Mendelian inheritance refers to patterns of inheritance that are typically characteristic of organisms that reproduce sexually. Gregor Mendel performed thousands of crosses with garden peas at his monastery during the middle of the 19th century. Mendel explained his results by describing two laws of inheritance that introduced the idea of dominant and recessive genes. This inheritance pattern occurs on a single gene with only two allele options.
Mendelian inheritance patterns are exceedingly rare, especially in humans. We now know that inheritance is much more complex, involving many genes that interact in varied ways. Nonetheless, a clear understanding of basic inheritance patterns that follow Mendel’s original observations will provide a springboard for understanding current scientific exploration.
Inheritance patterns that follow Mendelian rules are as follows:
- Traits are governed by single genes
- There are two alternate forms of a gene, known as alleles
- Alleles are expressed as dominant and recessive
Based on what you have learned in lecture, readings, and videos in class, please read through the following descriptions, and answer the questions based on Mendelian genetics problems.
Lab Vocabulary Words
(Hint for exams: by the end of this lab, you should be familiar with all these terms)
- Punnett Square
- Sex-linked traits
- Incomplete Dominance
Part 1: Terminology Practice
1. Describe the genotypes given (use your notes). The first two are done as an example.
- DD Homozygous dominant
- Dd heterozygous
- dd _________________________
- ss _______________________
- Yy _______________________
2. In humans, brown eye color (B), is dominant over blue eye color (b). What are the phenotypes of the following genotypes? In other words, what color eyes will they have?
- BB ______________________
- bb ______________________
- Bb ______________________
Part 2: The Five Steps of Solving a Genetics Problem
If you take the time to follow the steps below, you will be able to solve most genetics problems.
1. Determine the genotypes of the parents.
2. Set up your Punnett Square. This example is modeling the inheritance of a single gene. Each parent has two alleles for each gene, so the number of possible offspring outcomes is 4 (2 alleles x 2 parents= 4 offspring)
3. Fill in the boxes. This will show the possible combinations of offspring that could occur through a fertilization with parents of the given allele combinations.
4. Write out the genotypic ratio of the offspring. What are the ratios of the genotypes that result from this crossing?
5. Using the genotypic ratio, determine the phenotypic ratio for the offspring.
Part 2B: Practice Problem:
A heterozygous male, black eyed mouse is crossed with a red eyed, female mouse. Predict the possible offspring.
Step 1: Determine the genotype of the parents. The male parent is heterozygous which means he has one allele for black eyes and one allele for red eyes. Since his eyes are black, this means that black allele must be dominant over the red allele. So the male parents genotype is “Bb” (B = allele for black eye, b = allele for red eye). The female parent has red eyes, there is only one way to have this recessive phenotype, so she must to be homozygous recessive. Homozygous recessive means that her genotype must be “bb”. Therefore, genotype of the parents is Bb x bb.
Step 2: During meiosis (the formation of sex cells) one member (allele) of each gene pair separate. The male mouse (Bb) produces some sperm containing “B” (the allele for black eye) and some sperm containing “b” (the allele for red eyes). The female mouse has only “b” alleles, so even though the pair separate during meiosis, each ovum (egg cell) she produces will have the “b” allele.
On one axis of the Punnett Square you put the two possible gametes for the male.
On the other axis, put the two possible alleles of the female.
Step 3: During fertilization, sperm meets the ovum to form a zygote (fertilized egg). The zygote must have two alleles for the gene. It gets one from each parent. The Punnett Square show us the various possibilities during fertilization. The offspring must be one of these genotypes listed in the squares.
By repeating the process, the Punnett Square can show all the possible genotypes that this couple can produce for this gene.
Step 4: The genotypic ratio is the proportion of possible genotypes in the offspring. It is determined by counting each possible genotype. You’ll note that in this case there are two “Bb” for every two “bb”. Therefore, we write the ratio as:
Normally we reduce to the lowest terms:
Step 5: The Phenotypic Ratio is the proportion of ways that the individuals will look. The Bb will produce a black-eyed mouse and the bb will produce a red eyed mouse (phenotype)
The phenotypic ratio in this case is written as:
Ratios tell you there is an even chance of having offspring with black eyes as there is for having offspring with red eyes. That would be the same as a 50% probability of having red eyes, or a 50% probability of having black eyes.
Part 3 Monohybrid Cross
When we study the inheritance of a single gene it is called a monohybrid cross.
Answer the problems below:
- A heterozygous, smooth pea pod, plant is crossed with a wrinkled pea pod plant. There are two alleles for pea pod, smooth (S) and wrinkled (s). Predict the offspring from this cross.
- What is the genotype of the parents?
- Set up a Punnett Square with possible gametes.
Fill in the Punnett Square for the resultant offspring.
- What is the predicted genotypic ratio for the offspring?
- What is the predicted phenotypic ratio for the offspring?
- If this cross produced 50 seeds how many would you predict to have a wrinkled pod?
2. In humans, a condition called achondroplasia, a mutation on the FGFR3 gene, presents as “dwarfism”. When the mutation occurs, a “dwarfed” (D) condition is dominant over “non-dwarfed” (d). A homozygous dominant (DD) person dies before the age of one. A heterozygous (Dd) person has achondroplasia. A homozygous recessive individual does not have achondroplasia. A heterozygous male with achondroplasia mates with a heterozygous female with achondroplasia…
What is the probability that this mating pair will produce….
- A child who does not have achondroplasia?
- A child who has achondroplasia?
- A child who dies in infancy from this condition?
3. In humans, free earlobes (F) is dominant over attached earlobes (f). If one parent is homozygous dominant for free earlobes, while the other has attached earlobes, can they produce any children with attached earlobes?
4. In humans, widow’s peak (W) is dominant over straight hairline (w). A heterozygous male for this trait marries a female who is also heterozygous.
- Complete the Punnett Square above.
- List possible genotypes of their offspring.
- What is the genotypic ratio of the offspring?
- What is the phenotypic ratio of the offspring?
Part 4: Working Backwards
Sometimes we only know the offspring’s genotype and we want to know the parents’. When both parents are heterozygous the phenotypic ratio always comes out 3 to 1. If one parent is homozygous recessive and the other is heterozygous, the phenotypic ratio always comes out 1 to 1. Keeping this in mind see if you can solve the next two problems.
1. In pea plants, yellow seeds (Y) are dominant and green seeds (y) are recessive. A pea plant with yellow seeds is crossed with a pea plant with green seeds. The resulting offspring have about equal numbers of yellow and green seeded plants. What are the genotypes of the parents?
2. In another cross, a yellow seeded plant was crossed with another yellow seeded plant and it produced offspring of which about 25% were green seeded plants. What are the genotypes of both parents?
Part 5: Back Cross/Test Cross
When an organism has the dominant phenotype, then its genotype can be either heterozygous or homozygous dominant (you can’t tell by looking at it). We must do a test cross using a homozygous, recessive organism to know the right combination.
For example: In Dalmatian dogs, the gene for black spots is dominant to the gene for liver-colored spots. If a breeder has a black spotted dog, how can she find out whether it is homozygous (BB) or heterozygous (Bb) spotted dog? *B = black spots and b = liver spots
If the breeder finds a black spotted dog, whose ancestry is not known, she cannot tell by looking at the dog if it is BB or Bb. She should find a liver spotted dog, whose genotype must be “bb” and mate it with the black spotted dog in question.
This is the cross of a homozygous dominant (BB) individual. Notice that all of the offspring will be Bb and therefore, there is no possibility of having a liver spotted offspring.
*This would be the resultant Punnett Square for the heterozygous (Bb) individual.
If any of the breed offspring has liver spots, then she can say that she had a heterozygous black spotted dog. If all the offspring had black spots, then she can say that the suspect dog was homozygous.
You found a wild, black mouse. In mice, white fur is recessive.
- Explain how you would determine the genotype of this mouse.
- Draw Punnett Squares for your possible crosses. What are the possible parental genotypes?
- You have 24 offspring, 23 with black fur and 1 with white fur. What was the genotype of the mouse you found?
- If you only had 3 offspring and all were black, can you tell the genotype of the original mouse? Why or why not.
Part 6: Dihybrid Cross
When we study two traits on different chromosomes, at one time, we call this a dihybrid cross. You still follow the same five step process for monohybrid crosses but now there will be four times as many possibilities because we are studying two traits.
Example: A female guinea pig that is heterozygous for both fur color and coat texture is crossed with a male that has light fur color and is heterozygous for coat texture. What possible offspring can they produce? Dark fur color is dominant (D) and light fur (d) is recessive. Rough coat texture (R) is dominant, while smooth coat (r) is recessive.
Step 1: The guinea pig that is heterozygous for both color and texture this means it has one allele for each trait. Therefore its genotype would be “DdRr”. The other guinea pig has light fur, since that is a recessive trait the genotype for that trait must be “dd”. It is also heterozygous for fur texture, which means a genotype of “Rr”. All together its overall genotype must be “ddRr”.
Step 2 and 3: The Punnett Square will be larger now because there are more possible sperm and egg combinations. During the formation of sperm a “D” could go with a “R” producing a sperm “DR”, or a “D” could go with a “r” forming a sperm with “Dr”.
Filling-in the Punnett Square it should look like the one below.
Step 4: After filling-in the Punnett Square you should obtain the following genotypic ratio:
*remember the numbers should add up to the number of squares filled in: 4 DdRr : 2 DdRR : 4 ddRr : 2 ddRR : 2 Ddrr : 2 ddrr
You may choose to simplify this ratio: 2DdRr: 1DdRR: 2ddRr: 1ddRR: 1Ddrr: 1ddrr
Step 5: There will be only four different phenotypes because the 4 DdRr and the 2 DdRR will have dark fur with rough coat, and the 4 with ddRr and the 2 ddRR will have light fur with rough coat, while the 2 Ddrr will have dark fur with smooth coat and the 2 ddrr will have light fur with smooth coat.
Therefore the phenotypic ratio would be:
6 dark, rough : 6 light rough : 2 dark smooth : 2 light smooth.
Or reduced, 3 dark rough: 3 light rough: 1 dark smooth: 1 light smooth.
- In pea plants, the round seed allele is dominant over the wrinkled seed allele, and the yellow seed allele is dominant over the green seed allele. The genes for seed texture and those for seed color are on different chromosomes. A plant heterozygous for seed texture and seed color is crossed with a plant that is wrinkled and heterozygous for seed color. *R = round, r = wrinkled, Y= yellow, y = green
- Construct a Punnett Square for this cross.
- What is the expected phenotypic ratio for the offspring?
2. In humans there is a disease called Phenylketonuria (PKU) which is caused by a recessive allele. People with this allele have a defective enzyme and cannot break down the amino acid phenylalanine. This disease can result in mental retardation or death. Let “E” represent the non-PKU enzyme. Also in humans in a condition called galactose intolerance or galactosemia, which is also caused by a recessive allele. Let “G” represent the non-galactosemia allele for galactose digestion. In both diseases, not having the disease dominates over having the disease (recessive).
In the following example, both parents are heterozygous for these traits (EeGg). Fill in the Punnett Square and answer the following questions.
- What are the chances of having a child that does not have PKU?
- What are the chances of having a child with PKU?
- What are the chances of having a child with just galactosemia?
- What are the chances of having a child with both PKU and galactosemia?
Part 7: Incomplete Dominance or Co-dominance
In both incomplete and co-dominance neither allele is dominant over the other. In co-dominance, a heterozygous individual expresses both alleles simultaneously without blending. An example of co-dominance can be seen in cows — if a white cow (ww) and a black cow (bb) are crossed, the offspring will express both alleles (wb) and will be spotted. With incomplete dominance, the traits are mixed, as happens in snapdragon flowers, where white and red flowers produce pink offspring.
When we do incomplete or co-dominance Punnett Squares, each allele gets its own lower case letter. For snapdragons, the alleles for red and yellow color are indicated as “r” and “y”. All red and yellow flowers are homozygous. Orange flowers are heterozygous.
Step 1: Genotype of the parents. Red parent: rr Yellow parent: yy
Step 2: Prepare the Punnett Square with the parents’ alleles.
Step 3: Fill in the Punnett Square.
Step 4: Genotypic ratio.
Step 5: Phenotypic ratio. All are heterozygous red and yellow, so they will be phenotypically orange.
- If two orange snapdragons are crossed, what will their offspring look like?
- Complete the Punnett Square for this cross.
- What is the predicted genotypic ratio for the offspring?
- What is the predicted phenotypic ratio for the offspring?
Part 8: Multiple Allele
So far, we have studied traits or genes that are coded for by just two alleles. However, some traits are coded for by more than two alleles. One of these is blood type in humans.
In humans, there are four types of blood; type A, type B, type AB, and type O. The alleles A and B are codominant to each other and the O allele is recessive to both A and B alleles. So a person with the genotype AA or AO will have A type of blood.
- What possible genotypes will produce Type B blood?
- What is the only genotype that will produce Type O blood?
- What is the only genotype that will produce type AB blood?
- A person who is Type O blood mates with a person who has Type AB blood.
- Complete the above Punnett Square for this cross.
- List the possible blood types (phenotypes) of your offspring.
2. In the 1950’s, a young woman sued film star/director Charlie Chaplin for parental support of her illegitimate child. Charlie Chaplin’s blood type was already on record as type AB. The mother of the child had type A and her son had type O blood.
- Complete a Punnett Square for the possible cross of Charlie and the mother.
- The judge ruled in favor of the mother and ordered Charlie Chaplin to pay child support costs of the child. Was the judge correct in his decision based on blood typing evidence? Explain why or why not. *refer to any Punnett Squares to support your answer.
3. Suppose a newborn baby was accidentally mixed up in the hospital. In an effort to determine the parents of the baby, the blood types of the baby and two sets of parents were determined.
|Baby 1 had type O||Mrs. Brown had type B||Mr. Brown had type AB|
|Mrs. Smith had type B||Mr. Smith had type B|
- Draw Punnett Squares for each couple (you may need to do more than 1 square/ couple)
- To which parents does baby #1 belong? Why? Hint you may want to refer to your Punnett Squares.
Part 9: Sex-Linked Traits Explanation
In humans, sex is determined by the 23rd pair of chromosomes known as “sex chromosomes”. Two x-shaped (XX) chromosomes indicate a female, while one x and one y-shaped (XY) indicate male. Since the X and Y chromosomes carry different information, any genes found on the X chromosomes are referred to as sex-linked genes. (This is an overly simplified explanation, and we will discuss this in greater detail in class).
The sex chromosomes carry the information about the anatomical differences between males and females, but they also carry other information as well. Since X and Y chromosomes carry different information, any genes carried on either sex chromosome are referred to as sex-linked genes. Since only males have the Y chromosome, only males can inherit Y-linked traits. Both males and females can inherit X-linked traits because both males and females inherit X-chromosomes.
X-linked recessive traits that are not related to feminine body characteristics are more often expressed in the phenotype of males than they are in females. This is because males only have one X chromosome, and therefore no ability to mask a recessive allele with a dominant one. In females, a recessive allele on one X chromosome is often masked in their phenotype with a dominant allele the other. They must be homozygous recessive in order for the recessive allele to be expressed and the statistical chances of that are 3:1.
Some examples of sex-linked traits that are much more common in human males than females are baldness, color blindness, muscular dystrophy, and hemophilia. Females also get these conditions, but in much lower numbers than males.
Example: In fruit flies, the gene for eye color is carried on the X chromosome which is a sex chromosome (sex-linked). The allele for red eyes (XR) is dominant over the allele for white eyes (Xr). If a white-eyed female fruit fly is mated with a red-eyed male, predict the possible offspring.
Step 1: (genotype of the parents): Since the female has white eyes, she must be XrXr. The male is red-eyed and because he has only one X chromosome, he has only one allele for eye color. His eyes are red, so he must be XRY since the allele “R” is present on the X chromosome only, and there is no other allele for eye color because the male other sex chromosome is a Y chromosome.
Step 2/3 (set up and fill out Punnett Square): For sex-linked traits we need to list the genotype in a different fashion. We must identify the individual as being male or female according to their sex chromosomes. Females are XX, and males are XY.
Step 4: (genotypic ratio):
Step 5 (phenotypic ratio): The individual XRXr will be a female because she has two X chromosomes. She will have red eyes because she has Rr. The individual with XrY will be a male because he has the X and Y chromosomes. He will have white eyes because he has only one allele and it is “r”. So from this cross you would expect all of the females to have red eyes and all of the males to have white eyes.
- Hemophilia is a sex-linked trait. A person with hemophilia is lacking certain proteins that are necessary for normal blood clotting. Hemophilia is caused by a recessive allele so let “N” stand for the normal clotting allele, and “n” be the allele for hemophilia. Since hemophilia is sex-linked, remember a female will have two alleles (NN or Nn or nn) but a male will have only one allele (N or n).
- A female who is heterozygous (a carrier) for hemophilia marries a male who does not have hemophilia. What are the genotypes of the parents?
- Make a Punnett Square for the above cross.
- What is the probability that any of the offspring will have hemophilia
- What is the probability that a male offspring will have hemophilia?
- What is the probability that a female offspring will have hemophilia?
- What is the probability that a female offspring will be a carrier?
2. Can a color-blind female have a son that has normal vision? Color blindness is caused by a sex-linked recessive allele. *use XN = not color-blind and Xn = color-blind
3. Baldness is a sex-linked trait. What parental genotypes could produce a bald female? *use XH = normal hair, and Xh = bald
Once you’ve completed the answer sheet, please submit through Blackboard.