Question 1

Use the following to answer questions 
You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna or green shells and short antenna. Crossing these lines yields FS1U1B11S1U1B0 progeny with blue shells and long antenna. Crossing FS1U1B11S1U1B0 progeny with beetles that have green shells and short antenna yields the following progeny:
Blue shell, long antenna82
Green shell, short antenna78
Blue shell, short antenna37
Green shell, long antenna43
Total240
-Assuming that the genes are linked, what is the map distance between them in m.u.?

Accepted Answer

The map distance is calculated using the formula: (Number of recombinant offspring / Total number of offspring) x 100. The recombinant offspring are those with blue shell, short antenna and green shell, long antenna. So, (37 + 43) / 240 x 100 = 33.3 m.u.

Question 2

Genetic distances within a given linkage group: (Select all that apply.)

Accepted Answer

Genetic distances are based on crossover frequencies between chromatids and are measured in centiMorgans or map units.

Question 3

Use the following to answer questions : Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following makeup: 41with baby blue eyes and pink wings 207with baby blue eyes only 210with pink wings only 42with wild-type phenotype -Assuming the wild-type alleles for these two genes are b⁺ and pw⁺, what is the correct testcross of the F₁ flies?

Accepted Answer

C

Question 4

If the recombination frequency between genes (A) and (B) is 5.3%, what is the distance between the genes in map units on the linkage map?

Accepted Answer

The distance between genes in map units is equal to the recombination frequency between them. Therefore, the distance between genes (A) and (B) is 5.3 m.u.

Question 5

What major contribution did Barbara McClintock and Harriet Creighton make to the study of recombination?

Accepted Answer

Barbara McClintock and Harriet Creighton demonstrated that genetic recombination of alleles is associated with physical exchange between chromosomes, providing evidence for the physical basis of genetic recombination.

Question 6

Two linked genes, (A) and (B), are separated by 18 m.u. A man with genotype Aa Bb marries a woman who is aa bb. The man's father was AA BB. What is the probability that their first child will be Aa bb?

Accepted Answer

The answer of Two linked genes, (A) and (B), are...

Question 7

Linked genes always exhibit:

Accepted Answer

Linked genes exhibit recombination frequencies of less than 50% because they are located close to each other on the same chromosome, reducing the likelihood of crossing over between them.

Question 8

Assume that an individual of AB/ab genotype is involved in a testcross and four classes of testcross progeny are found in equal frequencies. Which of the following statements is TRUE?

Accepted Answer

Equal frequencies of testcross progeny suggest independent assortment, which occurs when genes are on different chromosomes or are far apart on the same chromosome, allowing for recombination.

Question 9

Linked genes:

Accepted Answer

Linked genes are located close together on the same chromosome and tend to be inherited together, or co-segregate, during meiosis.

Question 10

You are studying two linked genes in lizards. You have two females and you know that both are the same genotype, heterozygous for both genes (A/a and B/b). You testcross each female to a male that is fully homozygous recessive for both genes (a/a and b/b) and get the following progeny with the following phenotypes: $$\begin{array} { l l } 
\text { Female } 1 & \text { Female } 2 \
A B - 37 & A B - 5 \
a b - 33 & a b - 4 \
A b - 4 & A b - 35 \
a B - 6 & a B - 36
\end{array}$$ How can you explain the drastic difference between these two crosses?

Accepted Answer

The answer of You are studying two linked genes in...

Question 11

Use the following to answer questions 
You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue shells and long antenna or green shells and short antenna. Crossing these lines yields FS1U1B11S1U1B0 progeny with blue shells and long antenna. Crossing FS1U1B11S1U1B0 progeny with beetles that have green shells and short antenna yields the following progeny:
Blue shell, long antenna82
Green shell, short antenna78
Blue shell, short antenna37
Green shell, long antenna43
Total240
-A chi-square test is done to test for independent assortment. What is the resulting chi-square value and how many degree(s) of freedom should be used in its interpretation?

Accepted Answer

The answer of Use the following to answer questions 
You...

Question 12

Assume that A and B are two linked genes on an autosome in Drosophila. A testcross is made where AB/ab flies are crossed to ab/ab flies and the progeny are counted and shown below. However, it is known that the Aa/bb genotype is lethal before the flies hatch and does not appear among the testcross progeny counted. What is the most precise map distance between the two genes that can be calculated from these data? Aa Bb = 235
Aa bb = 225
Aa Bb = 20

Accepted Answer

The answer of Assume that A and B are two...

Question 13

Use the following to answer questions :
Dr. Disney has been raising exotic fruit flies for decades. Recently, he discovered a strain of fruit flies that in the recessive condition have baby blue eyes that he designates as bb. He also has another strain of fruit flies that in the recessive condition have pink wings that are designated as pw. He is able to establish flies that are homozygous for both mutant traits. He mates these two strains with each other. Dr. Disney then takes phenotypically wild-type females from this cross and mates them with double recessive males. In the resulting testcross progeny, he observes 500 flies that are of the following makeup:
41with baby blue eyes and pink wings
207with baby blue eyes only
210with pink wings only
42with wild-type phenotype
-What is the relationship with respect to location between the two genes?

Accepted Answer

The genes are linked because the number of recombinant phenotypes (baby blue eyes and pink wings, wild-type) is less than the number of parental phenotypes (baby blue eyes only, pink wings only). The recombination frequency is calculated as (41 + 42) / 500 = 16.6%, indicating they are 16.6 map units apart.

Question 14

Is it possible for two different genes located on the same chromosome to assort independently?

Accepted Answer

If two genes are far enough apart on the same chromosome, the likelihood of crossover events between them is high enough that they can assort independently, resulting in a recombination frequency of approximately 50%.

Question 15

A genetic map shows which of the following?

Accepted Answer

A genetic map shows the linear order of genes on a chromosome, indicating their relative positions based on recombination frequencies.

Question 16

Recombination frequencies can be calculated by:

Accepted Answer

Recombination frequencies can be calculated by counting the number of recombinant and parental offspring when an individual who is heterozygous for two genes is involved in a testcross.

Question 17

Crossing over occurs during:

Accepted Answer

The answer of Crossing over occurs during:
...

Question 18

Two linked genes, (A) and (B), are separated by 18 m.u. A man with genotype Aa Bb marries a woman who is aa bb. The man's father was AA BB. What is the probability that their first two children will both be ab/ab?

Accepted Answer

The correct answer is A.The probability of the first child being ab/ab is 0.084. The probability of the second child being ab/ab is also 0.084. The probability of both children being ab/ab is 0.084 * 0.084 = 0.007056.

Question 19

A testcross includes:

Accepted Answer

A testcross involves crossing an individual showing the dominant phenotype with one that is homozygous recessive to determine the genotype of the dominant individual.

Question 20

Recombination occurs through:

Accepted Answer

Recombination occurs through crossing over, where homologous chromosomes exchange genetic material, and independent assortment, where chromosomes are distributed randomly into gametes.

Quiz 7: Linkage, Recombination, and Eukaryotic Gene Mapping