Concept explainers
For all seven characters described in the data of Figure 2.5, Mendel allowed the
|
True-Breeding | Non-True-Breading | Ratio |
Round | 193 | 372 | 1:1.9 |
Yellow | 166 | 353 | 1:2.1 |
Gray | 36 | 64 | 1:1.8 |
Smooth | 29 | 71 | 1:2.4 |
Green | 40 | 60 | 1:1.5 |
Axial | 33 | 67 | 1:2.0 |
Tall | 28 | 72 | 1:2.6 |
Total: | 525 | 1059 | 1:2.0 |
When considering the data in this table, Page 45 keep in mind that they describe the characteristics of the
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Genetics: Analysis and Principles
- In Mendel’s 1866 publication as shown in Figure 1-4, he reports 705 purple-flowered (violet) offspring and 224 white-flowered offspring. The ratio he obtained is 3.15:1 for purple: white. How do you think he explained the fact that the ratio is not exactly 3:1?arrow_forwardFor all seven characters described in the data of Mendel allowed the F2 plants to self-fertilize. He found that whenF2 plants with recessive traits were crossed to each other, theyalways bred true. However, when F2 plants with dominant traitswere crossed, some bred true but others did not. A summary ofMendel’s results is shown to the right When considering the data in this table, keep in mind that theydescribe the characteristics of the F2 generation parents that haddisplayed a dominant phenotype. These data were deduced byanalyzing the outcome of the F3 generation. Based on Mendel’slaws, explain why the ratios were approximately 1:2arrow_forwardWhen Mendel examined the inheritance of seed colour, he crossed true-breeding green -seeded and true-breeding yellow-seeded plants. All of the plants produced were yellow- seeded. He then crossed the yellow-seeded plants and found that some of the offspring were green-seeded and some were yellow-seeded. What were the genotypes of the F2 offspring? Yy and Yy yellow green and Yy yellow and green Yy, YY and yyarrow_forward
- When Mendel examined the inheritance of seed colour, he crossed true-breeding green and true-breeding yellow plants. All of the plants produced were yellow seeded. He then crossed the yellow-seeded plants and found that some of the offspring were green-seeded and some were yellow-seeded. What were the genotypes of the original parents? Yy and Yy yellow and green There is not enough information to tell. YY and yy green and Yyarrow_forwardGregor Mendel examined the inheritance of two traits in pea plants: seed coat texture and colour. Seed coat texture can be represented as S-smooth and s-wrinkled, and seed coat colour can be represented as Y-yellow and y-green. SSYY plants were crossed with ssyy plants to yield F1 pea seeds that were all smooth and all yellow. By crossing plants grown from these F1 seeds, Mendel obtained four different phenotypes of F2 seeds: • smooth and green seeds wrinkled and green seeds smooth and yellow seeds wrinkled and yellow seeds ● Use the following information to answer the next question. ● The F2 phenotypic ratio that Mendel obtained upon crossing two heterozygous smooth and yellow F1 individuals would have been: smooth and green wrinkled and green : smooth and yellow: wrinkled and yellow Record only the numeric values associated with the phenotypes. (Do not include the colons, spaces, commas, etc.)arrow_forwardMendel crossed peas having green seeds with peas having yellow seeds. The F1 generation produced only yellow seeds. In the F2, the progeny consisted of 6022 plants with yellow seeds and 2001 plants with green seeds. Of the F2 yellow-seeded plants, 519 were self-fertilized with the following results: 166 bred true for yellow and 353 produced an F3 ratio of 3/4 yellow: 1/4 green. Explain these results by diagramming the crosses.arrow_forward
- Mendel crossed two Pea plants for plant height and flower color Tall plant (T) is dominant to Short Plant (t). Purple Flower (P) is dominant to white flower (p). Using the following information perform the dihybrid cross using punnett squares that will predict all possible genotypes of the offspring and list the number and description of the phenotypes of the offspring. A. One plant homozygous dominant for plant height and flower color crossed with another plant homozygous recessive for plant height and heterozygous for flower color.arrow_forwardWhen Mendel examined the inheritance of pod colour, he crossed true-breeding green and true-breeding yellow plants. All of the plants produced were yellow. He then crossed two of the F1 yellow plants and found that some of the offspring were green and some were yellow. What were the genotypes of the original parents? O yellow and green YY and yy There is not enough information to tell. O green and Yy O Yy and Yyarrow_forwardMendel crossed a wrinkle-seeded plant with pure round-seeded plant. Round is the dominant trait. Create a Punnett Square to represent this relationship. Using this information outlined in this Punnett Square, what percentage of the offspring do we anticipate will be wrinkle-seeded? (Chp 2) O 100% O 75% O 50% O 25% 0 0%arrow_forward
- A cross like this (between two individuals heterozygous for two traits) is often referred to as a "two-point test cross". The expected ratio of phenotypes if the two traits are caused by unlinked genes displaying simple Mendelian dominant inheritance is 9:3:3:1. For this ratio, match each term of the ratio with the appropriate phenotype Wrinkled and green peas 1. 9 Round and yellow peas 2. 3 Round and green peas 3. 1 Wrinkled and yellow peas > >arrow_forwardIn general terms, genes found on the same chromosome are linked, and will appear to defy Mendel’s Law of Independent Assortment. This law states that alleles (Links to an external site.) for different traits (Links to an external site.) are transmitted (Links to an external site.) to offspring (Links to an external site.) independently of one another. Functionally, this means that in a dihybrid testcross, in which a heterozygote is crossed to a double homozygous recessive individual, the expected 1:1:1:1 ratio will not be obtained. Instead, lower than expected numbers of non-parentals will result, because these non-parental flies are the result of recombination during synapsis. Interestingly, and functionally important in this exercise, synapsis only occurs in female fruit flies, requiring that the heterozygote in any study of linkage must be the female. Determining the relative positions of linked genes on a chromosome can be accomplished by calculating the frequency of…arrow_forwardA dihybrid cross is performed between two heterozygous individuals (heterozygous for two traits). The resulting offspring had 62 individuals that were dominant for trait 1 and 2 (D/D), 7 individuals that were R/R, 21 individuals that were R/D, and 25 individuals that were D/R. Using Mendelian inheritance as the null hypothesis, use χ2 analysis to determine if the trait follows Mendelian inheritance. A. How many D/D phenotype offspring are expected? B. How many R/D phenotype offspring are expected? C. How many degrees of freedom are there? D. What is your calculated χ2 value? E. What is the critical value if using a probability of 0.05? F. Does the trait follow Mendelian inheritance?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning