Q: Which formula correctly calculates the coefficient of coincidence (c.o.c.) in a three-point testcross?

$$\text{c.o.c.} = \frac{\text{observed double crossovers}}{\text{expected double crossovers}}$$

Question 1

Which of the following statements best explains why the number of linkage groups in an organism is typically equal to the number of chromosome pairs in its haploid genome?

Accepted Answer

Genes on the same chromosome are inherited together unless separated by recombination, so each chromosome acts as a linkage group.

Question 2

In a two-point testcross between genes w and m in Drosophila, 38% of the progeny are recombinants. What is the estimated map distance between these two genes?

Accepted Answer

$$38$$ map units

Question 3

A testcross between a heterozygote for two linked genes (A and B) and a homozygous recessive individual produces the following progeny: 62% parental types and 38% recombinants. What does this indicate about the relationship between genes A and B?

Accepted Answer

Genes A and B are linked and located close together on the same chromosome.

Question 4

If the recombination frequency between two genes is 50%, what can you conclude about their genetic relationship?

Accepted Answer

They are on different chromosomes or far apart on the same chromosome.

Question 5

In a three-point testcross, the following numbers of progeny are observed: 98, 92, 47, 43, 4, 6, 361, 349. What is the first step in determining the gene order?

Accepted Answer

Identify the most and least abundant progeny classes to determine parentals and double crossovers.

Question 6

Which formula correctly calculates the coefficient of coincidence (c.o.c.) in a three-point testcross?

Accepted Answer

$$	ext{c.o.c.} = \frac{	ext{observed double crossovers}}{	ext{expected double crossovers}}$$

Question 7

A three-point testcross yields 10 observed double crossovers, but 20 are expected based on single crossover frequencies. What is the interference value ($$I$$)?

Accepted Answer

$$I = 1 - \frac{10}{20} = 0.5$$

Question 8

Why do recombination frequencies underestimate the true physical distance between two genes when they are far apart on a chromosome?

Accepted Answer

Because double crossovers restore the parental combination, making some recombinants undetectable.

Question 9

Which of the following is a molecular genetic marker used in modern gene mapping techniques?

Accepted Answer

Single-nucleotide polymorphisms (SNPs)

Question 10

In a genome-wide association study (GWAS), a particular SNP is always found in individuals with a specific disease phenotype. What does this suggest?

Accepted Answer

The SNP is likely closely linked to the gene responsible for the disease.

Question 11

A Drosophila geneticist observes that the progeny of a testcross between two genes show a 1:1:1:1 ratio. What does this indicate about the genes?

Accepted Answer

The genes are unlinked and assort independently.

Question 12

If the recombination frequency between gene w and gene sn is 15%, and between gene sn and gene m is 23%, what is the map distance between w and m?

Accepted Answer

$$15 + 23 = 38$$ map units

Question 13

Which of the following best describes the effect of chromatid interference?

Accepted Answer

The occurrence of one crossover reduces the probability of another crossover nearby.

Question 14

In a three-point testcross, why is it important to identify the double crossover classes?

Accepted Answer

Double crossover classes reveal the gene in the middle, allowing correct gene order determination.

Question 15

Which of the following is NOT a reason why recombination frequencies plateau at 50%?

Accepted Answer

Recombination frequencies above 50% indicate complete linkage.

Question 16

A geneticist is mapping three genes (b, a, c) and finds the following recombination frequencies: b-a = 20%, a-c = 10%, b-c = 30%. Which gene is in the middle?

Accepted Answer

Gene a

Question 17

Which of the following best describes the difference between a genetic map and a physical map?

Accepted Answer

A genetic map is based on recombination frequencies, while a physical map is based on actual distances in base pairs.

Question 18

Which of the following is a key advantage of using molecular markers such as SNPs in gene mapping compared to traditional phenotypic markers?

Accepted Answer

Molecular markers are more abundant and can detect smaller genetic differences.

Linkage, Recombination, and Gene Mapping: Study Notes

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