Biology 205 Mendelian genetics: Some evil practice problems.

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1. X linkage. Hemophilia in humans is inherited as an X linked recessive trait. A woman whose father is hemophiliac marries a man with normal clotting ability. What is the probability that her first child will have hemophilia? Assume that the woman's mother is homozygous dominant.

2. Multiple Alleles. A,B,O Blood system. A person's blood type with respect to the ABO blood system is a multiple allele system where the person's blood type is determined by which two autosomal alleles from the set I^A, I^B or i an individual has.

A.

By way of review, fill out the following table: Consult your text if need be.

Hints: remember the 'i' allele is recessive to the other two and the I^A, I^B alleles are codominant

B.

Dr. Paul is blood type O. His father was blood type A and his mother was blood type B. What were the genotypes of his parents and what are the possible blood types and ratios expected for crosses involving these parental genotypes?

3. Independent assortment.

Pretend that a couple with the same blood type genotypes as Dr. Paul's parents want to have a child. The mother happens to be a carrier of hemophilia.

A. What is the probability that their first child will be type AB and have hemophilia?

B. The parents conceive, and amniocentesis reveals that the child is male. Recalculate your probability in A given this new information.

Hints: the hemophilia locus is on the X chromosome and the ABO blood system locus is on an autosome so we can use Mendel's law of independent assortment. Write the cross as a combination of two independent single locus crosses, one for blood type and one for the hemophilia locus.

4. Incomplete dominance and codominance.

A snapdragon pure breeding for red flowers is bred with one for white flowers. The F1 generation flowers are all pink. What would you predict for the phenotypic ratios for the F2 generation? Assume that this trait is controlled by two alleles at a single locus. Note typically the alleles are represented with superscripted capital letters similar to what we did for the ABO blood system.

Note that the genotype and phenotype ratio is the same!

5. Explain the difference between incomplete dominance and codominance.

Comments: Incomplete dominance-heterozygote has one allele that is not producing a functional product. In codominance the each allele in the heterozygote is producing a functional product. In both the heterozygote is detectable by the phenotype! In incomplete dominance the heterozygote phenotype is intermediate in appearance to either homozygote, but in theory this could also be true of codominance.

6. An odd case.

In mice, the yellow coat color never breeds true. When yellow mice are bred to non-yellow mice the progeny show a 1 : 1 ratio of yellow to non yellow mice. When two yellow mice are bred together a phenotypic ratio of about 2 yellow to 1 non-yellow is typically observed.. What might be going on here?

7. Independent assortment.

In humans achondroplastic dwarfism and neurofibromatosis are both extremely rare dominant conditions. If a woman with achondroplasia marries a man with neurofibromatosis, what phenotypes could be produced and in what proportions? Hint: why did I mention that both conditions are extremely rare?

Hints: Assume independence of the loci: The woman might be Aann and the man aaNn. The basic assumption here is that the conditions are rare enough that it is extremely unlikely that homozygous dominants will be found in randomly selected individuals.

8. Epistasis

In Labrador retrievers, coat color is controlled by two loci each with two alleles B,b and E,e respectively. When pure breeding Black labs with genotype BBEE are crossed with pure breeding yellow labs of genotype bbee the resulting F1 offspring are black .

When a test cross is done using the F1 to a double homozygous recessive individual, breeders typically get the following phenotypic ratios:

1 Black Lab : 1 Chocolate lab : 2 Yellow lab

Explain these results. This is an example of Epistasis.

Hint: First figure the rules for which genotypes lead to which phenotypes and then think of a simple biochemical model consistent with this rule.

9. What phenotypic ratios are expected from the cross BbEe X BbEe in the black lab example?

Hints: You can do this assuming independence using a Punnett Square. Or if you are clever, use a branching diagram remembering that ee individuals always are yellow regardless of what's happening at the 'B' locus.

10. Think about this one carefully. Different genes at different loci govern albinism and hair color. A recessively inherited form of albinism causes affected individuals to lack pigment in their skin hair and eyes. At the hair color locus, red hair is inherited as a recessive trait and brown hair as a dominant trait.

An albino woman whose parents both have red hair has two children with a man who is normally pigmented (non-albino) and has brown hair. The man has one parent who has red hair. The couple's first child is normally pigment and has brown hair. The second child is albino.

A. What would the hair color of the albino woman be if she had pigmented hair?

Hint: in other words if she produced pigment at all (Homozygous dominant or heterozygous at the 'albino' locus. Might help to draw a simple pedigree and find the genotypes.

B. What is her genotype with respect to the hair color locus?

Should be easy as you need to do this for part A!

C. What is the genotype of the brown haired male parent with respect to both loci?

Since one parent of this male has red hair..that parent is homozygous recessive, say bb. Thus the male must be Bb. Notice I didn't tell you anything about the male's other parent. You didn't really need that information. :-)

D. What is the genotype of the first child with respect to both loci?

Should be easy, given the phenotype of the child.

E. What are the possible genotypes of the second child for hair color? What should the phenotype be with respect to the hair color locus? Why is the child albino?

Note the child must be aa with respect to the 'albino locus'

11. In fruit flies as recessive mutation in either of two independently assorting genes, brn and prp which prevent the synthesis of wild type red eye pigment. Homozygotes for either of these mutations have brownish purple eyes. But heterozygotes for both these mutations have wild type eyes. If double heterozygotes are crossed with each other what are the phenotypes and expected ratios for the offspring? Hint let the wild type alleles of each gene be brn+ and prp+ respectively.

Assume, that the wild type alleles are dominant as we discussed in class.

This is really an epistasis problem.

12. Multiple unlinked loci.

Two individuals who are heterozygous at four loci are mated. This cross is:

13. Pedigree analysis with inbreeding.

Consider the accompanying pedigree. Male 1 in generation I has a rare form of albinism that is inherited as a recessive trait so let his genotype be AA Individuals with genotypes AA and AA have normally pigmented skin. The individuals shown in generation III want to marry. Based solely on the pedigree information, what is the probability that their first child will be albino?

Hints: This seems like tough problem. Notice I have labeled all the individuals in the pedigree so your first step ought to figure out the genotypes that you can given the information. For instance I1 must be homozygous recessive, I2 assume is homozygous dominant since I tell that the albinism allele is rare. etc.

The other trick to this problem is to note that III1 and III2, the cousins in generation three can be either homozygous dominant or heterozygous and that these are independent events. What is important here is the probability that both cousins are heterozygous. What is this probability? Once you figure this then given that the cousins are heterozygous what is the probability that the first child will be albino? The answer is 1/4 using a basic Punnett square.

But you have to multiply this probability by the probability that both cousins are heterozygous to answer the question because the information at hand does not allow you to determine the actual genotypes of the cousins! So what you really have is

Pr(First child is albino) - Pr(First child is albino given that at least one cousin is AA)*Pr(at least one cousin is AA) +

Pr(First child is albino given that at least one cousin is AA)*Pr(both cousins are AA)

0*3/4 + 1/4*1/4 = 1/16

14. Additive Inheritance (A challenge)

In plant, flower color is inherited as an additive trait with four unlinked loci each with two alleles. Call the alleles A¹ , A² B¹, B² and C¹ C² and D¹ D². Plants that are homozygous A¹ A¹ B¹ B¹ C¹ C¹ D¹ D¹ have dark red flowers and plants that are homozygous A² A² B² B² C² C² D² D² have white flowers. Assume that the genes are additive meaning that the shade of red the flower is depends solely on the number of alleles with the superscript "1". Many continuous traits behave as if they are inherited in this way. Think of this mode of inheritance as being like incomplete dominance extended to multiple loci!

The trick to this problem is to remember that what determines the phenotype in these additive situations is simply the number of '1' alleles or '2' alleles out of the total regardless of arrangement. This allows use of our expression for the terms of a binomial expansion.

A. If you mate a true breeding dark red flowered plant with a true breeding white flowered plat, what is the genotype and likely phenotype of the F1 plants?

B. If you mate two of the F1 plants from part A together, how many possible flower color classes are there? Think carefully about the mode of inheritance.

Note you can have 0, 1, 2,...,7, 8 '1 alleles. Plants with 0 '1' alleles are white while plants with 8 '1' alleles are dark red. So there are 9 phenotypes!

C. What fraction of the F2 offspring from the cross in B are going to be dark red?

Plug into the general formula for a term of the binomial expansion. Let N = 8 m = 8 and remember that (! means factorial) and that 0! = 1. Or you can remember that for each locus Pr(both alleles being the homozygote with '1' alleles) is 1/4. So assuming independence Pr(all loci being homozygous for '1') is (1/4)^4

pgd 09/231/03

hints added 10/06/03 revised 03/11/04