Biol/ MBios 301 (General Genetics) Spring 2002
Sample Midterm Examination (100 points possible)

10 Multiple Choice Questions-4 pts. each (Choose the best answer)

1. You observe a 1:2:1 ratio of visible types in the F2 generation of a cross between two homozygous lines of corn. Choose the correct statement(s) about this result.

    1. A single gene is involved in the cross
    2. Two phenotypes are observed in the F2 generation
    3. The alleles involved show complete dominance
    4. a and b
    5. a and c

2. You cross two mouse strains heterozygous for four genes (AaBbCcDd). What proportion of the total offspring would be heterozygous for all four genes?

    1. 1/256
    2. 1/32
    3. 1/16
    4. 1/8
    5. 1/4


3. Choose the correct statements about tetrad analysis.

    1. All of the products of a single meiosis can be studied at once in a tetrad.
    2. If nonparental ditypes equal parental ditypes, the genes are probably on the same chromosome.
    3. Tetrad analysis in animals is done by studying cases where more than one offspring is born at the same time.
    4. a and b
    5. a and c


4. Choose the correct ordering of the events in meiosis.


    1. homologous chromosomes separate
    2. chromosomes split at the centromere and sister chromatids separate.
    3. homologous chromosomes pair
    4. homologous chromosomes recombine

    1. (4), (3), (1), (2)
    2. (2), (3), (4), (1)
    3. (3), (4), (1), (2)
    4. (1), (2), (3), (4)
    5. (2), (4), (3), (1)


5. Choose the correct statement(s) about the packing of DNA in a chromosome.

    1. Histones are involved in the first level of folding, the nucleosome.
    2. Compaction is greatest at interphase of the cell cycle.
    3. Compaction makes it easier for chromosomes to be transmitted in mitosis and meiosis.
    4. a and b
    5. a and c


6. The ABO system in humans has three alleles. Choose the correct statements about this system.

    1. Two alleles are codominant, one is recessive
    2. Type O individuals can be described as "universal recipients" for transfusions.
    3. Antibodies against substances on the surface of the red blood cells of type A and B individuals are naturally present in individuals lacking those blood types.
    4. a and b
    5. a and c

7. Which of the following would represent "ideal" qualities of a research organism for genetic studies?

    1. Short generation time
    2. Large numbers of offspring per mating
    3. Inexpensive to raise
    4. b and c
    5. a, b and c


8. Choose the correct statement(s) about chromosome abnormalities in humans

    1. Chromosome abnormalities are rare among spontaneous abortions.
    2. The relatively high frequency of X chromosome abnormalities (such as XO and XXX) is thought to be related to the phenomenon of X chromosome inactivation.
    3. Trisomy for autosomes is more common than monosomy for autosomes among live births.
    4. b and c
    5. a, b and c


9. Which are the correct statements about the Chi-square test used for testing "goodness of fit" in genetic crosses?

    1. A high Chi-square value indicates a large deviation between observed and expected values under the model.
    2. Degrees of freedom corresponds to one more than the number of classes.
    3. If the Chi-square value could only be found less than one in twenty times due to chance (p (probability value less than .05)) the hypothesis is rejected.
    4. a and b
    5. a and c


10. A woman with normal vision who is heterozygous for colorblindness (X-linked recessive) is also heterozygous for another gene on the X chromosome (D), 10 map units from the colorblindness gene, which can be detected using DNA tests. The 76 allele of the D gene and the allele causing colorlindness both came to her from her father. She also is carrying the 53 allele of the D gene.

Assuming that her husband has normal vision and carries the 86 allele of the D gene, what proportion of all of their children will be colorblind boys carrying the 53 allele of the D gene?

    1. 0.025
    2. 0.05
    3. 0.10
    4. 0.45
    5. 0.475


10 Short Identification- 2 Pts. each

11. genotype- genetic makeup of an individual

12. testcross- cross to the homozygous recessive type

13. Hershey-Chase experiment- experiment in which the transfer of radioactive phosphorus (in DNA) rather than sulfur (in protein) during bacterial virus infection demonstrated that DNA was the genetic material.

14. allele- alternate form of a gene

15. telomere- tip of a chromsome, has special properties and a special DNA sequence.

16. heterochromatin- genetically inactive chromosomal material, darkly stained, often near centromeres and telomeres.

17. Klinefelter syndrome- XXY males, not fertile, higher frequency of mental retardation.

18. allopolyploid- polyploidy in an inter-species hybrid

19. satellite DNA- highly repeated DNA sequences associated with heterochromatin, often near centromeres.

20. incomplete penetrance- a given genotype may not always be expressed in the phenotype. Penetrance gives a percentage measure of the frequency with which it is expressed.


6 Problems (40 Points total) (Showing your work will make it easier for us to give partial credit).

21. (10 points) The following cross is performed in fruit flies: +++/+++ X abc/abc to give F1 progeny of the genotypes +++/abc. These F1s are crossed with abc/abc testers, giving the following cross:

+++/abc X abc/abc

The following testcross progeny are observed:



















  1. What is the map distance between the a and b genes?

    80+76+3+1 = 160/1000 = 16 cM

  2. What is the map distance between the b and c genes?

    104 + 92 + 3 + 1 = 200/ 1000 = 20 cM

  3. What is the frequency of double recombinants?

    4/1000 = .004

  4. What is the order of the a, b, and c genes on the chromosome?

    The b gene is in the middle, based on map distances and the class that are the double crossovers.

  5. What is the level of interference ( i )?

    I = 1-c = 1 - obs dco/ exp dco = 1 - 4/ (1000)(.2)(.16) = 1-4/32 =1-1/8 =7/8



Note: The table from #22 has been corrected. MSWord's automated capitalization feature messed up the genotypes

22. What is the map distance between the A, B and C genes in the cross below, and which are apparently located on the same chromosomes? The heterozyous parent arose from a cross of an AABBCC and an aabbcc individual. (10 points).

Parents AaBbCc X aabbcc

Progeny phenotypes

Numbers of progeny

















Map distances:
A-B = (44 +57 + 43 + 56 /1000)100 = 20 cM
B-C = (205 + 195 +44 + 56/1000)100 = 50 cM
A-C = (205 + 195 + 44 + 56/ 1000) 100 = 50 cM
A and B are apparently on the same chromosome, 20 CM apart. C assorts independently from those genes and may or may not be on the same chromosome as A and B.


23. The distance between genes based on recombination (genetic map distance) often does not correspond to the distance between genes based on the length of DNA between the genes on the chromosome. Explain why this is true. (5 points)

There seems to reduced recombination in some parts of chromosomes, such as heterochromatin near the centromeres and telomeres. This results in a small genetic map distance in those regions even when the physical distance may be large.


24. Given the following results from tests for a DNA marker (E) located at the centromere of chromosome 21 in the parents of a Down syndrome child and in the child herself, interpret which parent provided the extra chromosome and at which meiotic division the error occurred. (5 points).

Father has E 46 and E 63 alleles.

Mother has E 85 and E 32 alleles.

Down syndrome daughter has E 46, E 63 and E 32 alleles.

Because two of the alleles (46 and 63) came from the father, the nondisjunction must have occurred in the father. Because those two alleles are different from each other, the nondisjunction must have occurred in the first division of meiosis. Because the marker is at the centromere, it always shows first division segregation. Thus, both alleles will be passed to the same pole in meiosis I and will be included in the gametes after a normal meiosis II.


25. Briefly describe one DNA-based method for detecting polymorphisms that can be used for genetic mapping, paternity and criminal cases. (5 points)

3 Possible Answers:
RFLP- restriction fragment length polymorphism, detects variations based on variations at restriction enzyme cut sites.
STRP- simple tandem repeat polymorphism, detects variations based on varying numbers of simple tandem repeats (also termed microsatellites) which tend to vary among individuals.
SNP- single nucleotide polymorphism, detect single base changes which are present about once in every thousand bases.


26. White flower color is recessive to other flower colors in peas. Two strains of white-flower-colored pea plants are crossed and give purple-colored F1 hybrids. The F2 generation gives a 9:7 ratio of purple flowered to white flowered plants. Briefly explain this result. (5 points).

We are seeing complementation in the F1 generation indicating that the mutations are in two separate genes. Each is providing something that the other is biochemically lacking. The 9:7 ratio in the F2 confirms that two different genes showing independent assortment are involved. Dominant alleles must be present at both in order for the purple color to be present (9/16) Presence of only one dominant allele (6/16) or none (1/16) gives white flower color. This is a form of epistasis (gene interaction).