Amanda Olsen-Dufour

Numerade Educator

Biography

Hello! My name is Amanda and I am a recent grad of Dartmouth College, now working in neuroscience research in Washington DC. I love biology and I would love to help you understand and love it too!

Education

Amanda has not yet added their education credentials.

Educator Statistics

Numerade tutor for 4 years
85 Students Helped

Topics Covered

Mendelian Genetics: Understanding Inheritance Patterns
The Central Dogma: Understanding Gene Expression

Amanda's Textbook Answer Videos

1

Amanda's Quick Ask Videos

02:35
Biology

The mouse parents NnHhBb and nnhhbb were crossed and they produced the following offspring pups: 40 NHB; 40 nhb; 6 NHB; 6 nhB; 4 Nhb; and 4 nhB. What is the recombinant frequency between N and B?

Amanda Olsen-Dufour
05:37
Biology

In the fruit fly Drosophila melanogaster, vestigial wings, and
hairy body are produced by two recessive genes located on different
chromosomes. The normal alleles, long wings, and hairless body are
dominant.
a). Suppose a vestigial-winged hairy male is crossed with a
homozygous normal female. What types of progeny would be
expected?
b). If the F1 from this cross are permitted to mate randomly
among themselves, what progeny would be expected in F2? Show
complete genotypes, phenotypes, and ratios for each generation.
c). Suppose a hairy female, normal wing (heterozygous for wing)
is crossed with a vestigial-winged, normal body (heterozygous for
hairy body) male. Give offspring genotypic and phenotypic
ratios.

Amanda Olsen-Dufour
02:24
Biology

1) If the gene for helicase is mutated, what part of replication
will be affected?
A) If helicase is mutated, the DNA strands will not be separated
at the beginning of replication.
B) If helicase is mutated, the DNA strands will not be joined
together at the beginning of replication.
C) If helicase is mutated, replication will continue past the
replication fork.
D) if helicase is mutated, the DNA strands will not be joined
together at the end of replication.
2) Compare and contrast prokaryotic and eukaryotic DNA
replication.
A) A prokaryotic organism’s rate of replication is ten times
faster than that of eukaryotes. Prokaryotes have a single origin of
replication and use five types of polymerases, while eukaryotes
have multiple sites of origin and use fourteen polymerases.
Telomerase is absent in prokaryotes. DNA pol I is the primer
remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol
III performs strand elongation in prokaryotes and pol δ and pol ε
do the same in eukaryotes.
B) A prokaryotic organism’s rate of replication is ten times
slower than that of eukaryotes. Prokaryotes have a single origin of
replication and use five types of polymerases, while eukaryotes
have multiple sites of origin and use fourteen polymerases.
Telomerase is absent in eukaryotes. DNA pol I is the primer remover
in prokaryotes, while in eukaryotes it is RNase H. DNA pol III
performs strand elongation in prokaryotes and pol δ and pol ε do
the same in eukaryotes.
C) A prokaryotic organism’s rate of replication is ten times
faster than that of eukaryotes. Prokaryotes have five origins of
replication and use a single type of polymerase, while eukaryotes
have a single site of origin and use fourteen polymerases.
Telomerase is absent in prokaryotes. DNA pol I is the primer
remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol
III performs strand elongation in prokaryotes and pol δ and pol ε
do the same in eukaryotes.
D) A prokaryotic organism’s rate of replication is ten times
slower than that of eukaryotes. Prokaryotes have a single origin of
replication and use five types of polymerases, while eukaryotes
have multiple sites of origin and use fourteen polymerases.
Telomerase is absent in prokaryotes. DNA pol I is the primer
remover in eukaryotes, while in prokaryotes it is RNase H. DNA pol
III performs strand elongation in prokaryotes and pol δ and pol ε
do the same in eukaryotes.
3) What is the function of telomeres?
A) Telomeres are repeated DNA sequences that code for no gene.
They protect the genes from getting deleted as cells continue to
divide.
B) Telomeres are repeated DNA sequences that code for introns.
They mark the start and end of genes.
C) Telomeres are the inactive form of the enzyme telomerase, the
enzyme that links the fragments of the lagging strand.
D) Telomeres add the 3' poly-A tail, preventing information loss
during DNA replication.

Amanda Olsen-Dufour
03:25
Biology

A large population of land turtles on an isolated island has two alleles for a gene that determines shell thickness. The allele for thinner shells is dominant over the allele for thicker shells. The thinner-shell allele occurs at a frequency of 20%. Assuming there is no net advantage to thick or thin shells, what should the frequency (%) of homozygous thicker-shelled turtles be in the population? (hint: use a large Punnett square to calculate the answer).

Use the following information for the next 2 questions. Pigeons have two alleles of a single gene that determines whether they have feathers on their lower legs. The alleles are called no grouse (associated with the featherless phenotype) and grouse (associated with feathery legs).

2. Suppose a pigeon that is homozygous for the no grouse allele mates with a heterozygous pigeon. What is the expected frequency (%) of the homozygous no grouse genotype in the offspring?

3. The no grouse allele is dominant over the grouse allele. If a pigeon homozygous for the grouse allele mates with a heterozygous pigeon, what is the expected frequency (%) of the feathery-legged (grouse) phenotype in the offspring?

Amanda Olsen-Dufour
01:14
Biology

a large population of land turtles on an isolated island has two alleles for a gene that determines shell thickness go forth in their shells is dominant over the allele for sickle shares the thinner shell Elia appears at a frequency of 20% assuming there is no net advantage to fix a thin shell what should the frequency of homozygous sticker shelled turtles be in the population

Amanda Olsen-Dufour
02:05
Biology

Some bacteria may be able to respond to environmental stress by
increasing the rate at which mutations occur during cell division.
Might there be an evolutionary advantage to this ability?
Explain.
A. An increase in the mutation rate was useful for
ancestral forms of bacteria in the constantly changing environment,
but now it cannot be advantageous for species and for their
survival.
B. If the frequency of mutations increases, so does the
probability of adaptations to new conditions. Thus, this species is
more likely to survive.
C. The increase in DNA replication mistakes leads
to failures in bacterial reproduction, thus decreasing the chance
of the species' thriving. This cannot be advantageous.
D. If the frequency of mutations increases, the species
becomes less susceptible to viruses, because the targets of the
viruses also change due to mutations. Thus, this species is more
likely to survive.

Amanda Olsen-Dufour
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