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Chapter 3: Problem 7

Mendel crossed peas having round seeds and yellow cotyledons (seed leaves) with peas having wrinkled seeds and green cotyledons. All the \(F_{1}\) plants had round seeds with yellow cotyledons. Diagram this cross through the \(\mathrm{F}_{2}\) generation, using both the Punnett square and forked-line, or branch diagram, methods.

Short Answer

Expert verified
In Mendel's pea experiment, given the traits for seed shape (round or wrinkled) and cotyledon color (yellow or green), we determined the F2 generation genotypes using both Punnett square and forked-line methods. The F2 generation genotypes are distributed as follows: 1 RRYY (round seeds, yellow cotyledons), 2 RrYY (round seeds, yellow cotyledons), 1 rrYY (wrinkled seeds, yellow cotyledons), 2 RRYy (round seeds, green cotyledons), 4 RrYy (round seeds, green cotyledons), 2 rrYy (wrinkled seeds, green cotyledons), 1 RRyy (round seeds, green cotyledons), 2 Rryy (round seeds, green cotyledons), and 1 rryy (wrinkled seeds, green cotyledons).

Step by step solution

01

Determine the F1 generation genotypes

Using the Punnett square method, we will find the possible genotype combinations for the F1 generation. Since both traits are inherited independently, we only need to cross the genotypes for each trait separately and then combine the results. Seed shape (R and r) cross: ``` R R r Rr Rr r Rr Rr ``` Cotyledon color (Y and y) cross: ``` Y Y y Yy Yy y Yy Yy ``` Combine the two results: all F1 generation plants have the genotype RrYy (round seeds, yellow cotyledons).
02

Set up the F2 generation genotypes

For the F2 generation, we will need to cross the F1 generation plants (RrYy) with themselves. Again, we will first cross the genotypes for seed shape and cotyledon color separately and then combine the results.
03

Determine the F2 generation seed shape genotypes using the Punnett square method

Cross the Rr genotype for seed shape (for F1 plants) with itself: ``` R r R RR Rr r Rr rr ``` We have the genotypes: RR (1), Rr (2), and rr (1) for the F2 generation seed shape.
04

Determine the F2 generation cotyledon color genotypes using the Punnett square method

Cross the Yy genotype for cotyledon color (for F1 plants) with itself: ``` Y y Y YY Yy y Yy yy ``` We have the genotypes: YY (1), Yy (2), and yy (1) for the F2 generation cotyledon color.
05

Combine the F2 generation results using the forked-line method

Now we can combine the results from steps 3 and 4 using the forked-line method. * Seed shape: 1 (RR) : 2 (Rr) : 1 (rr) * Cotyledon color: 1 (YY) : 2 (Yy) : 1 (yy) Using forked-line method: ``` 1 (RRYY) : 2 (RrYY) : 1 (rrYY) // seed shape combined with YY 2 (RRYy) : 4 (RrYy) : 2 (rrYy) // seed shape combined with Yy 1 (RRyy) : 2 (Rryy) : 1 (rryy) // seed shape combined with yy ``` The F2 generation genotypes are distributed as follows: 1 RRYY, 2 RrYY, 1 rrYY, 2 RRYy, 4 RrYy, 2 rrYy, 1 RRyy, 2 Rryy, and 1 rryy.

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Most popular questions from this chapter

Dentinogenesis imperfecta is a rare, autosomal, dominantly inherited disease of the teeth that occurs in about one in 8000 people (Witkop 1957 ). The teeth are somewhat brown in color, and the crowns wear down rapidly. Assume that a male with dentinogenesis imperfecta and no family history of the disease marries a woman with normal teeth. What is the probability that (a) their first child will have dentinogenesis imperfecta? (b) their first two children will have dentinogenesis imperfecta? (c) their first child will be a girl with dentinogenesis imperfecta?

An alternative to using the expanded binomial equation and Pascal's triangle in determining probabilities of phenotypes in a subsequent generation when the parents' genotypes are known is to use the following equation: \(\frac{n !}{s ! t !} a^{s} b^{t}\) where \(n\) is the total number of offspring, \(s\) is the number of offspring in one phenotypic category, \(t\) is the number of offspring in the other phenotypic category, \(a\) is the probability of occurrence of the first phenotype, and \(b\) is the probability of the second phenotype. Using this equation, determine the probability of a family of 5 offspring having exactly 2 children afflicted with sickle-cell anemia (an autosomal recessive disease \()\) when both parents are heterozygous for the sickle-cell allele.

What is the basis for homology among chromosomes?

In this chapter, we focused on the Mendelian postulates, probability, and pedigree analysis. We also considered some of the methods and reasoning by which these ideas, concepts, and techniques were developed. On the basis of these discussions, what answers would you propose to the following questions: (a) How was Mendel able to derive postulates concerning the behavior of "unit factors" during gamete formation, when he could not directly observe them? (b) How do we know whether an organism expressing a dominant trait is homozygous or heterozygous? (c) In analyzing genetic data, how do we know whether deviation from the expected ratio is due to chance rather than to another, independent factor? (d) since experimental crosses are not performed in humans, how do we know how traits are inherited?

Dentinogenesis imperfecta is a tooth disorder involving the production of dentin sialophosphoprotein, a bone-like component of the protective middle layer of teeth. The trait is inherited as an autosomal dominant allele located on chromosome 4 in humans and occurs in about 1 in 6000 to 8000 people. Assume that a man with dentinogenesis imperfecta, whose father had the disease but whose mother had normal teeth, married a woman with normal teeth. They have six children. What is the probability that their first child will be a male with dentinogenesis imperfecta? What is the probability that three of their six chil- dren will have the disease?
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