BIOL 210 Problem Set 6 KEY

Week 6, Oct 7-11

This is the key for PS 6. Before reading this document, you should have completed the problems. Use this key to check and correct your work BEFORE submitting the corrected version via the google form.

You should compare your responses with this key and make any changes in another font color. Be sure to explain why you got things wrong (showing that now you understand) as well as providing corrected responses.

Question Key

1. Consider the four-o’clock plant (Mirabilis jalapa), in which there are two alleles for flower color, R (homozygotes have red flowers) and W (homozygotes have white flowers). A homozygous plant is also called “true-breeding” because all of its gametes will inherit the same allele. And if you cross two true-breeding red-flowered plants, then you know the offspring will “breed true”; the offspring will all have red flowers like their parents. If you cross a true-breeding red-flowered plant with a true-breeding white-flowered plant, the offspring have pink flowers.
1. Write out the genotypes of the two true-breeding parents and of the offspring. Which are homozygous? Heterozygous?

True-breeding parents crossed: RR × W W. Offspring will all be RW (and have pink flowers). The parents were homozygous but all of the offspring are heterozygous for this locus.

2. What is the name for the pattern of inheritance of the phenotypes encoded by the R and W alleles?

This inheritance pattern is incomplete dominance for flower color because the heterozygotes have a phenotype that is intermediate between the phenotypes of the two homozygotes.

3. Why does the heterozygote have pink flowers instead of red or white?

The explanation for why the heterozygote has pink flowers rather than red or white could be a few things. One example, maybe the R allele encodes the production of red pigment and the W allele results in no pigment production. However, one R allele does not result in enough pigment being produced to give a darker red color (it’s haplo-insufficient), thus we get a pink color instead.

4. Do you make any connections between this example and topics we learned earlier in the course?
Responses will vary… You should find some connections!




2. As discussed in the content video on Independent Assortment, phenotypic plasticity is the phenomenon in which a single genotype can produce different phenotypes, depending on external conditions. For example, water fleas (Daphnia spp., pictured) are zooplankton, microscopic aquatic animals commonly found in ponds and lakes. They eat algae and are eaten by juvenile fish. If a female Daphnia lives in a pond also inhabited by fish, she will produce clonal babies (by parthenogenesis) that grow head and tail spikes (on the left in image). But if that same mom (genetically identical) lives in a pond without fish, she will have clonal babies with rounded heads and short tails (on the right in image).

   

   1. How do you think Daphnia moms are able to change the way in which their babies develop to either have or lack head and tail spikes? (Think about how this relates to previous topics.)

   Since the babies develop inside the female Daphnia, they will be able to receive signals from her (and she will have deposited proteins in the cytoplasm of the egg cells she produced). These signals could include transcription factors that cause epigenetic changes in the genomes of the babies so that they “turn off” the round-head signal cascade and “turn on” the spiky-head signal cascade. This will be a type of epigenetic change that results in a permanent change in the babies (they can’t change back to round-heads if it turns out there aren’t any fish). However, if fish disappear from the environment, they will send those signals to their own babies and may have offspring with rounded heads.

   2. Throughout the course, we’ve actually seen quite a few examples of phenotypic plasticity, though I didn’t give them that name at the time. Find at least 2 examples that we have used earlier in the course which also happen to be examples of phenotypic plasticity. See if you can find one of them in a unicellular organism!

   Here are some examples:

   • From Mitosis and Binary fission video, the (unicellular) bacterium Caulobacter has two phenotypes: a flagellated swarmer cell and a stalked cell. The swarmer cell eventually settles down attached to a surface and switches phenotypes to be a stalked cell.

   • Also in Mitosis and Binary Fission video, the (unicellular) bacterium Bacillus licheniformis has two types of cells, vegetative and endospores. These are phenotypically and metabolically quite different (endospores can be dormant to survive bad conditions).

   • Temperature-dependent sex determination (like in the red-eared slider turtle, Transcription video) is another example: which gonads develop depends on the environmental conditions. So the same exact genome can develop as either male or female.

   • People who have the HbS allele for betaglobin have globin molecules that can be the usual protein complex that floats freely OR under low blood oxygen, their globin molecules polymerize to make rigid fibers that warp the shape of the red blood cell (to be a sickle-cell, Translation video).

You are a plant breeder studying two phenotypes in the trout lily plant: flower color and leaf pattern. Your studies lead you to believe that the flower color phenotype is determined by a single gene with two alleles (yellow, A, and white, a) and the leaf pattern phenotype is also determined by a single gene with two alleles (spotted, B, and plain, b). You have a true-breeding plant that has white flowers and spotted leaves (aaBB). You do not know the genes that are responsible for these two phenotypes but have inferred the genotype of your true-breeding plant with white flowers and spotted leaves.

3. A fellow lily-breeder sends you a true-breeding plant that has yellow flowers and plain leaves. What is the genotype of this plant? (Hint: what does true-breeding mean?)

Let A represent the allele for yellow flowers and b represent the allele for plain leaves. Then, a true-breeding plant with yellow flowers and plain leaves must have the genotype AAbb.


4. You cross your white and spotted plant with the yellow and plain plant. If yellow is dominant to white and spotted is dominant to plain, what will be the genotype and phenotype of the offspring (the F₁s)?

The cross is aaBB × AAbb so all of the offspring will inherit a white allele and a spotted allele from one parent and a yellow allele and a plain alllele from the other parent. Thus, the genotype of all offspring will be AaBb. Since yellow is dominant to white and spotted is dominant to plain, the phenotype of these plants will be yellow flowers and spotted leaves.


5. Imagine you wanted to figure out whether the flower color and leaf pattern genes are linked. What cross would you need to carry out? What phenotypic ratio would you expect to observe in the cross if the genes are NOT physically linked? (Hint: You must use the plants you already have available, which does not include an aabb individual.)

To figure out if they are linked, we should begin by rewriting our genotypes to reflect which alleles were inherited together from a given parent. The true-breeding white (a) and spotted (B) parent produces gametes that all have aB while the true- breeding yellow (A) and plain (b) parent’s gametes are all Ab. So the F₁ offspring of their cross can be written as aB/Ab to reflect which alleles came from which parental gamete.


To find out if the genes underlying the two traits are linked, we can cross two of the F₁ plants, all of which have the genotype aB/Ab. So the cross would be:


AaBb x AaBb


Or, if we wanted to show which alleles were inherited from which parental gamete:

aB/Ab × aB/Ab


If the genes for these two traits are NOT linked, then we could just look at equation (1), without worrying about which alleles were in which parental gamete. In this case of NOT linked, we could figure out the offspring distribution for the two traits separately like so:

For flower color:
25% aa (white), 50% Aa (yellow), 25% AA (yellow)

For leaf pattern:
25% BB (spotted), 50% Bb (spotted), 25% bb (plain)

We then combine these probabilities to get the multilocus genotype and phenotype probabilities. Assuming flower color and leaf pattern segregate independently (are not linked), the probability of the offspring of our cross being white flowered with spotted leaves will be the probability of having white flowers (0.25 or 25%) times the probability of having spotted leaves (0.25 + 0.50 = 0.75 or 75%) which is 0.25 x 0.75 = 0.1875 or 18.75% (3/16).

Overall, if the genes for the traits are not linked, we’d expect to observe these phenotype and genotype frequencies (make the 4 by 4 Punnett square to see these results):

• white, spotted will be 18.75% (3/16) where 1/16 are aaBB and 2/16 are aaBb
  

• white, plain will be 6.25% (1/16) where 1/16 are aabb
  

• yellow, spotted will be 56.25% (9/16) where 1/16 are AABB, 2/16 are AABb, 2/16 are AaBB, and 4/16 are AaBb.

• yellow, plain will be 18.75% (3/16) where 1/16 are AAbb and 2/16 are Aabb.

In contrast, if the genes are tightly linked, then the white/spotted alleles will almost always be inherited together and the yellow/plain alleles will also almost always be inherited together. In that case, the only types of gametes our F₁ plants can produce are aB or Ab, in equal frequencies. So, if we combine them to get our multilocus genotypes and phenotypes, we get the following expected phenotypic and genotypic ratios for tightly linked genes:

• white, spotted will be 25% (1/4), all aB/aB

• white, plain will be absent, 0% because no ab gametes are made

• yellow, spotted will be 50% (2/4), all aB/Ab where either gamete could come from either parent so this is 50% of the total zygotes created

• yellow, plain will be 25% (1/4), all Ab/Ab

If the genes are very closely physically linked, would you ever be able to produce a true-breeding plant with yellow flowers and spotted leaves? Explain.

If the genes are VERY closely physically linked, as described above, then the only way we could ever produce a true-breeding plant with the phenotype yellow/spotted would be if crossing-over occurred between the two genes, which would finally give us gametes that carried yellow/spotted alleles. Otherwise, while we could produce a plant that was yellow/spotted, only half of its offspring (crossed with itself) would actually be yellow/spotted because the plant is heterozygous for both genes. The closer together the genes are, the less likely this is to happen.

Linkage between genes arises because a single chromosome often contains a very long sequence of DNA that includes many genes. Given that humans have an estimated 20-25 thousand genes and n = 23 different chromosomes (2n = 46), we might expect that each chromosome is likely to contain around 1000 genes (in fact, some chromosomes only contain a few dozen genes, others have more than 2000).

Below are models of a few human chromosomes (in condensed form), noting along the length of the chromosome where different genes are located for human autosomes 5 through 8. The labeled genes are ones where some versions (alleles) of that gene are known to cause disease. The dark bands on the chromosomes indicate areas that are usually heterochromatin (highly condensed). Use the figure to answer the following questions.

7. On chromosome 7, the OB gene encodes the hormone leptin, which helps regulate energy balance (wild-type allele B). The b allele for this gene produces non-functional leptin molecules which increase risk of obesity in homozygotes (bb). Based on the figure, the OB (Obesity) gene is physically linked to which other genes?
The OB gene is closely linked to CFTR and Pendrin. It is possibly linked, but more distantly, to ELN and GCK.


8. Consider the CFTR gene on chromosome 7, which encodes a protein that transports chloride and thiocyanate ions across the cell membrane (wild-type allele F). The mutant f allele for this gene is a deletion that causes protein-misfolding, leading to cystic fibrosis in homozygotes (ff) but no effect in heterozygotes (Ff). If a person was heterozygous for both the OB gene and the CFTR gene (BbFf), and the genes were NOT linked, what would be the chance that this person’s child would inherit BOTH the b and f alleles from this parent?

The parent is BbFf and if the genes were not linked, we’d expect them to produce equal proportions of the following 4 gamete types: BF, bf, Bf, and bF. Thus, there is a 25% chance of this parent passing both the b and the f alleles to their child.



9. Because the OB and CFTR genes are linked, we need to think through some additional information to figure out how this affects the chance of an offspring inheriting mutant alleles for both genes. Consider 2 scenarios:


1. Scenario 1: Jack’s parents have the genotypes BBff (mom) and bbFF (dad), giving Jack the genotype BbFf. If this is Jack’s genotype, and no crossing-over occurs between the genes, can Jack’s child inherit both the b and f alleles from Jack? Explain.

If Jack is Bf/bF and the genes are tightly linked so no crossing-over occurs, then Jack will produce 2 types of gametes: Bf and bF. As a result, Jack CANNOT pass both b and f to one child.


2. Scenario 2: Alex’s parents have the genotypes bbff (mom) and BBFF (dad), giving Alex the BbFf genotype. If no crossing-over occurs between OB and CFTR, can Alex pass on both the b and f alleles to one child?
In this scenario, Alex will make only bf and BF gametes meaning there is a 50% chance that Alex passes both the b and f alleles to a child.




10. Now consider the children of parents with the genotypes Bf/bF and Bf/bF. Assuming that these genes are so tightly linked that crossing-over does not occur between them,
1. What’s the chance that their child has a high risk of obesity?

We only need to consider the OB gene here. There is a 50% chance of getting the b allele from each parent, yielding a 25% chance of the child being at high risk for obesity.

2. What’s the chance that their child inherits a predisposition for cystic fibrosis?

Here we need only consider the CFTR gene. Again, there is a 50% chance of each parent contributing the f allele to the child, yielding a 25% chance that the child is predisposed to cystic fibrosis.

3. What’s the chance that their child is at high risk for both obesity and cystic fibrosis? How would your response change if the genes were NOT linked (calculate the probability)?

To be at high risk for both conditions, the child would have to be bbff. In order to get that genotype, the child would have to inherit bf from both parents. Given the earlier information on the parents and linkage, it is impossible for either parent to contribute a bf gamete. Thus, the risk of the child being bbff is 0% because the obesity-risk allele b is inherited with the wild-type F allele (leaving the cystic fibrosis causing f allele inherited with the wild-type B allele).

What if the genes were not linked? In that case, we would consider each condition separately and then combine the probabilities. From parts (i) and (ii) we know the risk for each condition alone is 25% thus, we can say that the risk of inheriting both is 0.25 × 0.25 = 0.0625 or 6.25% (also known as 1/16).