Learning Objectives
By the end of this section, you will be able to:
- Define basic genetic terms
- Develop a Punnett square and calculate the expected proportions in a monohybrid cross
- Explain the purpose and methods of a test cross
- Identify several non-Mendelian inheritance patterns
The seven characteristics that Mendel evaluated in his pea plants were each expressed as one of two versions, or traits. The physical expression of characteristics is accomplished through gene expression carried on chromosomes. The genetic makeup of peas consists of two similar, or homologous, copies of each chromosome, one from each parent. Each pair of homologous chromosomes has the same order of genes. Diploid organisms utilize meiosis to produce haploid gametes, which contain one copy of each homologous chromosome that unite at fertilization creating a diploid zygote.
For cases in which a single gene controls a single characteristic, a diploid organism has two genetic copies that may or may not encode the same version of a characteristic. Alternate forms of a gene are known as alleles. An allele will have the same genes in same corresponding area(loci) on a pair of homologous chromosomes. Although Mendel only examined the inheritance of genes with just two allele forms, it is common to encounter more than two alleles for any given gene in a natural population.
Phenotypes and Genotypes
Two alleles for a given gene in a diploid organism are expressed and interact to produce physical characteristics. The observable traits expressed by an organism are referred to as its phenotype. An organism’s underlying genetic makeup, both physically visible and non-expressed alleles, is called its genotype. Mendel’s hybridization experiments demonstrate the difference between phenotype and genotype. When true-breeding plants, one with yellow pods and one with green pods, were cross-fertilized, all of the F1 hybrid offspring had yellow pods. The hybrid offspring were phenotypically identical to the true-breeding parent with yellow pods. We know that the allele donated by the parent with green pods was not lost because it reappeared in some of the F2 offspring. The F1 plants must have been genotypically different from the parent with yellow pods.
The P1 plants used in Mendel’s experiments were identical for the trait being studied. Diploid organisms that are homozygous at a given gene, or locus, have two identical alleles for that gene on their homologous chromosomes. Mendel’s parental pea plants always bred true because both of the gametes produced carried the same trait. When P1 plants with contrasting traits were cross-fertilized, all of the offspring were heterozygous for the contrasting trait. Their genotype reflected different alleles for the gene being examined.
Dominant and Recessive Alleles
In Mendel’s experiments, why were the F1 offspring identical to one of the parents? In all seven pea-plant characteristics, one of the two contrasting alleles was dominant, and the other was recessive. We know they are actually genes on homologous chromosome pairs. For a gene expression, homozygous dominant and heterozygous organisms will have the same phenotype but differ in genotype. The recessive allele will only be observed in homozygous recessive individuals, or when the dominant allele is absent. Table 1 shows a few of the disorders inherited by these dominant and recessive alleles.
Table 1. Human Inheritance in Dominant and Recessive Patterns | |
---|---|
Dominant Traits | Recessive Traits |
Achondroplasia | Albinism |
Brachydactyly | Cystic fibrosis |
Huntington’s disease | Duchenne muscular dystrophy |
Marfan syndrome | Galactosemia |
Neurofibromatosis | Phenylketonuria |
Widow’s peak | Sickle-cell anemia |
Wooly hair | Tay-Sachs disease |
Several methods exist for referring to genes and alleles. For our purposes, we will abbreviate genes using the first letter of the gene’s corresponding dominant trait. For example, violet is the dominant trait for a pea plant’s flower color, so the flower-color gene would be abbreviated as V. We will use uppercase and lowercase letters to represent dominant and recessive alleles, respectively. Therefore, we would refer to the genotype of a homozygous dominant pea plant with violet flowers as VV, a heterozygous pea plant with violet flowers as Vv, and a homozygous recessive pea plant with white flowers as vv.
The Punnett Square Approach for a Monohybrid Cross
When fertilization between two true-breeding parents differing in only one characteristic occurs, the process is known as amonohybrid cross with the resulting offspring being monohybrids. Mendel performed seven monohybrid crosses for each plant characteristic. Based on his results, Mendel believed that each parent contributed one of two paired factors to each offspring with every possible combination of factors equally likely.
Consider the case of true-breeding pea plants with yellow versus green pea seeds. The dominant seed color is yellow, so 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 applying the rules of probability to predict the possible outcomes of a genetic cross 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. This arrangement represents their meiotic segregation into haploid gametes. The combinations of egg and sperm are placed in the boxes to show which alleles are combining. Each box 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 understood. In a monohybrid cross of two true-breeding parents, each parent contributes one type of allele. In our example, only one genotype is possible. All offspring are Yy(genotype) and have yellow seeds(phenotype) (Figure 1).
![This illustration shows a monohybrid cross. In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy. Each parent produces one kind of gamete, resulting in an F_{1} generation with a dominant yellow phenotype and the genotype Yy. Self-pollination of the F_{1} generation results in an F_{2} generation with a 3 to 1 ratio of yellow to green peas. One out of three of the yellow pea plants has a dominant genotype of YY, and 2 out of 3 have the heterozygous phenotype Yy. The homozygous recessive plant has the green phenotype and the genotype yy.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184030/Figure_12_02_01.jpg)
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.
What result would occur is we self-pollinated the F1 offspring? 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 . 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 possibilities must be counted. The two possible heterozygous combinations produce offspring that are genotypically and phenotypically identical, despite the origin of the alleles. Because fertilization is rando, 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). 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. Working with large sample sizes, Mendel observed this ratio in every F2 generation resulting from crosses for individual traits.
Test Cross
Mendel developed a way to determine whether an organism that expressed a dominant trait was homozygous or heterozygous. Called a test cross, this technique is still used today 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). 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 Mendel’s belief that pairs of factors segregate equally.
Art Connection
![In a test cross, a parent with a dominant phenotype but unknown genotype is crossed with a recessive parent. If the parent with the unknown phenotype is homozygous dominant, all of the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, fifty percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184115/Figure_12_02_02.jpg)
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?
PEDIGREE CHARTS
Many human diseases are inherited through our genetics. What are my chances of inheriting a particular disease? Can I pass defective genes to my offspring? Geneticists use pedigree charts to study inheritance patterns and help answer these types of questions. Figure 3 shows a pedigree chart for alkaptonuria. Individuals with this disorder have dark skin, brown urine, and suffer joint damage. For our purposes, we will discuss three inheritance patterns in pedigree analysis.
(1) autosomal recessive – affected child may have neither parent affected(carriers); if both parents affected then all children affected
(2) autosomal dominant – unaffected child has affected parents; affected child has at least one affected parent; all heterozygotes affected
(3) X-linked recessive – affected son can have normal parents; if mom affected then all sons affected; more males than females affected
Art Connection
![This is a pedigree of a family that carries the recessive disorder alkaptonuria. In the second generation, an unaffected mother and an affected father have three children. One child has the disorder, so the genotype of the mother must be Aa and the genotype of the father is aa. One unaffected child goes on to have two children, one affected and one unaffected. Because her husband was not affected, she and her husband must both be heterozygous. The genotype of their unaffected child is unknown, and is designated A?. In the third generation, the other unaffected child had no offspring, and his genotype is therefore also unknown. The affected third-generation child goes on to have one child with the disorder. Her husband is unaffected and is labeled “3.” The first generation father is affected and is labeled “1.” The first generation mother is unaffected and is labeled “2.” The Art Connection question asks the genotype of the three numbered individuals.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184146/Figure_12_02_03.jpg)
Figure 3. In this pedigree analysis for alkaptonuria, 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.
Alkaptonuria is a recessive genetic disorder in which two amino acids are not properly metabolized. Affected individuals may have darkened skin, brown urine, and may suffer joint damage and other complications. Looking at Figure 3, we can see that it is often possible to determine a person’s genotype from the genotype of their offspring. 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.
What are the genotypes of the individuals labeled 1, 2 and 3?
BEYOND MENDEL
Incomplete Dominance
![Photo is of a snapdragon with a pink flower.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184218/Figure_12_02_04.jpg)
Figure 4. These pink flowers of a heterozygote snapdragon result from incomplete dominance. (credit: “storebukkebruse”/Flickr)
Mendel’s results contradicted the view that offspring exhibited a blend of their parents’ traits. But sometimes blending does exist. In the snapdragon(Figure 4), a cross between a homozygous parent with white flowers (CWCW) and a homozygous parent with red flowers (CRCR) will produce offspring with pink flowers (CRCW). Note that different genotypic abbreviations are used to distinguish these patterns from basic Mendelian genetics. This pattern of inheritance is described as incomplete dominance, where the heterozygote is an intermediate between the dominant and recessive traits. The allele for red flowers is incompletely dominant over the allele for white flowers. The results of a heterozygote self-cross can still be predicted. In this case, the genotypic ratio would be 1 CRCR:2 CRCW:1 CWCW, and the phenotypic ratio would be 1:2:1 for red:pink:white.
Codominance
A variation on incomplete dominance is codominance. Codominance allows both alleles for the same characteristic to be expressed in the heterozygote. An example of codominance is familial hypercholesterolemia. Individuals that inherit this as a heterozygote may exhibit cardiovascular disease early in life(age 30) due to high cholesterol deposits. Other health issues may arise as the cholesterol also deposits in the skin, tendons, and corneas. Most of us inherit as a homozygous dominant case. Recessive inheritance of this disorder results in early death, generally as an infant.
Multiple Alleles
Mendel thought that only two alleles could exist for a given gene. Although individuals can only have two alleles for a given gene, multiple alleles exist and are observed. An example of multiple alleles is in the coat color of rabbits (Figure 5). Four alleles exist for the c gene. The wild-type version, C+C+, is expressed as brown fur. The chinchilla phenotype, cchcch, is shown as black-tipped white fur. The Himalayan phenotype, chch, has white fur with black furred extremities . Finally, the albino, or “colorless” phenotype, cc, is expressed as white fur. The wild-type allele is dominant over all the others, chinchilla is incompletely dominant over Himalayan and albino, and Himalayan is dominant over albino. By observing the phenotypes of each possible heterozygote offspring, it is easy to understand this dominance hierarchy.
![This illustration shows the four different variants for coat color in rabbits at the c allele. The genotype CC produces the wild type phenotype, which is brown. The genotype c^{ch}c^{ch} produces the chinchilla phenotype, which is black-tipped white fur. The genotype c^{h}c^{h} produces the Himalayan phenotype, which is white on the body and black on the extremities. The genotype cc produces the recessive phenotype, which is white](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184300/Figure_12_02_05.jpg)
Figure 5. Four different alleles exist for the rabbit coat color (C) gene.
![This photo shows Drosophila that has normal antennae on its head, and a mutant that has legs on its head.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184330/Figure_12_02_06.jpg)
Figure 6. As seen in comparing the wild-type Drosophila (left) and the Antennapedia mutant (right), the Antennapedia mutant has legs on its head in place of antennae.
The complete dominance of one phenotype over all other often occurs as an effect of “dosage” of a specific gene product. In rabbits, one allele may supply a given dosage of fur pigment, while the others receive a lesser dosage or none at all. Interestingly, the Himalayan phenotype is the result of an allele producing a temperature-sensitive gene product that only produces pigment in the cooler extremities of the rabbit’s body.
Alternatively, one mutant allele can be dominant over all other phenotypes. This may occur when the mutant allele somehow interferes with the genetic message so that even a heterozygote expresses the mutant phenotype. The mutant allele can interfere by enhancing the function of a gene product or changing its distribution in the body. One example is in Drosophila, the fruit fly. In this case, the mutant allele expands the distribution of the gene product and the heterozygote develops legs on its head instead of antennae(Figure 6).
Evolution Connection
Multiple Alleles Confer Drug Resistance in the Malaria Parasite
Malaria is a parasitic disease in humans that is transmitted by infected female mosquitoes, including Anopheles gambiae (Figure7a), and is characterized by high fevers, chills, flu-like symptoms, and severe anemia. Plasmodium falciparum and P. vivax are the most common causative agents of malaria, and P. falciparum is the most deadly (Figure 7b). When promptly and correctly treated, P. falciparum malaria has a mortality rate of 0.1 percent. However, in some parts of the world, the parasite has evolved resistance to commonly used malaria treatments, so the most effective malarial treatments can vary by geographic region.
![Photo a shows the Anopheles gambiae mosquito, which carries malaria. Photo b shows a micrograph of sickle-shaped Plasmodium falciparum, the parasite that causes malaria. The Plasmodium is about 0.75 microns across.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184426/Figure_12_02_07.png)
Figure 7. The (a) Anopheles gambiae, or African malaria mosquito, acts as a vector in the transmission to humans of the malaria-causing parasite (b) Plasmodium falciparum, here visualized using false-color transmission electron microscopy. (credit a: James D. Gathany; credit b: Ute Frevert; false color by Margaret Shear; scale-bar data from Matt Russell)
X-linked Traits
In many life forms, the sex of the individual is determined by sex chromosomes. The sex chromosomes are one pair of non-homologous chromosomes. In our previous discussions, we have considered inheritance of the autosomes. Autosomes are all chromosomes of inheritance except sex. In addition to 22 homologous pairs of autosomes, human females have a homologous pair of X chromosomes. Human males have an XY chromosome pair. Although the Y chromosome contains a small region of similarity to the X chromosome allowing them to pair during meiosis, the Y chromosome is much shorter and contains fewer genes. When a gene being examined is present on the X chromosome, but not on the Y chromosome, it is said to be X-linked.
![Photo shows six fruit flies, each with a different eye color.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184500/Figure_12_02_08.jpeg)
Figure 8. In Drosophila, the gene for eye color is located on the X chromosome. Clockwise from top left are brown, cinnabar, sepia, vermilion, white, and red. Red eye color is wild-type and is dominant to white eye color.
Eye color in Drosophila was one of the first X-linked traits to be identified. In 1910, Thomas Hunt Morgan mapped this trait to the X chromosome. Like humans, Drosophila males have an XY chromosome pair, and females are XX. In flies, the wild-type eye color is red (XW) and is dominant to white eye color (Xw) (Figure 8). Because of the location of the eye-color gene, reciprocal crosses do not produce the same offspring ratios. Males are said to be hemizygous. They have only one allele for any X-linked characteristic. Hemizygosity makes the descriptions of dominance and recessiveness irrelevant for XY males. Drosophila males lack a second allele copy on the Y chromosome so their genotype can only be XWY or XwY. Females have two allele copies and can be XWXW, XWXw, or XwXw.
In an X-linked cross, the genotypes of F1 and F2 offspring depended on whether the recessive trait was expressed by the male or the female in the P1 generation. With regard to Drosophila eye color, when the P1 male expresses the white-eye phenotype and the female is homozygous red-eyed, all members of the F1 generation exhibit red eyes. The F1 females are heterozygous (XWXw)) and the males are all XWY. A subsequent cross between the XWXw female and the XWY male would produce only red-eyed females (XWXW or XWXw genotypes) and both red- and white-eyed males (XWY or XwY genotypes). Consider a cross between a homozygous white-eyed female and a male with red eyes(Figure 9). The F1 generation would exhibit only heterozygous red-eyed females (XWXw) and only white-eyed males (XwY). Half of the F2 females would be red-eyed (XWXw) and half would be white-eyed (XwXw). Similarly, half of the F2 males would be red-eyed (XWY) and half would be white-eyed (XwY).
Art Connection
![This illustration shows a Punnett square analysis of fruit fly eye color, which is a sex-linked trait. A red-eyed male fruit fly with the genotype X^{w}Y is crossed with a white-eyed female fruit fly with the genotype X^{w}X^{w}. All of the female offspring acquire a dominant W allele from the father and a recessive w allele from the mother, and are therefore heterozygous dominant with red eye color. All of the male offspring acquire a recessive w allele from the mother and a Y chromosome from the father and are therefore hemizygous recessive with white eye color.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184601/Figure_12_02_09.jpeg)
Figure 9. Punnett square analysis is used to determine the ratio of offspring from a cross between a red-eyed male fruit fly and a white-eyed female fruit fly.
What ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?
We can apply discoveries in fruit fly genetics to human genetics. When a female parent is homozygous for a recessive X-linked trait, she will pass the trait on to 100 percent of her offspring. Her male offspring are destined to express the trait, as they will inherit their father’s Y chromosome. In humans, the alleles for certain conditions, such as some forms of color blindness, hemophilia, and muscular dystrophy),are X-linked. Heterozygous females of a disease are said to be carriers and may not exhibit any phenotypic effects. These females will pass the disease to half of their sons and will pass carrier status to half of their daughters. For this reason, recessive X-linked traits appear more frequently in males than females.
Human Sex-linked Disorders
Sex-linkage studies provided the fundamentals for understanding X-linked recessive disorders in humans, including red-green color blindness, and Types A and B hemophilia. Because human males need to inherit only one recessive X allele to be affected, X-linked disorders are disproportionately observed in males. Females must inherit recessive X-linked alleles from both parents in order to express the trait. When they inherit one recessive X-linked allele and one dominant X-linked allele, they are carriers of the trait and are typically unaffected. However, female carriers can contribute the trait to their sons, resulting in the son exhibiting the trait. They can contribute the recessive allele to their daughters, resulting in the daughters being carriers of the trait (Figure 10). Although some Y-linked recessive disorders exist, typically they are associated with male infertility and are not transmitted to subsequent generations.
![A diagram shows an unaffected father with a dominant allele and an unaffected carrier mother with an x-linked recessive allele. Four figures of offspring are shown representing the various resulting genetic combinations: unaffected son, unaffected daughter, affected son, and unaffected carrier daughter.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184635/Figure_12_02_10.jpeg)
Figure 10. The son of a woman who is a carrier of a recessive X-linked disorder will have a 50 percent chance of being affected. A daughter will not be affected, but will have a 50 percent chance of being a carrier like her mother.
Link to Learning
Watch this video to learn more about sex-linked traits.
Lethality
Many genes in an individual’s genome are essential for survival. Occasionally, a nonfunctional allele for an essential gene can arise by mutation and be transmitted as long as individuals also have a functional copy. The functional allele works at a capacity sufficient to sustain life and is considered to be dominant over the nonfunctional allele. An inheritance pattern in which an allele is only lethal in the homozygous form is a recessive lethal. In this case, the heterozygote may be normal or have some altered non-lethal phenotype.
![Micrograph shows a neuron with nuclear inclusions characteristic of Huntington’s disease.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/198/2016/11/28184712/Figure_12_02_11.jpeg)
Figure 11. The neuron in the center of this micrograph (yellow) has nuclear inclusions characteristic of Huntington’s disease (orange area in the center of the neuron). Huntington’s disease occurs when an abnormal dominant allele for the Huntington gene is present. (credit: Dr. Steven Finkbeiner, Gladstone Institute of Neurological Disease, The Taube-Koret Center for Huntington’s Disease Research, and the University of California San Francisco/Wikimedia)
A single copy of the wild-type allele is not always sufficient for normal functioning or survival. The dominant lethal inheritance pattern is when an allele is lethal in the homozygote and the heterozygote This allele is transmitted if the lethality phenotype occurs after reproductive age. Individuals with mutations that result in dominant lethal alleles fail to survive, even in the heterozygote form. Dominant lethal alleles are very rare. As you might expect, the allele only lasts one generation and therefore not transmitted. However, just as the recessive lethal allele might not immediately manifest the phenotype of death, dominant lethal alleles might not be expressed until adulthood. Once the individual reaches reproductive age, the allele may be unknowingly passed on, resulting in a delayed death in both generations. In humans, Huntington’s disease, where the nervous system gradually wastes away, is an example of this dominant lethal allele(Figure 11). People, heterozygous for Huntington allele (Hh), will inevitably develop the fatal disease. The onset of Huntington’s disease may not occur until age 40, at which point the afflicted person may have already passed the allele to 50 percent of their offspring.
Section Summary
When true-breeding and homozygous recessive individuals that differ for a certain trait are crossed, all of the offspring will be heterozygotes for that trait. If these heterozygous offspring are self-crossed, the resulting F2 offspring will be equally likely to inherit gametes carrying the dominant or recessive trait. Using the product and sum rules, genetic outcomes can be approximated to help in our further understanding.
Alleles do not always behave in dominant and recessive patterns. Incomplete dominance describes situations in which the heterozygote exhibits a phenotype that is intermediate between the homozygous phenotypes. Codominance describes the simultaneous expression of both of the alleles in the heterozygote. Although diploid organisms can only have two alleles for any given gene, it is common for more than two alleles of a gene to exist.
In many life forms, females have two X chromosomes and males have one X and one Y chromosome. Genes that are present on the X but not the Y chromosome are said to be X-linked, such that males only inherit one allele for the gene, and females inherit two. Some alleles can be lethal. Recessive lethal alleles are only lethal in homozygotes, but dominant lethal alleles are fatal in both homozygotes and heterozygotes.
Additional Self Check Questions
- 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. The cross gives you all round peas. Can you tell if the round pea parent plant is homozygous dominant or heterozygous?
- What is the tile given to Mendel?
- What ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?
- The gene for flower position in pea plants exists as axial or terminal alleles. Given that axial is dominant to terminal, list all of the possible F1 and F2 genotypes and phenotypes from a cross involving parents that are homozygous for each trait. Express genotypes with conventional genetic abbreviations.
- Use a Punnett square to predict the offspring in a cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous). What is the phenotypic ratio of the offspring?
- Can a human male be a carrier of red-green color blindness?
Answers
- Based on your data, the round pea plant would be RR.
- Mendel is known as the “father of genetics”.
- Half of the female offspring would be heterozygous (XWXw) with red eyes, and half would be homozygous recessive (XwXw) with white eyes. Half of the male offspring would be hemizygous dominant (XWY) withe red eyes, and half would be hemizygous recessive (XwY) with white eyes.
- Because axial is dominant, the gene would be designated as A. F1 would be all heterozygous Aa with axial phenotype. F2 would have possible genotypes of AA, Aa, and aa; these would correspond to axial, axial, and terminal phenotypes, respectively.
- The Punnett square would be 2 × 2 and will have T and T along the top, and T and t along the left side. Clockwise from the top left, the genotypes listed within the boxes will be Tt, Tt, tt, and tt. The phenotypic ratio will be 1 tall:1 dwarf.
- No, males can only express color blindness. They cannot carry it because an individual needs two X chromosomes to be a carrier.