Practce questions quiz Flashcards

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1
Q

How can you differentiate a male from a female Drosophila melanogaster?

A

The females have a striped pattern on their abdomen while males have a solid black lower
abdomen

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2
Q

How do you find standard deviation?

A

sqrt of sum of (data point 1, 2, 3…-average)^2/n-1

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3
Q

The more spread out the data, the ____ the standard deviation.

A

higher

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4
Q

geneticist is breeding a species of mice to study a particular fur texture. In a large breeding experiment, wild-
type mice are crossed, and the offspring are observed. Out of 400 offspring, 300 exhibit the wild-type fur texture,
while 100 exhibit a mutant fur texture. When a reciprocal cross was performed, the results were the same. Given this
phenotypic ratio, how should the geneticist classify the trait responsible for the mutant fur texture?

A

Autosomal recessive

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5
Q

Why is the principle of segregation important?

A

it explains how the genotypic ratios in the haploid
gametes are produced

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6
Q

The map distance between two genes can be calculated using:

A

a. The number of recombinant offspring divided by the total number of offspring,
multiplied by 100

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7
Q

How are the distances in genetics maps measured?

A

Map units and centimorgans

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8
Q

Consider a case where a geneticist is studying two genes in capybaras and observes that the
number of nonrecombinant offspring is significantly higher than the number of recombinant
offspring. What does this suggest about the two genes?

A

They’re probably linked

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9
Q

SNP marker is found linked to the cystic fibrosis gene. Cystic fibrosis is an autosomal
recessive disease. A couple plans to have children together and both are carriers for the cystic
fibrosis gene but do not have the disease themselves. They are both A1/A2 at the SNP and
the A1 allele is linked with the allele causing cystic fibrosis. Assuming no crossing over
between the SNP and the disease gene, what genotypes in the offspring could result that
would cause cystic fibrosis?

A

A1 – cf, A1 – cf

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10
Q

In a genetic study of pea plants, researchers are examining the recombination frequency
between two genes: gene A (flower color) and gene B (seed shape). In a test cross between plants
heterozygous for both traits (AaBb) and plants homozygous recessive for both traits (aabb), they
observed the following offspring (Purple is dominant flower color, white is recessive flower
color, fusiform is dominant seed shape, and clavate is recessive seed shape):
460 purple, fusiform plants
440 white, clavate plants
40 purple, clavate plants
60 white, fusiform plants
Based on this data, what is the recombination frequency between gene A and gene B, and how
many map units (centiMorgans, cM) apart are these genes on the chromosome?

A

10% recombination frequency; 10 cM apart

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11
Q

To test for linkage between two genes, what specific cross is recommended?

A

c. A test cross between a heterozygous individual for both traits and another
individual homozygous recessive for those traits.

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12
Q

In cucumbers, heart-shaped leaves (h) are recessive to normal leaves (H), and having
numerous fruit spines (n) is recessive to having few fruit spines (N). The genes for leaf
shape and number of spines are located on the same chromosome; mapping
experiments indicate that they are 32.6 map units apart. A cucumber plant having
heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for
normal leaves and few spines. The resulting F1 are crossed with plants that have heart-
shaped leaves and numerous spines. What proportion of the progeny are expected to
have both of the dominant traits?

A

33.7%

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13
Q

What is the primary reason for conducting the IPTG/ONPG assay at 37°C?

A

To optimize the activity of β-galactosidase

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14
Q

In the IPTG/ONPG assay, why is IPTG used?

A

To induce the lac operon

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15
Q

Why is ONPG used in the IPTG/ONPG assay?

A

To serve as an artificial substrate for β-galactosidase, producing a colorimetric
change upon cleavage

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16
Q

T biotech company is developing a new synthetic biology tool to enhance the production of a
novel antibiotic in E. coli. The antibiotic gene is inserted into a bacterial operon that is
engineered to be highly responsive to specific inducible conditions. For optimal expression of the
antibiotic gene, where should the company’s scientists design the polymerase to bind in the
operon?

A

To the promoter region to ensure efficient initiation of transcription

17
Q

What are the products of an operon after it is transcribed and translated

A

a. A single mRNA transcript that is translated into multiple proteins

18
Q

Why is gene regulation important for bacterial cells?

A

Gene regulation allows for biochemical flexibility and rapid responses to changes in the environment while
maintaining energy efficiency by the bacterial cells.

19
Q

What does an operon look like?

A

Promoter, operator, structural genes, regulatory gene (outside operator)

20
Q

The repressor binds at
the operator region and inhibits transcription of the lac operon by

A

preventing RNA
polymerase from binding to the promoter

21
Q

When lactose is present in the cell, the enzyme β-galactosidase __________________.

A

converts some of it
into allolactose.

22
Q

A mutant strain of E. coli produces β-galactosidase in both the presence and the
absence of lactose. Where in the operon might the mutation in this strain be located?

A

Within the operon, the operator region is the most probable location of the mutation.
If the mutation prevents the lac repressor protein from binding to the operator, then
transcription of the lac structural genes will not be inhibited. Expression will be
constitutive. Outside of the operon, a mutation in the lacI gene that inactivates the
repressor or keeps it from binding to the operator could also lead to constitutive
expression of the structural genes.

23
Q

OC mutation

A

Beta galactosidaese and permease Always transcribed

24
Q

I and P mutation

A

Nothing EVER transcribed.

25
Q

I and Y mutation

A

Only Beta galactosidase transcribed, which is all the time

26
Q

Allolactose ________, altering its shape and reducing
the repressor’s affinity for the operator.

A

binds to the lac repressor

27
Q

Why does lactose turn off repressor and allow for gene transcription?

A

An allolactose-bound repressor does not
occupy the operator, so RNA polymerase can initiate transcription of the lac structural
genes from the lac promoter