QUESTION 5
When determining the relative genetic distance between two genes, why is dihybrid back-cross preferable over traditional dihybrid cross?
a. 9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1 ratio.
b. Genotypes of the offspring can be determined based on their phenotype.
c. Unlike the traditional dihybrid cross, dihybrid back-cross produces all possible combinations of genotypes.
d. All of the above
Why do we map genes?
a. To understand how genes interact with each other
b. Comparative genomics analysis
c. To determine the genotype of an organism
d. All of the above
QUESTION 5
When determining the relative genetic distance between two genes, why is dihybrid back-cross preferable over traditional dihybrid cross?
| a. | 9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1 ratio. | |
| b. | Genotypes of the offspring can be determined based on their phenotype. | |
| c. | Unlike the traditional dihybrid cross, dihybrid back-cross produces all possible combinations of genotypes. | |
| d. | All of the above |
Why do we map genes?
| a. | To understand how genes interact with each other | |
| b. | Comparative genomics analysis | |
| c. | To determine the genotype of an organism | |
| d. | All of the above |
Related textbook solutions
Related questions
1. When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?
A. 9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio. | |
B. Genotypes of the offspring can be determined based on theirphenotype. |
C. If the genes are independently assorted, the dihybridback-cross would result in only 2 genotypes in the F1generation. | |
D. B and C |
Which of these statements is incorrect?
Syntenic genes are located on the same chromosome. |
Independent assortment results in recombinant chromosomes. |
You can reliably predict the relative genetic distance fromgenesâ physical distance on a chromosome. |
Linked genes are always syntenic. |
What is the relative genetic distance between two linked genesif the recombination frequency is 0.49?
0.49 cM |
4.9 cM |
49 cM |
490 cM |
What statement best explains the distortion in Mendelian ratiosobserved by Bateson & Punnett in 1905? (Reminder: they found anoverrepresentation of F2 offspring showing both dominant orrecessive phenotypes, and an underrepresentation of offspringdisplaying one dominant and one recessive phenotype)
| Human error: they should have been more careful about theirexperimental setup. |
| Gene linkage: Genes for flower color and pollen shape arephysically close on the same chromosome, leading to a breakdown inthe independent assortment of the alleles for these traits. |
| Chromosome crossover: Homologous recombination of twochromatids during meiosis caused the alleles to shuffle, resultingin a breakdown of the independent assortment of the alleles forthose genes. |
Random variation: No two situations are alike. In finitepopulations, you are going to get some variation across a mean. |
When determining the relative genetic distance between twogenes, why is dihybrid back-cross preferable over traditionaldihybrid cross?
| 9:3:3:1 phenotypic ratio is easier to work with than 1:1:1:1ratio. |
| Genotypes of the offspring can be determined based on theirphenotype. |
If the genes are independently assorted, the dihybrid back-crosswould result in only 2 genotypes in the F1 generation. |
B and C |
Why do we map genes?
| To understand how genes interact with each other |
| Comparative genomics analysis |
| To determine the genotype of an organism |
| All of the above |
In rats, several independently assorting autosomal genes affectcoat color. Gene A controls the distribution of yellowpigment in hair, and gene B causes black pigmentation. Thetwo genes interact as follows: AâBâ (gray), Aâbb(yellow), aaBâ (black), and aabb (cream). Thesegenotypes are only expressed in the presence of the dominant alleleof a third gene, C; rats with genotype cc arealbino.
a. Deduce the genotype of each albino mice, to the extent that ispossible, in the following table. Explain your answers.
True-breeding parents | F1 | F2 offspring |
Gray x albino-1 | All gray | 3/4 gray : 1/4 albino |
Gray x albino-2 | All gray | 9/16 gray : 3/16 yellow : 4/16 albino |
Part a. |
| Genotype of albino-1 parent is ⦠Explanation: |
| Genotype of albino-2 parent is ⦠Explanation: |
b. Deduce the genotype and phenotype of each parent in the followingtable. Explain your answers.
Parents | Numbers of offspring |
Cross 1 | 135 gray, 83 albino, 47 yellow, 44 black, 16 cream |
Cross 2 | 103 albino, 74 black, 25 cream |
Part b. | Parental phenotypes | Parental genotypes |
Cross 1 | x | x |
Explanation: | ||
Part b. | Parental phenotypes | Parental genotypes |
Cross 2 | x | x |
Explanation: | ||
c. A gray-colored rat ismated with one that is yellow. The offspring include an albino ratand a cream-colored rat. Diagram this cross. Be sure to include thePunnett square and the phenotypic ratio in the offspring.
Part c. |
| Cross is ⦠x Explanation: |
Punnett Square
Fill the punnet sq | ||||
Thank you
