Question

Three independently assorting genes are known to control the following biochemical pathway that provides the basis for flower color in a hypothetical plant: Colorless \(\stackrel{A^{-}}{\longrightarrow}\) yellow \(\stackrel{B^{-}}{\longrightarrow}\) green \(\stackrel{C-}{\longrightarrow}\) speckled Three homozygous recessive mutations are also known, each of which interrupts a different one of these steps. Determine the phenotypic results in the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) generations resulting from the \(P_{1}\) crosses of true-breeding plants listed here: (a) speckled \((A A B B C C) \times\) yellow \((A A b b C C)\) (b) yellow \((A A b b C C) \times \operatorname{green}(A A B B c c)\) (c) colorless \((a a B B C C) \times\) green \((A A B B c c)\)

Short answer

Answer: a) Speckled x Yellow: F1: Green F2: Speckled (1/16), Green (10/16), Yellow (2/16), Colorless (3/16) b) Yellow x Green: F1: Green F2: Speckled (1/64), Green (12/64), Yellow (6/64), Colorless (9/64) c) Colorless x Green: F1: Green F2: Speckled (1/64), Green (12/64), Yellow (6/64), Colorless (9/64)

Step-by-step solution

F1 Generation:

In this cross, the parent plants are homozygous dominant for gene A, homozygous recessive for gene B in the Yellow parent, and homozygous dominant for gene C. The F1 progeny will inherit one allele from each parent for each gene. They'll inherit A and C from both parents and B from the Speckled parent and b from the Yellow parent. Thus, the F1 generation will have the following genotype: AaBbCC. This corresponds to the Green phenotype because both A and C are functional (at least one dominant allele), but B is recessive and a single dominant allele in gene A does not produce the yellow color.

F2 Generation:

To find the phenotypic results for the F2 generation, we need to cross the F1 plants with each other (AaBbCC x AaBbCC). Since we have three independently assorting genes, we can look at each gene separately and then combine the results. Independent inheritance leads to the following possibilities for each gene: A: AA, Aa, Aa, and aa B: BB, Bb, Bb, and bb C: CC, CC, CC, and CC (all F2 plants will have C since parents are both homozygous) Phenotypes: - Speckled: AABBCC (1/16) - Green: AABbCC, AAbbCC, AaBBCC, aaBbCC, AABbcc, AaBbCC (10/16) - Yellow: AABBcc, AAbbcc (2/16) - Colorless: aaBBCC, aabbCC, aaBbcc (3/16) #a) Speckled (AABBCC) x Yellow (AAbbCC) Result#: F1: Green (AaBbCC) F2: Speckled (1/16), Green (10/16), Yellow (2/16), Colorless (3/16) #b) Yellow (AAbbCC) x Green (AABBcc)#

F1 Generation:

In this cross, both parent plants are homozygous for gene A, but the Yellow parent is homozygous recessive for gene B, and the Green parent is homozygous recessive for gene C. The F1 progeny will inherit one allele from each parent for each gene. The F1 genotypes will be AaBbCc. This corresponds to the Green phenotype as both A and B are functional, with at least one dominant allele.

F2 Generation:

We need to cross F1 plants with each other (AaBbCc x AaBbCc). Using the same approach as before, we look at each gene separately and then combine the results. Independent inheritance possibilities for each gene: A: AA, Aa, Aa, and aa B: BB, Bb, Bb, and bb C: CC, Cc, Cc, and cc Phenotypes: - Speckled: AABBCc (1/64) - Green: AABBcc, AABbCc, AAbbCc, aaBBcc, aaBbCc (12/64) - Yellow: AABBCc, AAbbCC, aaBBCC, aaBbCc (6/64) - Colorless: AAbbCC, aaBBCC, aabbCC, aaBbCC, aabbCc (9/64) #b) Yellow (AAbbCC) x Green (AABBcc) Result#: F1: Green (AaBbCc) F2: Speckled (1/64), Green (12/64), Yellow (6/64), Colorless (9/64) #c) Colorless (aabbCC) x Green (AABBBc)#

F1 Generation:

In this cross, the Colorless parent is homozygous recessive for genes A and B, while the Green parent is homozygous dominant for genes A and B but homozygous recessive for gene C. The F1 progeny will inherit the following genotype: AaBbCc, which corresponds to the Green phenotype as all genes have at least one dominant allele.

F2 Generation:

Cross F1 plants with each other (AaBbCc x AaBbCc). Using the same approach as before, we look at each gene separately and then combine the results. Independent inheritance possibilities for each gene: A: AA, Aa, Aa, and aa B: BB, Bb, Bb, and bb C: CC, Cc, Cc, and cc Phenotypes: - Speckled: AABBCc (1/64) - Green: AABBcc, AABbCc, AAbbCc, aaBBcc, aaBbCc (12/64) - Yellow: AABBCc, AAbbCC, aaBBCC, aaBbCc (6/64) - Colorless: AAbbCC, aaBBCC, aabbCC, aaBbCC, aabbCc (9/64) #c) Colorless (aabbCC) x Green (AABBBc) Result#: F1: Green (AaBbCc) F2: Speckled (1/64), Green (12/64), Yellow (6/64), Colorless (9/64)

Key concepts

Independent Assortment

Independent assortment is a key principle of genetics stating that different genes and their alleles are distributed independently of each other during the formation of gametes. This concept was first discovered by Gregor Mendel through his work on pea plants. In this exercise, three genes are involved, rendering it an excellent example of independent assortment in action. Each gene requires independent consideration when predicting the outcome of crosses.

• Gene A affects the progression from colorless to yellow.
• Gene B controls the change from yellow to green.
• Gene C determines the transition from green to speckled.

Due to independent assortment, the combinations of alleles that form a gamete are random. For example, when plants with genotypes AaBbCc are crossed, the alleles can independently assort to give many different combinations. Each gene pairing results in a variety of phenotypic outcomes depending on dominance and recessiveness interactions.

Phenotype Ratios

The phenotype ratio is an important aspect helping us predict the appearance of offspring from genetic crosses. In this scenario, several phenotype outcomes are possible depending on the combinations of alleles from the parent plants. For each set of organisms, the phenotype ratio explains how likely each resulting phenotype is in the F2 generation.
In the case of the cross between speckled and yellow plants, different phenotypes can appear:

• Speckled (1/16)
• Green (10/16)
• Yellow (2/16)
• Colorless (3/16)

These ratios are derived through the independent assortment of genes A, B, and C along with their combinations occurring during gamete formation. Understanding phenotype ratios allows scientists and students alike to predict how features, such as flower color, will manifest in future generations.

Biochemical Pathway

A biochemical pathway describes the series of chemical reactions that occur within a biological system, in this instance influencing flower coloration. Each step in the pathway is controlled by a gene that codes for an enzyme to convert one intermediate to the next compound in the sequence.
For these plants:

• Gene A produces an enzyme converting colorless intermediate to yellow pigment.
• Gene B regulates the transformation of yellow to green pigment.
• Gene C ultimately directs the formation of speckled from green pigment.

Each gene's expression is vital to the progression through the pathway. If any step is interrupted, it alters the final appearance, or phenotype, of the plant. For example, if a plant lacks the necessary B gene product, the plant's flowers cannot progress past yellow. This insight aids our understanding of how genetic information translates into physical traits through a complex network of biochemical interactions.