How Do We Pass On Our Genes?

Use the concepts of Mendelian and non-Mendelian genetics to explain inheritance. For example, incomplete dominance, independent assortment, sex-linked traits and linkage.


Inheritance

Once upon a time there was a monk named Gregor Mendel. He was born in 1822 in (what is now) the Czech Republic, in Europe. He came from a farming family and was very interested in not only the family business, but also in beekeeping. When he joined a monastery in order to become a monk, he was sent to college to learn more science.

While he was at the University of Vienna, he was inspired to perform a few experiments on pea plants. He was particularly interested in how it was that some pea plants were different from one another.

Mendel in his garden

Mendel had a lot of time on his hands – and peas. So he separated the pea plants that he had into several groups. Two of these groups were tall pea plants and short pea plants. The tall ones were actually taller than him, about 6 feet tall! The short ones were only about a foot tall, so it was easy to tell which was which. The interesting thing that he had noticed, though, was that there were no pea plants that were neither one nor six feet tall. This observation meant to Mendel that there had to be something else going on inside the pea plant.

Remember, in 1822, nobody knew anything about DNA, genes or chromosomes. So, Mendel took his two groups of pea plants, the tall plants and the short plants, and separated them completely. From his work in beekeeping, he knew that bees could carry the pollen from one plant to another, so he made sure there was no way the tall plants could share pollen with the short plants. After a few generations, there was nothing but tall plants in the one group and nothing but short plants in the second group. Mendel had successfully made two pure groups of pea plants.

After this, he started a third group. He brought together the tall plants and short plants in this third group, making sure that every new plant was a combination of a tall and a short plant. What surprised him was that every single one of the resulting pea plants was tall! What was going on here? Wouldn’t you expect to see a mixture of tall and short pea plants?

Mendel needed to know more. He called these new tall plants the F1 generation (after the Latin for the first children). He wanted more information: specifically, he wanted to know what would happen when these tall plants had offspring with each other. Would they have all tall offspring? Was the shortness of the one parent completely lost?

Mendel then combined the F1 generation (the tall plants) with each other. The results were incredible: 75% of the offspring were tall, and 25% of them were short! Somehow, these plants had “remembered” their short grandparents – but how? So Mendel made a hypothesis that traits are carried from generation to generation in genes. Each individual has one copy from each of their parents, and the trait can either be dominant or recessive. In this case, he thought that the trait for being tall (for a pea plant) was the dominant trait and being short was recessive. But here’s the tricky part of what he figured out.

Mendel figured that each plant in the F1 generation got one tall trait from one parent and one short trait from the other parent. Since the tall trait (or allele) was dominant, then it hid the recessive trait for being short. Even though he didn’t use a Punnett square, he figured out that the cross between the two F1 plants went like this (T = tall, t = short):

T t
T TT Tt
t Tt tt

A diagram of Mendel's pea plant experiments


He repeated this experiment with many more traits and many more plants, coming up with roughly the same results each time.

Mendel looked at a total of seven traits of pea plants. One of the other traits was whether the seed of the pea was smooth or wrinkled. He observed that when he did the same experiment with this trait, the results were the same. So he decided to take the experiment one step further. He then combined the two traits: he took tall plants that made smooth seeds, tall plants that made wrinkled seeds, short plants that made smooth seeds, and short plants that made wrinkled seeds. He combined them in every way that he could think, but he found that no matter what he did, when he combined purely smooth seeds with purely wrinkled seeds, the offspring were all smooth. It didn’t matter if they were tall or short at all. After trying this with several more traits, he found a pattern. One trait didn’t affect any of the others. This came to be called the Law of Independent Assortment. All of the genes seemed to mix themselves up, and it didn’t matter what the other traits were.

Actual X and Y chromosomes side-by-side

A karyotype of a male human showing all of the chromosomes

Later, scientists came to find that things weren’t so simple. They actually saw that there were some traits that did depend on other traits. For instance, plants with yellow flowers were usually tall and plants with blue flowers were usually short. They explained this by saying that when genes are close together on a chromosome, they can sometimes show linkage. This means that some genes are “linked” together and are not independent.

There’s another type of linkage, called sex-linked traits, that Mendel did not describe. These traits are not necessarily traits that have anything to do with sex organs or sex cells. Traits that are sex-linked are on the sex chromosomes. It’s important to understand that male humans and female humans have one major difference in their chromosomes: the 23rd and final pair of chromosomes is “XX” in females and “XY” in males. The “Y” in males is actually just a small chromosome (see picture) and contains much less information than the “X” chromosome. Because of this, there are alleles on the X chromosome that are not on the Y chromosome. For the alleles that are on the X chromosome but not the Y, they will always show up, dominant or recessive! Examples of sex-linked traits include hemophilia and color blindness. These traits, since they can be dominated in a female but not a male, often show up much more often in males than in females.

Well after Mendel passed, other scientists looked at his work and figured out that it fit in with their theories of inheritance. In fact, it wasn’t until the 1930’s that Mendel was recognized for his efforts and people began to accept that genes could be responsible for evolution! But Mendel’s work still didn’t explain quite a few things about genetics.

Questions
Remember
1. Why did the F2 generation of pea plants include short plants?
2. Define a sex-linked trait in your own words.
3. List all of the possible crosses among plants in the F2 generation.
Put it together
4. Paraphrase the main experiment that Mendel performed in one paragraph.
5. Compare and contrast the Law of Independent Assortment and linkage in at least two ways.
Think about it
6. Ask two thoughtful (not just factual) questions about Mendel's life to three other people. Record your results and underline your most difficult question.
Review
7. Identify the role of a cell to an organism.
8. What human cells are haploid?
9. What are the four bases in DNA? What are they in RNA?
10. What does dominant mean? Recessive?
Dragon Genetics

In this activity you will study the patterns of inheritance of multiple genes in (imaginary) dragons. For this activity, we will only consider one gene on each chromosome. These genes are described in the following table.

Dominant Alleles Recessive Alleles
Chromosome 1 W = has wings w = no wings
Chromosome 2 H = big horns h = small horns

The mother dragon is heterozygous for the wing gene (Ww) and the horn gene (Hh). The father is homozygous recessive for the wing gene (ww) and the horn gene (hh).

  1. What phenotypic traits will each parent have? Phenotypic traits are the observable bodily characteristics.
  2. Draw the appropriate characteristics for each parent below in your book:

Father

Mother

  1. On average, what percentage of the baby dragons will have big horns? _______

To predict the inheritance of the wing and horn genes, you first need to determine the genotypes of the eggs produced by the heterozygous (WwHh) mother dragon and the sperm produced by the homozygous (wwhh) father dragon. Use the figure below to answer the next questions.

  1. Considering both the wing and horn genes, what different genotypes of eggs could the heterozygous mother dragon produce?
  2. What genotypes or genotype of sperm can the homozygous (wwhh) father dragon produce?

The next step in predicting the inheritance of the wing and horn genes is to predict the outcome of fertilization between these eggs and sperm. In the following chart, label the gene on each chromosome in each type of zygote that could be produced by a mating between this mother and father. Then, fill in the genotypes of the baby dragons that result from each zygote and sketch in the characteristics of each baby dragon to show the phenotype for each genotype.

This type of mating involving two different genes is more typically shown as a Punnett square with four rows and four columns (see below). Notice that, because the father is homozygous for both genes, all his sperm have the same genotype, so all four rows are identical.

Mother (WwHh)
wh wH Wh WH
Father (wwhh) wh wwhh wwHh Wwhh WwHh
wh wwhh wwHh Wwhh WwHh
wh wwhh wwHh Wwhh WwHh
wh wwhh wwHh Wwhh WwHh
  1. Considering only the baby dragons with wings, what fraction do you expect to have big horns? (To answer this question, it may be helpful to begin by shading in the two columns of the above Punnett square that include all the baby dragons with wings.)
  2. Considering only the baby dragons without wings, what fraction do you expect to have big horns?
  3. Do you expect that baby dragons with wings and without wings will be equally likely to have big horns?

Procedure to Test Inheritance of Two Genes on Different Chromosomes

To test whether baby dragons with wings and baby dragons without wings will be equally likely to have big horns, you will carry out a simulation of the simultaneous inheritance of the genes for wings and horns. Since the father is homozygous (wwhh), you know that all of the father’s sperm will be wh. Therefore, to determine the genetic makeup of each baby dragon produced in your simulation, you will only need to determine the genetic makeup of the egg which is fertilized to become the zygote that develops into the baby dragon. During meiosis, each egg randomly receives one from each pair of homologous chromosomes. Your simulation will mimic this process.

For this simulation, each of the mother’s pairs of homologous chromosomes will be represented by a popsicle stick with the genes of one chromosome shown on one side and the genes of the other homologous chromosome shown on the other side. Since the mother dragon is heterozygous for both genes (WwHh), you will have one Popsicle stick representing a pair of homologous chromosomes which are heterozygous for the wing gene (Ww) and another Popsicle stick representing a pair of homologous chromosomes which are heterozygous for the horn gene (Hh).

  1. Hold one Popsicle stick in each hand about 6 inches above the desk. Hold each Popsicle stick horizontally with one side facing toward you and the other facing away (with one edge of the Popsicle stick on the bottom and the other edge on the top). The two Popsicle sticks should be lined up end-to-end, simulating the way pairs of homologous chromosomes line up in the center of the cell during the first meiotic division. Simultaneously drop both Popsicle sticks on the desk. The side of each Popsicle stick that is up represents the chromosome that is contained in the egg. This indicates which alleles are passed on to the baby dragon. Put a I in the appropriate box in the chart below to record the genotype of the resulting baby dragon.
    Mother (WwHh)
    wh wH Wh WH
    Fatherwwhh wh Genotype of baby = wwhhNumber of babies with this genotype =____ Genotype of baby = wwHhNumber of babies with this genotype =____ Genotype of baby = WwhhNumber of babies with this genotype =____ Genotype of baby = WwHhNumber of babies with this genotype =____
  2. Repeat step 1 three times to make and record three more baby dragons.

Summary and Interpretation of Data

  1. Compile the data for the baby dragons produced by all students in the following chart.
    Mother (WwHh)
    wh wH Wh WH
    Fatherwwhh wh Genotype ofbaby =________

    Number of

    babies with this genotype =___

    Phenotype:

    Wings __

    or no wings __

    Horns big __

    or small __

    Genotype ofbaby =________

    Number of

    babies with this genotype =___

    Phenotype:

    Wings __

    or no wings __

    Horns big __

    or small __

    Genotype ofbaby =________

    Number of

    babies with this genotype =___

    Phenotype:

    Wings __

    or no wings __

    Horns big __

    or small __

    Genotype ofbaby =________

    Number of

    babies with this genotype =___

    Phenotype:

    Wings __

    or no wings __

    Horns big __

    or small __

  2. Do any of the baby dragons with wings have small horns?
  3. Does either parent have the combination of wings and small horns?
  4. Considering only the baby dragons with wings, what fraction has big horns?
  5. Considering only the baby dragons without wings, what fraction has big horns?
  6. Are baby dragons with wings and without wings about equally likely to have big horns?
  7. Explain these results, based on what happens during meiosis and fertilization.
Reebops

By Patti Soderberg

Reebop

During the Reebop activity, you will observe all of the offspring produced by one set of Reebop parents. You will sort Mom and Dad Reebop’s chromosomes, select the new baby Reebop’s chromosomes, decode the baby’s chromosomes, and construct the baby Reebop according to the code. After all of the babies are “born,” the Reebop family will be assembled so the offspring can be compared to one another. Mom and Dad each have two antennae (small nails), a head (white large marshmallow), a neck (two toothpicks), two eyes (thumb tacks), an orange nose (an orange miniature marshmallow), three body segments (white large marshmallows), two green humps (green miniature marshmallows), four blue legs (blue push pins), and a curly tail (a pipe cleaner).

  1. Get a set of Mom and Dad Reebop’s chromosomes. Chromosomal analysis has revealed that Reebops have seven pairs, or fourteen total chromosomes. The 14 red chromosome are Mom’s chromosomes and the 14 green ones are Dad’s.
  2. One member in each pair will take the red chromosomes, and the other the green. Turn the chromosomes face down on the table so that no letters are visible, and sort them by length.
  3. Randomly take one chromosome of each length and place it in a separate “baby pile.” This will be your Reebop baby’s chromosomes. The remaining chromosomes can be returned to the envelope. Each Reebop baby will have 14 chromosomes, half red and half green.
  4. Construct your baby according to their “secret code” below.
    Reebop Alleles Homozygous recessive Heterozygous Homozygous dominant
    A (antennae) No antennae 2 antennae 1 antenna
    M (green humps) 3 green humps 2 green humps 1 green hump
    Q (nose color) Yellow nose Orange nose Red nose
    T (tail) Straight tail Curly tail Curly tail
    E (eyes) 3 eyes 2 eyes 2 eyes
    L (leg color) Red legs Blue legs Blue legs
    D (body segments) 2 body segments 3 body segments 3 body segments
  5. Place the completed Reebop babies in the designated “nursery.”
  6. How many chromosomes does your Reebop baby have?
  7. Were there any problems with your Reebop baby? If so, what?
  8. Was your Reebop baby different than all of the other babies? Why or why not?
  9. Was your Reebop baby exactly like either its mother or father? Why not?
Chi Square Lab

Introduction

My friend The FDC (Federal Department of Candy) is investigating Mars Co. There have been allegations that Mars has been fraudulent in their advertising! The accusations relate to Mars’ claims about the color distribution of their flagship product: Milk Chocolate M&Ms! Some consumers have complained about NOT getting the promised distribution of colors; now the FDC and Mars are mired in a bitter lawsuit that threatens the entire US candy industry. As usual, we’ve been hired to investigate and present our results at the trial, whose outcome hinges ENTIRELY on our work. Let us do well. Below is M&M’s CLAIMED color distribution, taken directly from their website http://www.m-ms.com.

Orange – 4.98 pieces – 23%
Blue – 4.27 pieces – 20%
Green – 3.93 pieces – 18%
Yellow – 3.75 pieces – 17%
Red – 2.66 pieces – 12%
Brown – 2 pieces – 9%

Background

The Chi Square test (X2) is often used in science to test if data you observe from an experiment is the same as the data that you would predict from the experiment. Calculating X2 values allow you to determine if test results are due to randomness or not. If the data is significantly different from random chance, other factors must be influencing your results. This investigation will help you to use the Chi Square test by allowing you to practice it with M&Ms.

Biologists generally accept a “p” value (probability) of 0.05 as a cutoff. If you get a probability (p) below 0.05, then this generally means that your observed results are significantly different than what was expected. This means that there is some explanation as to why you got the results that you did, and that it was not random.

Procedure

  1. Place about 200 candies in a cup or bowl. Record just the different colors in Table 1 and Table 2. Do not count the individual candies or tally them by color.
  2. Without counting, estimate the number (percentage out of 100%) of the different colors of each color of the candies. Record the estimates in Table 1 under “Percentage Expected by Estimate.”
  3. Copy the percentages from the introduction under “Percentage Expected from M&M Company.”
  4. Write a hypothesis which predicts the percentage of the different colors of the candies.
  5. Count the number of each color of candy and record the number in Table 1 under “Number Observed.”
  6. Calculate the number of each color expected by your estimate in Table 1 and record under “Number Expected by Estimate.”
  7. Calculate the number of each color expected by M&Ms estimate in Table 1 and record under “Number Expected by M&M Company.”
  8. Record the numbers expected by estimate (e), and the numbers observed (o) in Table 2.
  9. Record the numbers expected by the M&M company (e), and the numbers observed (o) in Table 3.
  10. Subtract the number expected from the number observed for both Table 2 and Table 3 and place those numbers in the column “o – e”.
  11. Square the column “o – e” and for both Table 2 and Table 3 and place those numbers in the column “(o – e)2“.
  12. Finally, divide column “(o – e)2” by the number expected (e) for both Table 2 and Table 3 and place those numbers in the last column. Add up all of the values in this column. This is your Chi Square value.
  13. You will now convert this to a probability, using Table 4. The Degrees of Freedom is the number of colors of M&Ms, minus one. Then find your Chi Square value in the middle of the table. The probability is the bold number at the top of the column. Write this down.
  14. What is the X2 value for your data?
  15. What is the p value for your data?
  16. Given your p-value, create a statement that describes how significant your data is.
  17. Given a p-value limit of 0.05, is your hypothesis accepted or rejected? Why or why not?
  18. If the hypothesis is rejected, propose an alternate hypothesis.
  19. Suppose you were to obtain a Chi Square value of 7.82 or greater in your data analysis, with 2 degrees of freedom. What would this indicate?

Table 1

Color of Candy Number Observed (o) Percentage Expected by Estimate Number Expected by Estimate Percentage Expected by M&M Number Expected by M&M
Total # of candies:

Table 2

Color of Candy Number Expected by Estimate (e) Number Observed (o) o – e (o – e)2 (o – e)2 / e

Table 3

Color of Candy Number Expected by M&M (e) Number Observed (o) o – e (o – e)2 (o – e)2 / e

Table 4

Degrees of

Freedom

(df)

Probability (p)

0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001

1

0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83

2

0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82

3

0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27

4

0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.28 18.47

5

1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52

6

1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46

7

2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32

8

2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12

9

3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88

10

3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59

Nonsignificant

Significant

 

 

Crossing Over Lab

Click here for Crossing Over Chromosomes

Background

Crossing over is a unique event of meiosis. It occurs during Prophase I, which is the first cell division of meiosis. In this stage, homologous pairs of duplicated chromosomes pair up together in groups called tetrads. Sister chromatids wrap up in each other, break off, then fuse back together onto the chromatid of their homologous pair. By doing so, they exchange alleles between chromosomes. This “crossing over” happens very frequently, in every chromosomes, during every meiosis.

Procedure

  1. Get the 4 chromosomes. Two chromosomes are green (father), and two are red (mother). Each parent provided a homozygous genotype; one parent only provided dominant alleles, the other parent only recessive.
  2. Cut out all 8 chromosomes.
  3. Set the chromosomes up in homologous pairs by taping them side-by-side at their centromeres. Align them in tetrads as they would be in Prophase 1.
  4. By cutting and taping back together your chromosomes, perform a crossover event for each arm of each chromatid in the center of the tetrad. This will mean you will perform four crossovers. Choose the cross over site at random, but do not cut in the middle of a gene.
  5. Complete meiosis and record the genotype of the gametes produced.
  6. What would the genotypes have been without crossing over?
  7. Why is meiosis important for sexual reproduction?
  8. What was the difference between the gametes produced without crossing over and the ones produced with crossing over?
  9. Did your cross-over event produce the same gametes as the other lab groups in the class?
  10. In what way is crossing over important for sexual reproduction?
  11. Was there a greater chance to cross over between some pairs of alleles than between others? Explain.
Hardy-Weinberg Lab

Background

A population is in Hardy-Weinberg equilibrium when evolution is not actively occurring in that population. We say that Hardy-Weinberg equilibrium happens when there are seven conditions that are met:

  1. Mutation is not occurring
  2. Natural selection is not occurring
  3. The population is infinitely large
  4. All members of the population breed
  5. All mating is totally random
  6. Everyone produces the same number of offspring
  7. There is no migration in or out of the population

This equilibrium is calculated by using the formula p2 + 2pq + q2 .

Procedures for an “Ideal” Population

  1. Everyone in the classroom will start as a heterozygote. You will receive one dominant allele and one recessive allele. Find a partner at random.
  2. Randomly choose one of your alleles and combine it with your partner’s allele. Write down the genotype of this produced offspring.
  3. Return your alleles, then repeat step #2.
  4. Now that you have two offspring, you and your partner will become those two offspring. Decide which of you will become which offspring. If you need a new allele in order to represent your new genotype, trade what you don’t need with the teacher.
  5. Find a new partner at random in the class. Repeat steps #2 – 4 four more times with different partners so you have a total of five generations.
  6. Combine your data with the remainder of the class and fill out Tables 1 & 2.
  7. Complete Table 3 with the equilibrium values. Fill out the second row with an ideal version of what should happen. Basically, you need to figure out what would happen if 25% of the individuals were AA, 50% were Aa and 25% were aa.
  8. Do the class results for the p and q values of the 5th generation agree with the predicted values?
  9. What does this mean about the population?
  10. Which one of the assumptions (from the introduction) is not followed?

Table 1

Parental 1st 2nd 3rd 4th 5th
My Genotype

Table 2

Generation # AA Aa aa Allele A Allele a Total Alleles
Parental
1st
2nd
3rd
4th
5th

Table 3

Generation # Frequency of AA Frequency of Aa Frequency of aa Frequency of A Frequency of a
Parental
Theoretical Results in 5th generation
Actual Results in 5th generation

Procedures for Selection Pressure

  1. You will now repeat the previous experiment, except with one major difference. If you produce an offspring that is homozygous recessive, it immediately dies. You must replace it with a new offspring in order to continue.
  2. Combine your data with the remainder of the class and fill out Tables 4 & 5.
  3. Complete Table 6 with the equilibrium values. Fill out the second row with an ideal version of what should happen. Basically, you need to figure out what would happen if 25% of the individuals were AA, 50% were Aa and 25% were aa.
  4. Do the class results for the p and q values of the 5th generation agree with the predicted values?
  5. What does this mean about the population?
  6. Predict what might happen to these frequencies if you simulated for another five generations.
  7. Since homozygous recessives are strongly selected against, would you expect the recessive allele to be completely removed from the population?

Table 4

Parental 1st 2nd 3rd 4th 5th
My Genotype

Table 5

Generation # AA Aa aa Allele A Allele a Total Alleles
Parental
1st
2nd
3rd
4th
5th

Table 6

Generation # Frequency of AA Frequency of Aa Frequency of aa Frequency of A Frequency of a
Parental
Theoretical Results in 5th generation
Actual Results in 5th generation

Procedures for Heterozygous Advantage

  1. You will now repeat the previous experiment, except with one major difference. If you produce an offspring that is homozygous recessive, it immediately dies and if you produce an offspring that is homozygous dominant, you flip a coin to determine if it survives or dies. If the offspring dies, you must replace it with a new offspring in order to continue.
  2. Combine your data with the remainder of the class and fill out Tables 7 & 8.
  3. Complete Table 9 with the equilibrium values. Fill out the second row with an ideal version of what should happen. Basically, you need to figure out what would happen if 25% of the individuals were AA, 50% were Aa and 25% were aa.
  4. Do the class results for the p and q values of the 5th generation agree with the predicted values?
  5. What does this mean about the population?
  6. Compare this case to the first two. Explain how the changes in frequencies show evolution to be occurring.
  7. Do you think the recessive allele will be eliminated? Why or why not.

Table 7

Parental 1st 2nd 3rd 4th 5th
My Genotype

Table 8

Generation # AA Aa aa Allele A Allele a Total Alleles
Parental
1st
2nd
3rd
4th
5th

Table 9

Generation # Frequency of AA Frequency of Aa Frequency of aa Frequency of A Frequency of a
Parental
Theoretical Results in 5th generation
Actual Results in 5th generation

Procedures for Genetic Drift

  1. You will now repeat the first experiment (“ideal”), except with two major differences. Now you will be placed in small groups of 3 – 4 students and you must mate within those groups. You will also do this for 10 generations instead of 5.
  2. Combine your data with the remainder of your group and fill out Tables 10 & 11.
  3. Complete Table 12 with the equilibrium values. Fill out the second row with an ideal version of what should happen. Basically, you need to figure out what would happen if 25% of the individuals were AA, 50% were Aa and 25% were aa.
  4. Do the class results for the p and q values of the 5th generation agree with the predicted values?
  5. What does this mean about the population?
  6. Compare your results with other groups. Did genetic drift happen in other groups? Did it happen in the same way?
  7. What is the relationship between group size and genetic drift?

Table 10

Parental 2nd 4th 6th 8th 10th
My Genotype

Table 11

Generation # AA Aa aa Allele A Allele a Total Alleles
Parental
2nd
4th
6th
8th
10th

Table 12

Generation # Frequency of AA Frequency of Aa Frequency of aa Frequency of A Frequency of a
Parental
Theoretical Results in 10th generation
Actual Results in 10th generation
The Secrets of Bedrock: Sex-linked Traits with Fred and Wilma

By Ms. Jenny Holt

 

Background

Geneticists have succeeded in sequencing the genes on the sex chromosomes of Bedrock’s most famous couple, Fred and Wilma Flintstone. Shocking discoveries have been made – the secrets of Bedrock can now be revealed.

 

Part I: Flintstone Family Secret Analysis:

Use the data for Fred & Wilma’s sex chromosome in the tables below to answer questions 1-15 in the spaces provided.

 

Traits on the X chromosome (in the order they appear from top to bottom)

 

Dominant

 

Recessive

O predisposed to obesity o not predisposed to obesity
N Normal vision (can see red and green) n red-green colorblindness
B Normal hair growth b baldness
H Normal blood clotting h hemophilia (blood does not clot)
D Normal hearing d deafness
P Pigmented eyes (brown, blue, or green) p red eyes (no pigment)
E Faulty tooth enamel e normal tooth enamel
S Sweat glands present s sweat glands absent
M Not predisposed to migraines m predisposition to migraines

 

Trait on the Y chromosome

 

Dominant

 

Recessive

H hair growth in ears absent h hair growth in ears present

 

  1. Use the genotype for Fred and Wilma to figure out their phenotype. Write each phenotype in the space provided in the table.

 

X-Linked Traits

Trait

Fred

Wilma

Genotype

Phenotype

Genotype

Phenotype

Obesity

XOY

Obesity

XOXo

Color Vision

XnY

XnXn

Hair Growth

XBY

XbXb

Blood Clotting

XHY

XHXh

Hearing

XDY

XdXd

Eye Pigment

XpY

XPXp

Tooth Enamel

XEY

XeXe

Sweat Glands

XsY

XSXs

Migraines

XmY

XMXM

 

Y-Linked Traits

Trait

Fred’s Genotype

Fred’s Phenotype

Wilma’s Genotype

Wilma’s Phenotype

Ear Hair

XYh

XX

 

  1. Can Wilma tell the difference between red and green Christmas lights? Explain.
  2. Does Wilma have faulty tooth enamel? Explain.
  3. Do you think Wilma has to get her hair cut & colored often?  Explain.
  4. Are there traits for which Wilma is a carrier?  If so, list all of them or explain why not.
  5. Are there traits for which Fred is a carrier?  If so, list all of them or explain why not.
  6. Is Fred lying when he tells Wilma that he thinks her hair is a gorgeous shade of red? Explain.
  7. A dedicated pet owner, Fred walks Dino once a day, but has a hard time cooling his body down.  Explain why this is true.
  8. Does Fred need to wear a toupee (which is a wig for a man)? Explain.

10. Does Pebbles (their daughter) need to be treated for faulty tooth enamel?  Set up a Punnett square to show your answer and identify the possible genotypes for Pebbles.

11. Does Pebbles wear a wig?  Set up a Punnett square to illustrate your answer.  Set up a Punnett square to show your answer and identify the possible genotypes for Pebbles.

12. What percent chance does Pebbles have to be predisposed to obesity?  Show work.

13. If Fred and Wilma had a son, what are his chances of having normal vision?  Show work.

14. Does Wilma get angry when Fred screams…”WILMAAAAAAAAAA”? Explain.

15. What chance does Pebbles have to develop hairy ears?  Explain.

 

Part II:  Family or Fraud?

A young man named “Fredrick” has just arrived in Bedrock, and he claims to be the son of one of the Flintstones!!  Who fathered/mothered Fredrick?  It’s up to you to solve this mystery.  Fortunately Fredrick agreed to DNA sequencing and the results are now in.

 

Procedures:

Use Fredrick’s X and Y chromosomes to answer the following questions.

 

  1. Fill in Fredrick’s genotype for each sex-linked trait in the table below.

Trait

Fredrick’s Genotype

Fredrick’s Phenotype

Obesity

No obesity

Color Vision

Red-green colorblind

Hair Growth

Bald

Blood Clotting

Normal blood clotting

Hearing

Normal hearing

Eye Pigment

Red eyes

Tooth Enamel

Normal tooth enamel

Sweat Glands

Sweat glands absent

Migraines

Predisposition to migraines

Ear Hair

Hair growth in ears absent

 

  1. Could Fred be the father of Fredrick?  Provide genetic evidence to support your answer.
  2. Could Wilma be Fredrick’s mother?  Provide genetic evidence to support your answer.
  3. Which parent determines the gender of the child?  Explain why.
  4. Why are males affected by recessive sex-linked diseases more often than females?
  5. If a male has a disease that is Y-linked, what percentage of his sons will inherit the disease?
  6. If a male has a disease that is Y-linked, what percentage of his daughters will inherit the disease?  What percentage will be carriers?