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Mendelian Genetics Practice Problems

Mendelian Genetics Practice Problems

For the problems listed below, you are to solve the type of inheritance and explain the rationale of your choice.

  1. Mice I

    • a) cross 1: red-eyed mouse X white-eyed mouse
      gives F1: all red-eyed

      cross 2: red-eyed F1 X red-eyed F1
      gives F2:
      36 red-eyed
      13 white-eyed

    • b) cross 1: long-eared mouse X short-eared mouse
      gives F1:
      12 long-eared
      10 short-eared

      cross 2: long-eared F1 X long-eared F1
      gives F2:
      36 long-eared
      13 short-eared

  2. Flowers

    cross 1: blue-flowered plant X white-flowered plant
    gives F1: all pale-blue-flowered

    cross 2: pale-blue F1 X pale-blue F1
    gives F2:
    27 blue
    49 pale-blue
    24 white

  3. Blood Type

    • a) cross 1: person, type A blood X person with type B
      gives F1: all type AB blood

      cross 2: type AB F1 X type AB F1
      gives F2:
      2 type A
      4 type AB
      1 type B

    • b) cross 1: type A blood X type B
      gives F1:
      2 type A blood
      3 type AB blood
      1 type B blood
      2 type O blood

  4. Mice II

    cross 1: tail-less mouse X normal mouse
    gives F1:
    10 tail-less
    9 normal

    cross 2: tail-less F1 X tail-less F1

    gives F2:
    10 normal
    21 tail-less
    9 dead

  5. Flies

    cross 1: red-eyed female X red-eyed male
    gives F1:
    50 red-eyed female
    25 red-eyed male
    25 white-eyed male

    cross 2: white-eyed male F1 X red-eyed female F1
    52 crosses give:
    30 red male
    33 red female

    48 crosses give:
    22 red male
    24 red female
    21 white male
    23 white female

  6. Pedigrees

  7. Mice III

    • a) cross 1: blue-eyed, long-toothed mouse X brown-eyed, short-toothed mouse
      gives F1: all blue-eyed, short-toothed

      cross 2: blue-eyed, short-toothed F1 X blue-eyed, short-toothed F1
      gives F2:
      92 blue-eyed short-toothed
      31 blue-eyed long-toothed
      29 brown-eyed short-toothed
      9 brown-eyed long-toothed

      (NOTE how this result is different from the blood-type result because here we are dealing with two genes and there we were dealing with multiple alleles of one gene.)

    • b) cross 1: tall, green plant X short, yellow plant
      gives F1:
      20 tall green
      20 short green

      cross 2: tall, green X short, yellow (different plants than (b))
      gives F1:
      19 tall green
      21 tall yellow

  8. Pathways I

    • a) assume the following pigment-producing pathway in a plant:
      	      enz A			enz B
      compound 1 ----------> compound 2 ------------> compound 3
      (colorless)		      (red)			(purple)
      Also assume that the A and B alleles produce functional enzyme, while the a and b alleles produce no functional enzyme. Assume that one functional copy of an enzyme is sufficient to catalyze the reaction.

      cross 1: AaBb X AaBb

      predict the colors of the progeny.

    • b) what if the pathway were:

                   enz B			enz A
      compound 1 ----------> compound 2 ------------> compound 3
      (red)		      	(colorless)			(purple)

    • c) Assume two parallel pigment pathways:
      		enz A				enz B
      red pigment <----------colorless compound --------------> blue pigment

      cross 1: Aabb X aaBb
      gives F1:
      33 white
      32 red
      35 blue

  9. Linkage:

    cross 1: rough, soft pea X smooth, hard pea

    gives F1: all rough, hard

    cross 2: rough, hard F1 X smooth, soft (note: different mating than usual)

    gives F2: 115 rough soft
    110 smooth hard
    8 rough hard
    12 smooth soft

  10. A pedigree that illustrates multiple possible models is:

  11. A man with a Y-linked disorder has three sons and three daughters by the same mother. His first son has two sons and two daughters by another woman. Draw the pedigree for this family. How could you tell if this disorder was Y linked or sex-linked recessive?

  12. Three genes in Drosophila with the recessive alleles l, m, and n are known to be linked. You wish to determine the map distances between them and perform the cross shown below. (Note: The gene order drawn here is arbitrary - you need to determine the order of these genes on the chromosome)

    Progeny with the following phenotypes are seen:

    Determine the order of these markers on the chromosome and the distances between them in centimorgans.

  13. This problem appeared on a previous 7.012 Exam.

    Alkaptonuria is an extremely rare disease. The gene for Alkaptonuria (ALK) has recently been shown to lie on human chromosome 9 and to be linked to the gene encoding the ABO blood group, with a recombination frequency of 11% between the loci.

    A pedigree of a family with the disease is shown below, with affected individuals indicated in black. In addition, the blood type of family members is given.

    The two alleles at the ALK locus will be denoted ALK+ and ALK-. The three alleles at the ABO blood group locus will be denoted IA, IB (which are co-dominant) and i (which is recessive to IA and IB).

    (a) What is the genotype of individual 1 at the ALK and ABO loci?

    (b) What is the genotype of individual 2 at the ALK and ABO loci?

    (c) What is the genotype of individual 3 at the ALK and ABO loci? Which alleles of each gene are carried on the chromosome he inherited from his father and which alleles are carried on the chromosome he inherited from his mother?

    (d) Individuals 3 and 4 are expecting their fifth child. A physician draws a prenatal blood sample and determines that the child has blood type B. What is the probability that the child will have alkaptonuria? Explain your answer.

You can check your answers on the Solutions Page.
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