BIO 311C FINAL QUESTIONS AND ANSWERS A+ GRADED
Two black guinea pigs were mated and they produced 14 black and 6 white offspring. Based on the results what are the genotypes of the parent pigs?
A) Homozygous dominant
B)Heterozygous dominant
C) Homozygous recessive
D) Heterozygous rece...
Two black guinea pigs were mated and they produced 14 black and 6 white
offspring. Based on the results what are the genotypes of the parent pigs?
A) Homozygous dominant
B)Heterozygous dominant
C) Homozygous recessive
D) Heterozygous recessive
E) Hemizygous dominant
B)Heterozygous dominant
almost 3/4 of offspring were black so this is the dominant trait and since white offspring
occurred (~25%) we know that the parents must both have the recessive gene (if one
parent was homozygous dominant for black, then there would be no white offspring)
In certain plants, tall is dominant to short trait. If a heterozygous plant is crossed
with another heterozygous (tall) plant, what is the probability that the offspring
will be short?
A) 1/2
B) 1/4
C) 0
D) 1
E) 1/6
B) 1/4
The parents' genotype is Tt because they are both heterozygous tall. The punnet
square would indicate 3/4 chance that the offspring will be tall (TT, Tt, Tt) and 1/4
chance they will be short (tt)
A cross between pea plants with red flowers and plants with white flowers
resulted in all offspring with pink flowers. This experimental result demonstrates
A) True-breeding.
B) Blending model of genetics.
C) Dominance.
D)Incomplete dominance
E) Epistasis
D)Incomplete dominance
This is correct. The heterozygotes show an intermediate phenotype between the parent,
homozygotes.
,A couple without any cystic fibrosis symptoms have their first child showing
symptoms of cystic fibrosis.
What is the probability that their next child will be a carrier for cystic fibrosis ?
A) 0%
B)50%
C) 25%
D) 100%
E) 75%
B)50%
CF is recessive, so in order to have it, the child must be homozygous for the gene.
When you do a test cross, you can see that 50% will be a carrier (heterozygous - Aa)
while 25% will be homozygous dominant (non-carrier - AA) and 25% will have CF
(homozygous recessive - aa)
In crossing a homozygous recessive with a heterozygote, what is the chance of
getting an offspring with the homozygous dominant?
A)0%
B) 75%
C) 25%
D) 100%
E) 50%
A)0%
This is correct. Each parent donates one copy of the allele, so if one parent is
homozygous recessive, they do not have the dominant allele so cannot pass it along.
Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to
long tails (t).
What fraction of the progeny of the cross BbTt × BBtt will have black fur and long
tails?
A) 1/16
B) 3/16
C) 3/8
D)1/2
E) 9/16
D)1/2
Since one parent is homozygous for black fur (BB) you know all offspring will have black
fur. now just look at the percent that will have long tails when you do the cross.
Which of the following experimental treatments proved the "transforming
principle" (later identified as DNA) is the genetic material?
A) Infection of mice with pathogenic strain resulting in no disease.
, B) Infection of mice with E. coli strain showing it can cause death.
C)Infection of live pathogenic strain mixed with dead non-pathogenic strain,
causing disease.
D)Infection of dead pathogenic strain mixed with live non-pathogenic strain,
causing disease.
E) Infection of dead pathogenic strain mixed with dead E.coli strain, causing
disease.
D)Infection of dead pathogenic strain mixed with live non-pathogenic strain, causing
disease.
Review this experiment. rough (non-pathogenic) bacteria was mixed with dead, smooth
(pathogenic) bacteria, and this caused disease. This does not necessarily show that
DNA is the genetic material, but shows that "something" from the dead strain was
passed to the live, non-pathogenic strain.
Which of the following will not be able to transform non-pathogenic bacteria to
become pathogenic bacteria?
A) Cell extract from pathogenic bacteria treated with protease
B) Cell extract from pathogenic bacteria treated with RNase
C) Cell extract from pathogenic bacteria treated with DNase
D) Cell extract from pathogenic bacteria not treated with any enzymes
E) Pathogenic bacteria
C) Cell extract from pathogenic bacteria treated with DNase
If a RNA virus with protein capsid was allowed to replicate inside a bacterial cell
in the presence of radioactive sulfur. The resulting virus was purified and allowed
to infect non-radioactive bacteria. Then the cells were centrifuged to separate the
virus from bacteria. Which fraction will contain the highest radioactivity?
A) The bacteria in the pellet.
B)RNA inside bacterial pellet.
C)Bacteria and RNA in the virus.
D)Protein coat of virus in the supernatant
E) Both protein coat and RNA of the virus
D)Protein coat of virus in the supernatant
The protein coat of the virus would have the radioactivity. When it was replicating in the
radioactive bacteria, the radioactive sulfur would be incorporated into the viral proteins.
All of the following are true about DNA structure except __________.
A) The purines are H-bonded to pyrimidines.
B) 5' refers to the phosphate group attached to the 5th Carbon of deoxyribose.
C)3' refers to the Nitrogenous base attached to the 3rd C of dexyribose.
D) The two strands run antiparallel to each other.
E)The 2nd C of deoxyribose contains only H and no oxygen.
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