Chromosome Theory and the Castle and Morgan Debate. Discovery and Types of Genetic Linkage. Genetics and Statistical Analysis. Thomas Hunt Morgan and Sex Linkage. Developing the Chromosome Theory. Genetic Recombination. Gregor Mendel and the Principles of Inheritance. Mitosis, Meiosis, and Inheritance. Multifactorial Inheritance and Genetic Disease. Non-nuclear Genes and Their Inheritance. Polygenic Inheritance and Gene Mapping. Sex Chromosomes and Sex Determination.
Sex Determination in Honeybees. Test Crosses. Biological Complexity and Integrative Levels of Organization. Genetics of Dog Breeding. Human Evolutionary Tree. Mendelian Ratios and Lethal Genes. Environmental Influences on Gene Expression. Epistasis: Gene Interaction and Phenotype Effects. Genetic Dominance: Genotype-Phenotype Relationships. Phenotype Variability: Penetrance and Expressivity.
Citation: Miko, I. Nature Education 1 1 Why can you possess traits neither of your parents have? The relationship of genotype to phenotype is rarely as simple as the dominant and recessive patterns described by Mendel.
Aa Aa Aa. Complete versus Partial Dominance. Figure 1. Figure Detail. Multiple Alleles and Dominance Series. Summarizing the Role of Dominance and Recessivity. References and Recommended Reading Keeton, W.
Heredity 35 , 85—98 Parsons, P. Nature , 7—12 link to article Stratton, F. Article History Close. Within a population of butterflies, the color brown B is dominant over the color white b.
Given this simple information, which is something that is very likely to be on an exam, calculate the following: The percentage of butterflies in the population that are heterozygous. The frequency of homozygous dominant individuals. Answers: The first thing you'll need to do is obtain p and q. So, since white is recessive i. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q 2 which works out to be 0.
Now then, to answer our questions. First, what is the percentage of butterflies in the population that are heterozygous? Well, that would be 2pq so the answer is 2 0.
Second, what is the frequency of homozygous dominant individuals? That would be p 2 or 0. A rather large population of Biology instructors have red-sided individuals and tan-sided individuals.
Assume that red is totally recessive. Please calculate the following: The allele frequencies of each allele. Therefore, q the square root of q 2 is 0. The expected genotype frequencies. The number of heterozygous individuals that you would predict to be in this population. Answer: That would be 0. The expected phenotype frequencies. Conditions happen to be really good this year for breeding and next year there are 1, young "potential" Biology instructors.
Since different loci are involved, you can't use the term multiple allele inheritance. So geneticists have devised the term "multiple gene" or "polygenic inheritance. Because there are different genes on different loci involved, numerous genotypes and phenotypes appearances are possible.
The Rh factor is a good example of polygenic inheritance. The disease sickle-cell anemia is a good example of a genetic mutation in which the gene for the vital protein hemoglobin has mutated. The sickle-cell gene has an altered DNA base pattern so that it codes for the amino acid valine instead of glutamic acid at a precise location in the hemoglobin molecule.
This results in a change in the structure of the molecule resulting in sickle-shaped rather than normal disk-shaped red blood cells. These abnormal cells do not flow as well through minute capillaries, forming painful "log jams" that impede blood circulation. A human male and female each have 23 pairs of homologous chromosomes per cell, a total of 46 chromosomes. A male or female with Down's syndrome has the 21st chromosome autosome in triplicate.
Instead of the normal homologous pair, there are three 21 chromosomes. In Klinefelter's syndrome, there are three 23 chromosomes X-Y chromosomes rather than the normal pair. In this case the individual has two X chromosomes and one Y chromosome. Because the Y chromosome carries the male-determining factor, the individual is a phenotypic male with a penis, although there may be some breast enlargement. In both of these syndromes, the total number of chromosomes per cell is raised by one compared with normal somatic cells.
See the following table: The Rh factor is an interesting example of polygenic inheritance. Unlike the A-B-O blood types where all the alleles occur on one pair of loci on chromosome pair 9, the Rh factor involves three different pairs of alleles located on three different loci on chromosome pair 1. Possible genotypes will have one C or c, one D or d, and one E or e from each chromosome. In order to determine how many different genotypes are possible, you must first determine how many different gametes are possible for each parent, then match all the gametes in a genetic checkerboard.
This number of gametes is based on all the total possible ways these genes can be inherited on each chromosome of homologous pair 1. The formula was actually devised by several of my general biology students.
It may occur somewhere in a textbook, but the students came up with it independently. See the following diagram showing one pair of homologous chromosomes, each with a single locus.
Only one allele can occur at each locus, but there are 4 possible alleles per locus. Questions 32 - See the following table showing the number of different gametes due to independent assortment of chromosomes during meiosis and random combination of gametes. Matching Questions 35 - 1. The following illustration shows a highly magnified cell membrane containing two kinds of embedded proteins, a carrier protein and a cell recognition protein.
The cell recognition protein contains a carbohydrate "antenna" composed of polysaccharide subunits. Vaccines Resulting In Active Immunity 2. Serums Resulting In Passive Immunity 3.
Genetics Extra Credit Problems Questions 61 - Remember that the gene allele for taster T is dominant over the gene allele for nontaster t : Questions 63 - Human skin color is a good example of polygenic inheritance in people. The offspring contain seven different shades of skin color based on the number of capital letters in each genotype. The words dominant and recessive are placed in quotation marks because these pairs of alleles are not truly dominant and recessive as in some of the garden pea traits that Gregor Mendel studied.
A genotype with all "recessive" small case genes aabbcc has the lowest amount of melanin and very light skin. Each "dominant" capital gene produces one unit of color, so that a wide range of intermediate skin colors are produced, depending on the number of "dominant" capital genes in the genotype.
For example, a genotype with three "dominant" capital genes and three small case "recessive" genes AaBbCc has a medium amount of melanin and an intermediate skin color.
This latter genotype would be characteristic of a mulatto. In the above cross between two mulatto genotypes AaBbCc x AaBbCc , each parent produces eight different types of gametes and these gametes combine with each other in 64 different ways resulting in a total of seven skin colors. The skin colors can be represented by the number of capital letters, ranging from zero no capital letters to six all capital letters.
The approximate shades of skin color corresponding to each genotype are shown in the above table. Note: Skin color may involve at least four pairs of alleles with nine or more shades of skin color. Each term in the expression represents the number of offspring with a specific skin color phenotype based on the number of capital letters in the genotype.
For example, 20 offspring have three capital letters in their genotype and have a skin color that is intermediate between very dark with all caps AABBCC and very light with no caps aabbcc. Questions 69 - These questions refer to the Rh types of Chrissy and John, and their baby boy named Cinco. Questions 63 - Questions 78 - Remember that the A and B alleles are dominant over the O allele. The type O blood phenotype must be homozygous for the O allele.
Type AB blood phenotype must be heterozygous for the A and B alleles. For these questions, use the process of elimination. Then eliminate the only parents that could have an AB baby, and so forth. Questions 87 - To determine the fractional probability for a taster boy with type B blood, you must make a cross between John and Mary using a genetic checkerboard Punnett square.
A probability of one percent for some event indicates that it is guaranteed to occur, whereas a probability of zero 0 percent indicates that it is guaranteed to not occur, and a probability of 0.
To demonstrate this with a monohybrid cross, consider the case of true-breeding pea plants with yellow versus green 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. A Punnett square, devised by the British geneticist Reginald Punnett, is useful for determining probabilities because it is drawn to predict all possible outcomes of all possible random fertilization events and their expected frequencies.
Figure 8. To prepare a Punnett square, all possible combinations of the parental alleles the genotypes of the gametes are listed along the top for one parent and side for the other parent of a grid. The combinations of egg and sperm gametes are then made in the boxes in the table on the basis of which alleles are combining. Each box then represents the diploid genotype of a zygote, or fertilized egg.
Because each possibility is equally likely, genotypic ratios can be determined from a Punnett square. If the pattern of inheritance dominant and 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 in the F 1 offspring. All offspring are Yy and have yellow seeds. When the F 1 offspring are crossed with each other, each has an equal probability of contributing either a Y or a y to the F 2 offspring.
The result is a 1 in 4 25 percent probability of both parents contributing a Y , resulting in an offspring with a yellow phenotype; a 25 percent probability of parent A contributing a Y and parent B a y , resulting in offspring with a yellow phenotype; a 25 percent probability of parent A contributing a y and parent B a Y , also resulting in a yellow phenotype; and a 25 percent probability of both parents contributing a y , resulting in a green phenotype.
When counting all four possible outcomes, there is a 3 in 4 probability of offspring having the yellow phenotype and a 1 in 4 probability of offspring having the green phenotype. Using large numbers of crosses, Mendel was able to calculate probabilities, found that they fit the model of inheritance, and use these to predict the outcomes of other crosses. Observing that true-breeding pea plants with contrasting traits gave rise to F 1 generations that all expressed the dominant trait and F 2 generations that expressed the dominant and recessive traits in a ratio, Mendel proposed the law of segregation.
This law states that paired unit factors genes must segregate equally into gametes such that offspring have an equal likelihood of inheriting either factor. For the F 2 generation of a monohybrid cross, the following three possible combinations of genotypes result: homozygous dominant, heterozygous, or homozygous recessive.
The equal segregation of alleles is the reason we can apply the Punnett square to accurately predict the offspring of parents with known genotypes. 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 F 1 offspring will be heterozygotes expressing the dominant trait Figure 8. Alternatively, if the dominant-expressing organism is a heterozygote, the F 1 offspring will exhibit a ratio of heterozygotes and recessive homozygotes Figure 8. The cross between the true-breeding P plants produces F1 heterozygotes that can be self-fertilized. The self-cross of the F1 generation can be analyzed with a Punnett square to predict the genotypes of the F2 generation.
Given an inheritance pattern of dominant—recessive, the genotypic and phenotypic ratios can then be determined. 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 parent plant is homozygous dominant or heterozygous? You cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present. Independent assortment of genes can be illustrated by the dihybrid cross, a cross between two true-breeding parents that express different traits for two characteristics.
Consider the characteristics of seed color and seed texture for two pea plants, one that has wrinkled, green seeds rryy and another that has round, yellow seeds RRYY.
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