Concept explainers
This problem examines possible biochemical explanations for variations of Mendel’s 9:3:3:1 ratio. Except where indicated, compounds 1, 2, 3, and 4 have different colors, as do mixtures of these compounds. A and B are enzymes that catalyze the indicated steps of the pathway. Alleles A and B specify functional enzymes A and B, respectively; these are completely dominant to alleles a and b, which do not specify any of the corresponding enzyme. If functional enzyme is present, assume that the compound to the left of the arrow is converted completely to the compound to the right of the arrow. For each pathway, what
a. | Independent pathways![]() |
b. | Redundant pathways![]() |
c. | Sequential pathway![]() |
d. | Enzymes A and B both needed to catalyze the reaction indicated.![]() |
e. | Branched pathways (assume enough of compound 1 for both pathways)![]() |
f. | Now consider independent pathways as in (a), but the presence of compound 2 masks the colors due to all other compounds. |
g. | Next consider the sequential pathway shown in (c), but compounds 1 and 2 are the same color. |
h. | Finally, examine the pathway that follows. Here, compounds 1 and 2 have different colors. The protein encoded by A prevents the conversion of compound 1 to compound 2. The protein encoded by B prevents protein A from functioning. |

a.
To determine:
The phenotypic ratio expected among the progeny if enzymes A and B follow independent pathways.
Introduction:
The pathways in which one enzyme converts a particular compound into another compound without interfering the catalytic action of another enzyme is termed as an independent pathway. For example, the reaction in which an enzyme A converts compound 1 into compound 2 and enzyme B converts compound 3 into compound 4 is an independent reaction.
Explanation of Solution
The genotype of both parents is as follows:
Parent 1: AaBb
Parent 2: AaBb
The alleles obtained from parent 1 are as follows:
AB, Ab, aB, and ab
The alleles obtained from parent 2 are as follows:
AB, Ab, aB, and ab
The cross between both the parents 2 is as follows:
♂/ ♀ | AB | Ab | aB | ab |
AB | AABB | AABb | AaBb | AaBb |
Ab | AABb | AAbb | AaBb | Aabb |
aB | AaBB | AABb | aaBB | aaBb |
ab | AaBb | Aabb | aaBb | aabb |
The alleles A and B are responsible for the activity of enzymes A and B. However, the alleles a and b are not responsible for the activity of an enzyme.
The above table represents that 9 progenies of parent 1 and 2 will contain enzyme A as they have both alleles A and B. 3 progenies will have enzyme A, 3 will have enzyme B and only 1 progeny would have no enzyme as it has only a and b alleles.
Thus, the phenotypic ratio expected among the progeny if enzyme A and B follow independent pathways is 9:3:3:1.

b.
To determine:
The phenotypic ratio expected among the progeny if enzymes A and B follow redundant pathways.
Introduction:
Redundant pathways are defined as the pathways in which a chemical reaction is catalyzed by more than one enzyme. In this type of pathways, two different enzymes are responsible for the conversion of one compound into another compound.
Explanation of Solution
The cross between both the parents 2 is as follows:
♂/ ♀ | AB | Ab | aB | ab |
AB | AABB | AABb | AaBb | AaBb |
Ab | AABb | AAbb | AaBb | Aabb |
aB | AaBB | AABb | aaBB | aaBb |
ab | AaBb | Aabb | aaBb | aabb |
Enzyme A and B follow the redundant pathways. They both convert compound 1 into compound 2. The presence of either allele A or B in the progeny can lead to the conversion of compound 1 into 2. There are 14 progenies that have either allele A or B, and there is one allele that has neither allele A nor B.
Thus, phenotypic ratio expected among the progeny if enzymes A and B follow redundant pathways is 15:1.

c.
To determine:
The phenotypic ratio expected among the progeny if enzymes A and B follow sequential pathways.
Introduction:
The pathway in which the conversion of one compound into another compound is a sequential process is termed as a sequential pathway. In this type of pathway, the enzyme A converts compound 1 into compound 2, and then the enzyme B convert compound 2 into compound 3. In this case, compound 1 is the reactant, while compound 3 is the product.
Explanation of Solution
The cross between both the parents 2 is as follows:
♂/ ♀ | AB | Ab | aB | ab |
AB | AABB | AABb | AaBb | AaBb |
Ab | AABb | AAbb | AaBb | Aabb |
aB | AaBB | AABb | aaBB | aaBb |
ab | AaBb | Aabb | aaBb | aabb |
The sequential pathway requires both the enzymes, enzyme A and enzyme B. This indicates that the progenies that have both the alleles A and B can have sequential pathways. According to the cross made between parents 1 and 2, there are 9 progenies that have both A and B alleles. These 9 progenies produce compound 3. However, the remaining 7 progenies will not produce compound 3. Out of these 7 progenies, 3 will produce compound 2 as they have allele A and 4 progenies will produce no compound.
Thus, the phenotypic ratio expected among the progeny if enzymes A and B follow sequential pathways is 9:4:3.

d.
To determine:
The phenotypic ratio expected among the progeny if enzymes A and B both are required to catalyze a reaction.
Introduction:
The set of alleles that provide characteristics to an organism are known as genotype. The characteristics that are expressed by the genotype are termed as phenotypic traits. The genotype controls the phenotypes. The ratio of all the phenotypes that are produced in the progenies is termed as the phenotypic ratio.
Explanation of Solution
The cross between both the parents 2 is as follows:
♂/ ♀ | AB | Ab | aB | ab |
AB | AABB | AABb | AaBb | AaBb |
Ab | AABb | AAbb | AaBb | Aabb |
aB | AaBB | AABb | aaBB | aaBb |
ab | AaBb | Aabb | aaBb | aabb |
According to the above table, there are only 9 progenies that can have both A and B enzymes. This is because only these 9 progenies have alleles A and B. The enzyme A is produced by the allele A while the enzyme B is produced by the allele B. However, the rest 7 progenies cannot produce both enzymes A.
Thus, the phenotypic ratio expected among the progeny if enzymes A and B both are required to catalyze a reaction is 9:7.

e.
To determine:
The phenotypic ratio expected among the progeny if enzymes A and B follow branched pathways.
Introduction:
The pathway in which a reactant can be converted into two different pathways by the action of two different enzymes is termed as a branched pathway. For example, compound 1 is a reactant. The enzyme A acts on it and converts it into compound 2. Another enzyme called B acts on this same reactant and produced another compound 3.
Explanation of Solution
The active enzyme A acts on compound 1 and produces compound 2. Similarly, the active enzyme B leads to the production of compound 3 from compound 1. In case both the enzymes are present, then both the compounds 2 and 3 are produced.
There are 9 progenies that produce the compounds, 3 produce only compound 2 and 3 progenies produce compound 3. There is only one progeny that do not produce any compound.
Thus, the phenotypic ratio expected among the progeny if enzymes A and B follow branched pathways is 9:3:3:1.

f.
To determine:
The phenotypic ratio expected among the progeny if enzymes A and B follow independent pathways and compound 2 masks color due to all compounds.
Introduction:
The enzyme A acts on compound 1 and converts it into another compound 2. In case the compound 2 masks the color due to the presence of all other compounds, then the presence or absence of enzyme B is immaterial.
Explanation of Solution
The cross between both the parents 2 is as follows:
♂/ ♀ | AB | Ab | aB | ab |
AB | AABB | AABb | AaBb | AaBb |
Ab | AABb | AAbb | AaBb | Aabb |
aB | AaBB | AABb | aaBB | aaBb |
ab | AaBb | Aabb | aaBb | aabb |
The above table represents a total of 16 progenies. There are 12 progenies that have active enzyme A as they have allele A. However, the rest four progenies do not have an active enzyme. This is because they lack the allele A. There are 3 progenies out of these 4 progenies that can produce active enzyme B as they have allele B. The remaining one progeny does not produce any active enzyme as it lacks both alleles A and B.
Thus, the phenotypic ratio expected among the progeny if enzymes A and B follow independent pathways and compound 2 masks color due to all compounds is 12:3:1.

g.
To determine:
The phenotypic ratio expected among the progeny if enzymes A and B follow sequential pathways given that compound 1 and 2 are of the same color.
Introduction:
The same color of compounds 1 and 2 indicate that the presence or absence of enzyme A that does not affect the sequential pathway. The enzyme responsible for the conversion of compound 2 into compound 3 is enzyme B.
Explanation of Solution
The genotype of both parents is as follows:
Parent 1: AaBb
Parent 2: AaBb
The alleles obtained from parent 1 are as follows:
AB, Ab, aB, and ab
The alleles obtained from parent 2 are as follows:
AB, Ab, aB, and ab
The cross between both the parents 2 is as follows:
♂/ ♀ | AB | Ab | aB | ab |
AB | AABB | AABb | AaBb | AaBb |
Ab | AABb | AAbb | AaBb | Aabb |
aB | AaBB | AABb | aaBB | aaBb |
ab | AaBb | Aabb | aaBb | aabb |
The above table represents that there are 9 progenies that have both enzymes A and B as they have alleles A and B. Out of these 9 progenies, 3 have enzyme A but not B. This reflects that compound 2 will be formed from these progenies. The rest 7 progenies will not have the same color as they do not have both enzymes A and B.
Thus, the phenotypic ratio expected among the progeny if enzymes A and B follow sequential pathways given that compound 1 and 2 are of the same color is 9:7.

h.
To determine:
The phenotypic ratio of the pathway in which protein A and B are associated with the conversion of compound 1 into compound 2.
Introduction:
The protein A is the inhibitory factor that prevents the conversion of compound 1 into compound 2. Another protein named B inhibits the functioning of protein A. The protein A can perform its function only in the absence of protein B.
Explanation of Solution
The protein A inhibits the formation of compound 2. This reflects that all the progenies that have protein A should not produce compound 2. The progenies that lack the allele A cannot produce the protein A. However, if the progenies have protein B along with protein A, then compound 2 can be produced. This reflects that the progenies that have both the proteins A and B can produce the same colored compounds. Similarly, the progenies that have proteins B and not A will also produce a colored compound 2. The progenies with proteins A but not B will produce a particular colored compound.
Thus, the phenotypic ratio of the pathway in which protein A and B are associated with the conversion of compound 1 into compound 2 is 12:3:1.
Want to see more full solutions like this?
Chapter 3 Solutions
GENETICS:FROM GENES TO GENOMES(LL)-PKG
- What is this?arrow_forwardMolecular Biology A-C components of the question are corresponding to attached image labeled 1. D component of the question is corresponding to attached image labeled 2. For a eukaryotic mRNA, the sequences is as follows where AUGrepresents the start codon, the yellow is the Kozak sequence and (XXX) just represents any codonfor an amino acid (no stop codons here). G-cap and polyA tail are not shown A. How long is the peptide produced?B. What is the function (a sentence) of the UAA highlighted in blue?C. If the sequence highlighted in blue were changed from UAA to UAG, how would that affecttranslation? D. (1) The sequence highlighted in yellow above is moved to a new position indicated below. Howwould that affect translation? (2) How long would be the protein produced from this new mRNA? Thank youarrow_forwardMolecular Biology Question Explain why the cell doesn’t need 61 tRNAs (one for each codon). Please help. Thank youarrow_forward
- Molecular Biology You discover a disease causing mutation (indicated by the arrow) that alters splicing of its mRNA. This mutation (a base substitution in the splicing sequence) eliminates a 3’ splice site resulting in the inclusion of the second intron (I2) in the final mRNA. We are going to pretend that this intron is short having only 15 nucleotides (most introns are much longer so this is just to make things simple) with the following sequence shown below in bold. The ( ) indicate the reading frames in the exons; the included intron 2 sequences are in bold. A. Would you expected this change to be harmful? ExplainB. If you were to do gene therapy to fix this problem, briefly explain what type of gene therapy youwould use to correct this. Please help. Thank youarrow_forwardMolecular Biology Question Please help. Thank you Explain what is meant by the term “defective virus.” Explain how a defective virus is able to replicate.arrow_forwardMolecular Biology Explain why changing the codon GGG to GGA should not be harmful. Please help . Thank youarrow_forward
- Stage Percent Time in Hours Interphase .60 14.4 Prophase .20 4.8 Metaphase .10 2.4 Anaphase .06 1.44 Telophase .03 .72 Cytukinesis .01 .24 Can you summarize the results in the chart and explain which phases are faster and why the slower ones are slow?arrow_forwardCan you circle a cell in the different stages of mitosis? 1.prophase 2.metaphase 3.anaphase 4.telophase 5.cytokinesisarrow_forwardWhich microbe does not live part of its lifecycle outside humans? A. Toxoplasma gondii B. Cytomegalovirus C. Francisella tularensis D. Plasmodium falciparum explain your answer thoroughly.arrow_forward
- Select all of the following that the ablation (knockout) or ectopoic expression (gain of function) of Hox can contribute to. Another set of wings in the fruit fly, duplication of fingernails, ectopic ears in mice, excess feathers in duck/quail chimeras, and homeosis of segment 2 to jaw in Hox2a mutantsarrow_forwardSelect all of the following that changes in the MC1R gene can lead to: Changes in spots/stripes in lizards, changes in coat coloration in mice, ectopic ear formation in Siberian hamsters, and red hair in humansarrow_forwardPleiotropic genes are genes that (blank) Cause a swapping of organs/structures, are the result of duplicated sets of chromosomes, never produce protein products, and have more than one purpose/functionarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage LearningHuman Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage LearningBiology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
- Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStaxConcepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax CollegeBiology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning





