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Monday, September 19, 2016


When fertilization occurs between two true-breeding parents that differ in only one characteristic, the process is called a monohybrid cross, and the resulting offspring are monohybrids. Mendel performed seven monohybrid crosses involving contrasting traits for each characteristic. On the basis of his results in F1 and F2 generations, Mendel postulated that each parent in the monohybrid cross contributed one of two paired unit factors to each offspring, and every possible combination of unit factors was equally likely.
To demonstrate a monohybrid cross, consider the case of true-breeding pea plants with yellow versus green pea seeds. The dominant seed color is yellow; therefore, the parental genotypes were YY for the plants with yellow seeds and yy for the plants with green seeds, respectively. A Punnett square, devised by the British geneticist Reginald Punnett, can be drawn that applies the rules of probability to predict the possible outcomes of a genetic cross or mating and their expected frequencies. To prepare a Punnett square, all possible combinations of the parental alleles are listed along the top (for one parent) and side (for the other parent) of a grid, representing their meiotic segregation into haploid gametes. Then the combinations of egg and sperm are made in the boxes in the table to show which alleles are combining. Each box then represents the diploid genotype of a zygote, or fertilized egg, that could result from this mating. Because each possibility is equally likely, genotypic ratios can be determined from a Punnett square. If the pattern of inheritance (dominant or recessive) is known, the phenotypic ratios can be inferred as well. For a monohybrid cross of two true-breeding parents, each parent contributes one type of allele. In this case, only one genotype is possible. All offspring are Yy and have yellow seeds (Figure 1).

 Figure 1 In the P generation, pea  plants  that are true-breeding for the dominant yellow phenotype are crossed with plants  with the  recessive green phenotype. This cross produces F1  heterozygotes with a yellow phenotype. Punnett square analysis can be used to predict the genotypes of the F2 generation.

A self-cross of one of the Yy heterozygous offspring can be represented in a 2 × 2 Punnett square because each parent can donate one of two different alleles. Therefore, the offspring can potentially have one of four allele combinations: YY, Yy, yY, or yy (Figure 1). Notice that there are two ways to obtain the Yy genotype: a Y from the egg and a y from the sperm, or a y from the egg and a Y from the sperm. Both of these possibilities must be counted. Recall that Mendels pea- plant characteristics behaved in the same way in reciprocal crosses. Therefore, the two possible heterozygous combinations produce offspring that are genotypically and phenotypically identical despite their dominant and recessive alleles deriving from different parents. They are grouped together. Because fertilization is a random event, we expect each combination to be equally likely and for the offspring to exhibit a ratio of YY:Yy:yy genotypes of 1:2:1 (Figure 1). Furthermore, because the YY and Yy offspring have yellow seeds and are phenotypically identical, applying the sum rule of probability, we expect the offspring to exhibit a phenotypic ratio of 3 yellow:1 green. Indeed, working with large sample sizes, Mendel observed approximately this ratio in every F2 generation resulting from crosses for individual traits.
Mendel validated these results by performing an F3 cross in which he self-crossed the dominant- and recessive-expressing F2 plants. When he self-crossed the plants expressing green seeds, all of the offspring had green seeds, confirming that all green seeds had homozygous genotypes of yy. When he self-crossed the F2 plants expressing yellow seeds, he found that one-third of the plants bred true, and two-thirds of the plants segregated at a 3:1 ratio of yellow:green seeds. In this case, the true-breeding plants had homozygous (YY) genotypes, whereas the segregating plants corresponded to the heterozygous (Yy) genotype. When these plants self-fertilized, the outcome was just like the F1 self-fertilizing cross.
The Test Cross Distinguishes the Dominant Phenotype

Beyond predicting the offspring of a cross between known homozygous or heterozygous parents, Mendel also developed a way to determine whether an organism that expressed a dominant trait was a heterozygote or a homozygote. Called the test cross, this technique is still used by plant and animal breeders. In a test cross, the dominant-expressing organism is crossed with an organism that is homozygous recessive for the same characteristic. If the dominant-expressing organism is a homozygote, then all F1 offspring will be heterozygotes expressing the dominant trait (Figure 2). Alternatively, if the dominant expressing organism is a heterozygote, the F1 offspring will exhibit a 1:1 ratio of heterozygotes and recessive homozygotes (Figure 2). The test cross further validates Mendels postulate that pairs of unit factors segregate equally.

Figure  2  A test  cross can  be  performed to determine whether an  organism expressing a dominant trait is a homozygote or a heterozygote.

In pea  plants,  round peas (R) are dominant to wrinkled peas (r). You do a test cross between a pea  plant with wrinkled peas (genotype rr) and  a plant of unknown  genotype that has  round peas. You end  up with three plants,  all which have  round peas. From this data, can  you tell if the round pea  parent plant is homozygous dominant or heterozygous? If the  round  pea  parent plant  is heterozygous, what  is the  probability  that  a random sample of 3 progeny peas will all be round?
Many human diseases are genetically inherited. A healthy person in a family in which some members suffer from a recessive genetic disorder may want to know if he or she has the disease-causing gene and what risk exists of passing the disorder on to his or her offspring. Of course, doing a test cross in humans is unethical and impractical. Instead, geneticists use pedigree analysis to study the inheritance pattern of human genetic diseases (Figure 3).

 Figure  3  Alkaptonuria  is a recessive genetic disorder in which two amino  acids,  phenylalanine and  tyrosine, are  not properly metabolized. Affected individuals  may have  darkened skin and  brown urine, and  may suffer joint damage and other complications. In this pedigree, individuals with the disorder are indicated in blue and have  the genotype aa. Unaffected individuals  are  indicated in yellow and  have  the genotype AA or Aa. Note that it is often possible to determine a persons genotype from the genotype of their offspring. For example, if neither  parent has the disorder but their child does, they must be heterozygous. Two individuals on the pedigree have  an unaffected phenotype but unknown  genotype. Because they do not have  the  disorder, they must  have  at least  one  normal allele, so their genotype gets  the A? designation.

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