Hardy-Weinberg principle to estimate the frequency of the dominant allele. Round to 4 decimal places
ANS: 0.9917
Solution
i) q^2 = 1/14600 (b/c involves gene not allele)
ii) q = (1/14600)^0.5 = 0.008276
iii) p = 1 - q = 1 - 0.008276 = 0.9917
(Note: its p not p^2 becuase it is looking for frequency of dominant allele not gene.)
