Download MCDB 1041 Activity 3: Thinking about how “linkage” affects the

MCDB 1041
Activity 3: Thinking about how “linkage” affects the probabilities of inheritance
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Objectives:
• Review how meiosis generates unique gametes.
• Calculate inheritance probabilities of genotypes and phenotypes of two genes which are located
physically close to each other on the same chromosome (linked)
• Practice using a Punnett Square, and the variation for calculating probability with linked genes
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PART I. Comparing inheritance of genes on the same chromosome (Linkage) to when they are on
different chromosomes.
In the 1930’s, scientists had the idea that they could figure out where all the genes were on the human
chromosomes by following patterns of inheritance. They looked for diseases or traits that were commonly
inherited together as an indication that the two genes causing the diseases were located close together on the
same chromosome. By looking at large pedigrees of families with genetic diseases, they were able to see some
patterns (for example, finding that hemophilia and color blindness are both on the X chromosome). However,
this turned out to be an inefficient way of “mapping” the locations of genes because it required very large
families with two specific genetic traits that could be followed. With the sequencing of the human genome and
other molecular techniques, we can find out much more quickly where genes are located on chromosomes.
Nevertheless, the idea of linkage is important; if two diseases are carried in a family, parents might want
to know the chance of an offspring getting these two diseases.
How can you figure this out?
Look at the comparison at left. In A., the two genes
represented by alleles P/p and L/l are on different
chromosomes (like 1 and 3, for example), while in B. the
two genes represented by alleles P/p and L/l are on the same
chromosome (for example, chromosome 10).
Say in case A., the AaFf individual has children with
another person with AaFf genotype.
1. Draw out the Punnett Square to show the genotypes
of the possible children from this mating.
Now imagine you are looking at the situation shown in B, where the two genes of interest are linked on the
same chromosome.
2. NOT taking into consideration crossing over, draw the Punnett Square to show the genotypes of the
possible children from a mating of two parents that are both heterozygous, PpLl, where the genes are on
the same chromosomes.
See the difference between the two Punnett Squares? In the linkage situation, certain genotypes (and
thus certain phenotypes) are going to be more common than in the unlinked situation.
3.
Which gametes are missing in the linked situation compared to the unlinked situation? How would this
impact the possible genotypes of the offspring? Is there a chance that those genotypes could result from
this mating?
4. Now draw out the possible genotypes for the linked chromosomes taking into account crossing over, and
make a Punnet square reflecting all possible gametes and genotypes of the children.
Part II
Chromosomes are frequently broken and repaired by a similar process; in the example of recombination during
meiosis, the exchange normally happens only between homologous chromosomes. Homologous chromosomes
are very similar: they contain the exact same genes arrayed in the exact same order along the chromosome.
Because their DNA sequences are so similar, when they are pairing together to undergo meiosis, rather than just
sticking together, they can actually break and reseal, and in the process, swap sequences. This can happen
anywhere along the chromosome, and does happen in every meiotic event.
Modeling recombination: Pop beads
Everyone should have two sets of pop beads, one yellow and one red. Each represents a replicated homologous
chromosome (ie, the red chromosome is from your father and the yellow is from your mother). The beads
represent genes along the chromosome. Each should have 12 beads. Look at the lower half of each
chromosome (containing 6 beads). Say you are interested in the alleles represented by bead 6 and 12. Usually,
when crossing over occurs, it only involves two of the 4 strands of DNA: in other words only one strand of red
will trade information with one strand of yellow.
1. How many different ways could you break the DNA strands between beads 6 and 12 to exchange the red
and yellow beads?
2. How many different ways could you break the DNA strands between beads 11 and 12?
3. SO: If two genes are far apart on the chromosome, is it more or less likely that a recombination event will
occur between them?
How recombination affects gamete frequency
The more possibilities for recombination between two genes, the more likely a gamete will be produced that is
the result of a recombination event. Over the last 80 years, scientists have figured out the location of most
known human genes, such that we now know about how many genes are located on each of the chromosomes.
To describe the distance between the genes, they use a term called “map units”. Map units define how far
apart genes are on a chromosome by how likely they are to recombine. So, a 1% chance of crossing over means
that the genes in question are 1 map unit apart on the chromosome. Thus, if two genes are 12 map units apart,
there is a 12% chance of crossing over between them.
Practice using the above information:
Three genes are known to be on the same chromosome. The % recombination between the genes is:
X and Y: 15%
X and Z : 8%
Y and Z: 7%
4. Draw how these three genes are positioned on the chromosome.
5. What is the chance of recombination between X and Z?
What about this scenario:
% recombination between A and B: 10%
% recombination between C and A: 20%
6. Can you place all three genes?
Probability calculations with linked genes
Let’s make this more real with an example. Remember the Roloff family, from earlier this week? Zach, who
had achondroplasia, was also a carrier (Dd) for the other form of dwarfism, diastrophic dysplasia. Say that
Zach has children with another little person named Kate. Kate has diastrophic dysplasia (so she’d be dd)
Diastrophic dysplasia is on chromosome 5. On this same chromosome, there is another gene of interest called
the adenomatous polyposis gene. This gene, when mutated, causes growths of thousands of polyps in the colon,
which often leads to colon cancer and can be fatal. This disease is inherited in an autosomal dominant fashion.
We’ll represent the gene with Q/q alleles. Both Zach and Kate have this disease. The two genes are 16 map
units apart.
The two cells below represent Zach’s spermatogonium, entering meiosis (on the left), and Kate’s oogonium,
entering meiosis (on the right). Use the diagrams to help you visualize what is happening in the production of
gametes when genes are linked together on the same chromosome.
The short-hand way to represent linked
genes is the following (each line represents a
chromosome):
Zach : d q
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D Q
Kate: d Q
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d q
6. Let’s consider Zach first. Without a crossing over event, what gametes could Zach produce?
7. If there were a crossing over event somewhere between the D/d and Q/q alleles during meiosis in Zach, what
gametes could Zach produce?
8. Now list all together the gametes that Zach can produce. Which of the gametes would occur more
frequently? Why?
9. The D/d and Q/q alleles are 16 map units apart, which determines their probability of recombination. What
are all the frequencies of the different gametes that Zach can produce? (Hint, the parental gametes will be seen
LESS than 50% of the time, which is what their frequency would have been if there hadn't been recombination).
10. Now look at Kate. She is homozygous recessive for one of the alleles. What are all the possible gametes
she can produce? If there is a recombination event between the genes on her chromosome, will you be able to
tell from the eventual outcome of the children’s genotypes?
When this occurs (Kate’s genotype), and you can see that the recombinant gametes will be indistinguishable
from the parental gametes, you no longer have to consider them. She will have a 50% chance of each possible
combination: d Q or d q
Knowing all this, proceed to drawing out a Punnett Square for the inheritance of these two linked traits.
Follow the steps below!
The two genes discussed above: Diastrophic dysplasia (D/d) is autosomal recessive and Adenomatous Polyposis
(Q/q) is autosomal dominant. Zach and Kate decide to have kids.
A. Put each possible gamete for both Zach and Kate on the sides of the Punnettt square
B. Write down next to each gamete the frequency with which it will be made (you worked this out already,
above)
C. Fill in the possible genotypes of each of the possible kids.
D. Calculate the frequency of each genotype (possible kid) in your square by multiplying the frequency of
each gamete.
Now to use the information in the square:
• How many different ways can Kate and Zach have a child that has BOTH diastrophic dysplasia and
adenomous polyposis?
• Calculate the probability of this child being born to Kate and Zach. If you don't have a calculator, you
can just write out what you’d multiply and add to get the answer.
More Practice Questions
1. Sam and his wife Maggie are concerned about the inheritance of two disease causing genes that are 8 map
units apart that run in their family, and are located on the same chromosome. They are both carriers for the b
allele, but are normal with respect to the A allele.
Sam's genotype is B A while Maggie's is b A
b A
B A
What is the chance they'll have a child affected with the disease caused by the b allele?
a. 50%
b. 25%
c. 8%
d. 4%
e. 2%
2. The Y and A genes are closely linked on the same chromosome. Jim (genotype YYAA) and his wife Elise
(genotype yyaa) have a son, Pat. Which of the following is true about the gametes that Pat will produce
(hint: think carefully about what you know and don’t know):
a. yA & Ya gametes will outnumber YA & ya gametes
b. YA & ya gametes will outnumber yA & Ya gametes
c. yA, Ya, YA, & ya gametes will be present in equal numbers
d. need more information
3. Eileen and Max are each heterozygous for two disease genes linked on the same chromosome, AaBb.
Eileen’s alleles are:
A b while Kevin’s chromosomes are A B. The two genes are 10 map units apart.
a B
a b
They have Devin who is ab
Ab
What was the chance for Riley to be born with this genotype? You may show how you would work it out
without giving a final mathematical answer if you don’t have a calculator.