Genetics: Analysis and Principles
Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 6, Problem 16EQ

Three recessive traits in garden pea plants are as follows: yellow pods are recessive to green pods, bluish green seedlings are recessive to green seedlings, creeper (a plant that cannot stand up) is recessive to normal. A true-breeding normal plant with green pods and green seedlings was crossed to a creeper with yellow pods and bluish green seedlings. The F 1 plants were then crossed to creepers with yellow pods and bluish green seedlings. The following results were obtained:  

2059 green pods, green seedlings, normal   

151 green pods, green seedlings, creeper   

281 green pods, bluish green seedlings, normal    

15 green pods, bluish green seedlings, creeper  

2041 yellow pods, bluish green seedlings, creeper   

157 yellow pods, bluish green seedlings, normal   

282 yellow pods, green seedlings, creeper    

11 yellow pods, green seedlings, normal

Construct a genetic map that indicates the map distances between these three genes.

Expert Solution & Answer
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Summary Introduction

To review:

A genetic map indicating map distances between the genes for pod color, seedling color, and plant stature.

Introduction:

Thepair of alleles ofa gene determines the protein encoded by the genes, and results in the phenotype of the trait. True breeders have the monomorphic condition for the alleles of a trait. A genetic map helps in describing the arrangement of the genes on a particular chromosome.

Explanation of Solution

A normal pea plant, which is truebreeding for green seedling and green pods is crossed with a creeper pea plant, having bluish-green seedlings and yellow pods. The F1 offspring are then crossed with a creeper having bluish-green seedlings and yellow pods.

Different characters of the pea plant can be denoted as follows:

Dominant characters: G for green pods, S for green seedling, and C for normal plants

Recessive characters: g for yellow pods, s for bluish green seedlings, and c for the creeper.

Given, the true breeding normal plant, having green seedlings as well as green pods is crossed with true breeding creeper, having bluish green seedling and yellow pods. The genotype of true breeding plants is GGSSCC and ggsscc. Thus, the gametes will be GSC and gsc. The plants in the F1 generation obtained by crossing these true breeders will be GgSsCc.

When three genes are linked then G, S, and C alleles will be linked together whereas g, s, and c alleles will be linked together, on a homologous chromosome. Given, the F1 plants GgSsCc are crossed with ggsscc (creepers having yellow pods and bluish green seedlings) and following results were obtained:

Number of plants Phenotype
2059 Green pods, green seedlings, normal
151 Green pods, green seedlings, creepers
281 Green pods, bluish green seedlings, normal
15 Green pods, bluish green seedlings, creepers
2041 Yellow pods, bluish green seedlings, creepers
157 Yellow pods, bluish green seedlings, normal
282 Yellow pods, green seedlings, creepers
11 Yellow pods, green seedlings, normal

The genetic map for the three genes can be constructed by using the below-mentioned formula for map distance:

Map distance = Number of recombinant offspringTotal number of offsping×100mu (map units)

The distance between the three genes can be measured by separating the data of the offspring and phenotypic categories into gene pair, and then calculating the map distance between the two genes.

For the map distance between the genes for plant stature and pod color,

the number of offsprings is calculated for each pair of plant stature and color of pods.

green pods, normal = 2059+281= 2340green pods, creeper = 151+15 = 166

yellow pods, normal = 157+11= 168yellow pods, creeper = 2041+282 = 2323

Non-recombinant offspring are 2340 normal, green pods, and 2323 creeper, yellow pods.

Recombinant offspring are 166 creeper, green pods, and 168 normal, yellow pods.

Therefore, the map distance will be calculated as,

Map distance =166+1682340+2323+166+168×100 muMap distance =3344997×100 muMap distance = 6.7 mu

For the map distance between the genes for plant stature and seedling color, the number of offspring having genes for plant stature and color of the seedlingis given as follows:

green seedling, normal = 2059+11 = 2070green seedling, creeper = 151+282 = 433

bluish green seedling, normal = 281+157 = 438bluish green seedling, creeper =15+2041 = 2056

Nonrecombinant offspring are 2070 normal, green seedlings, and 2056 creeper, bluish green seedlings, and the recombinant offspring are 433 creeper, green seedlings, and 438 normal, bluish green seedlings.

Therefore, the map distance will be calculated as.

Map distance = 433+4382070+2056+433+438×100muMap distance = 8714997×100muMap distance = 17.4 mu

For the map distance between the genes for seedling color and pod color, the number of offspring having genes for the color of seedling and pods is given as follows:

green seedlings, green pods = 2059+151= 2210bluish green seedlings, green pods = 15+281 = 296

green seedlings, yellow pods = 282+11= 293bluish green seedlings, yellow pods = 2041+157 = 2198

Non-recombinant offspring are 2210 green seedling, green pods, and 2198 bluish green seedling, yellow pods, and the recombinant offspring are 296 bluish green seedling, green pods, and 293 green seedling, yellow pods.Therefore, the map distance will be calculated as,

Map distance = 296+2932210+2198+296+293×100 muMap distance =5894997×100 muMap distance = 11.8 mu

Thus, the distance between the genes for seedling and pods colors is 11.8 mu, and that between the genes for plant stature and seedling color is 17.4 mu and for the genes for plant stature and pod color, it is 6.7 mu. Therefore, the order of the three genes, according to the map distance in-between them will be seedling color, pod color, plant stature. Gene for pod color is present in the middle. Genetic map is shown below:

Genetics: Analysis and Principles, Chapter 6, Problem 16EQ

Conclusion

Therefore, it can be concluded that according to the map distance calculated in-between the genes for pod color, plant stature, and seedling color, the order of the three genes will be seedling color, pod color, plant stature(or the opposite order). The distance between the genes responsible for seedling and pods colors is 11.8 mu, and between the genes for plant stature and pod color, it is 6.7 mu. The genes that code for plant stature and seedling color are 17.4 mu apart.

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Students have asked these similar questions
Three recessive traits in garden pea plants are as follows: yellowpods are recessive to green pods, bluish green seedlings are recessiveto green seedlings, creeper (a plant that cannot stand up) isrecessive to normal. A true-breeding normal plant with green podsand green seedlings was crossed to a creeper with yellow pods andbluish green seedlings. The F1 plants were then crossed to creeperswith yellow pods and bluish green seedlings. The following resultswere obtained: 2059 green pods, green seedlings, normal151 green pods, green seedlings, creeper281 green pods, bluish green seedlings, normal15 green pods, bluish green seedlings, creeper2041 yellow pods, bluish green seedlings, creeper157 yellow pods, bluish green seedlings, normal282 yellow pods, green seedlings, creeper11 yellow pods, green seedlings, normal Construct a genetic map that indicates the map distances betweenthese three genes?
A researcher at ASU is studying the exotic Unicorn Fairy plant. The plant has three genes of interest with the dominant alleles being T-tall, H=hairy leaves and P=purple flowers. The three genes are located in different chromosomes. A trihybrid plant is crossed with a plant that is heterozygous for T and homozygous recessive for the other two genes. What is the probability of getting offspring that is tall, has hairy leaves and white flowers? 9/64 5/8 8/64 O 3/64 O 3/16
J.W. McKay crossed a stock (true-breeding) melon plant that produced tan seeds with a plant that only produced red seeds and obtained the following results (J.W. McKay. 1936. Journal of Heredity 27:110-112). Cross F1 F2 Tan x red 13 tan 93 tan, 24 red a) Explain the inheritance of tan seeds and red seeds in this plant. b) Assign symbols for the alleles in this cross and draw out the Punnett Squares for the initial cross and the F1 cross.

Chapter 6 Solutions

Genetics: Analysis and Principles

Ch. 6 - 1. What is the difference in meaning between the...Ch. 6 - 2. When a chi square analysis is applied to solve...Ch. 6 - 3. What is mitotic recombination? A heterozygous...Ch. 6 - 4. Mitotic recombination can occasionally produce...Ch. 6 - 5. A crossover has occurred in the bivalent shown...Ch. 6 - A crossover has occurred in the bivalent shown...Ch. 6 - A diploid organism has a total of 14 chromosomes...Ch. 6 - If you try to throw a basketball into a basket,...Ch. 6 - 9. By conducting testcrosses, researchers have...Ch. 6 - In humans, a rare dominant disorder known as...Ch. 6 - 11. When true-breeding mice with brown fur and...Ch. 6 - Though we often think of genes in terms of the...Ch. 6 - 13. If the likelihood of a single crossover in a...Ch. 6 - 14. In most two-factor crosses involving linked...Ch. 6 - Researchers have discovered that some regions of...Ch. 6 - 16. Describe the unique features of ascomycetes...Ch. 6 - Figure 6.1 shows the first experimental results...Ch. 6 - In the experiment of Figure 6.6, Stern followed...Ch. 6 - 3. Explain the rationale behind a testcross. Is it...Ch. 6 - 4. In your own words, explain why a testcross...Ch. 6 - Explain why the percentage of recombinant...Ch. 6 - 6. If two genes are more thanapart, how would you...Ch. 6 - 7. In Morgan’s three-factor crosses of Figure 6.3,...Ch. 6 - Two genes are located on the same chromosome and...Ch. 6 - 9. Two genes, designated A and B, are locatedfrom...Ch. 6 - 10. Two genes in tomatoes areapart; normal fruit...Ch. 6 - In the tomato, three genes are linked on the same...Ch. 6 - A trait in garden peas involves the curling of...Ch. 6 - Prob. 13EQCh. 6 - 14. In the garden pea, several different genes...Ch. 6 - A sex-influenced trait is dominant in males and...Ch. 6 - Three recessive traits in garden pea plants are as...Ch. 6 - In mice, a trait called snubnose is recessive to a...Ch. 6 - 18. In Drosophila, an allele causing vestigial...Ch. 6 - 19. Three autosomal genes are linked along the...Ch. 6 - 20. Let’s suppose that two different X-linked...Ch. 6 - Prob. 21EQCh. 6 - In mice, a dominant allele that causes a short...Ch. 6 - 2. In Chapter 3, we discussed the idea that the X...Ch. 6 - Mendel studied seven traits in pea plants, and the...
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