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ADVANCED PLACEMENT BIOLOGY

GENETIC PROBLEMS

 

 

1.     Prove Mendel's Law of Segregation

 

2.     Prove Mendel's Law of Independent Assortment.

 

3.     Suppose you learned that "shmoos" may have long, oval, or round bodies and that matings of "shmoos" gave the following offspring:

      long  X  oval           52 long : 48 oval

      long  X  round         99 oval

      oval  X  round         51 oval : 50 round

      oval  X  oval           24 long : 53 oval : 27 round

What hypothesis about the inheritance of "shmoo" shape would be consistent with these results?

 

4.     Color‑blindness is a sex‑linked trait in humans.

(a)    A normal woman whose father was color‑blind marries a normal man. What will be the proportion of color‑blind children?

(b)    How could you ever get a color‑blind woman?

(c)    What name is given to such a person as the female referenced in problem (b) above?

 

5.     Probability and Genetic Events

(a)    What are the chances that a family with 4 children will have 2 boys and 2 girls?

(b)    What are the chances that a family with 4 children will have a boy, a girl, a boy, a girl; in that order.

(c)    What are the chances that a family with 4 children will have at least 3 boys?

 

6.     By backcrossing the following dihybrids, it is shown that each produces recombinant gametes in the percentages indicated:

        AaBb → 31%          XxBb → 5%          AaYy → 14%

        XxYy → 22%          AaXx → 36%

From this data determine the linear sequence and relative spacing of each gene locus on the chromosome and draw the proper chromosome map.

 

In 1790, several British mutineers from the HMS Bounty, together with a few men and women native to Tahiti, established a settlement on Pitcairn Island. This new island population remained largely isolated for several generations. The preceding statements represent historical fact; the following are, in part, deviations from the true history of Pitcairn, as far as it is known, for the purposes of these problems. Assume that:

a.               Families were founded by six men from the Bounty, of whom three were blue‑eyed, two brown‑eyed (but heterozygous for blue) and one homozygous brown‑eyed; and that there were two Tahitian men and eight women, all homozygous for brown.

b.              The blue‑eyed, brown alternative depends on a single autosomal allelic pair with brown (B) dominant and no modifiers of eye color capable of obscuring this alternative were segregated in this population.

c.               Intermarriage among the residents of Pitcairn Island was at random with respect to eye color, with approximately equal numbers of descendant surviving from each type of marriage.

 

7.     Gene frequencies in this population:

(a)    What were the gene frequencies for the eye color alleles among the six white men, two Tahitian men, and eight Tahitian women; who, we have assumed, established families on Pitcairn Island?

(b)    Was this population at equilibrium in this regard at the time of its establishment?

(c)    What genotypic and phenotypic proportions would you expect to prevail in this population after it had reached equilibrium?


8.     The following table illustrates the predicted results of the first generation of marriages on Pitcairn Island. Fill in the blank spaces (indicated by question marks <?>) on the table, then answer the questions below.

 

     Type of marriage      Number of     Number of progeny*

      female X male         marriages           BB     Bb     bb

 

      BB X BB                       3                    6        0        0

      BB X Bb                        2                    2        2        0

      BB X bb                       <?>               <?>    <?>    <?>

                                           ‑‑‑‑‑               ‑‑‑‑‑   ‑‑‑‑‑   ‑‑‑‑‑

                     TOTALS         8                   8      <?>    <?>

*Assuming two children per marriage.

 

(a)    Is the population at equilibrium after one generation? Explain.

(b)    Is this inconsistent with the statement that equilibrium is achieved in one generation of random matings?  Why?

 

9.     In fowl, genotype rrpp gives single‑comb; R–P– gives walnut‑comb; rrP– gives pea‑comb; and R–pp gives rose‑comb.

(a)    What comb types will appear, and in what proportions, in F1 and F2, if single‑combed birds are crossed with birds of a true‑breeding walnut‑combed strain?

(b)    What are the genotypes of the parents in a walnut X rose mating from which the progeny are 3 rose: 3 walnut: 1 pea: 1 single?

 

In man, Hemophilia is the "bleeders disease", a disease in which the time required for blood to clot is greatly prolonged; depends on the recessive allele of a sex‑linked gene. There are at present about 40,000 cases of the disease in the United States. In the following questions, let h+ be the allele for normal clotting time; h the allele for hemophilia. (Note: In all problems involving sex linkage, list the phenotypes of sons and daughters separately.)

 

10.   A man whose father was hemophilic, but whose own blood‑clotting time is normal, marries a normal woman with no record of hemophilia in her ancestry. What is the chance of hemophilia in their children?

 

11.   A woman whose father was hemophilic, but who is not herself a "bleeder", marries a normal man.

(a)    What is the chance of hemophilia in their children?

(b)    If these parents have normal son, what is the probability that their next son will also be normal?

(c)    If this family has two sons, what is the probability that they will both be normal?

(d)    If this family has six sons, what is the probability that three will be normal and three hemophilic?

(e)    This family is expecting a child. What is the probability that it will be normal? That it will be hemophilic?

 

12.   What is the chance of hemophilia among the sons of a daughter of the marriage in question 11, if she marries a normal man?

 

13.   Why, in all of these hemophilia problems, were there no hemophilic females considered?


In mice, there is a set of multiple alleles of the gene for albinism. Four of these alleles, listed in order of decreasing amount of color in the hair of homozygotes, are: C = full color (wild type); cch = chinchilla; cd = extreme dilution; and c = albino. This gene is not sex‑linked and you may assume that each allele is dominant to those below it in the list.

 

14.   Diagram a cross between a wild‑type mouse heterozygous for extreme dilution and a chinchilla mouse homozygous for albinism.

 

15.   Another nonsex‑linked gene affecting hair color in mice has alleles B– for black hair (wild type) and bb for brown hair (recessive type). Expand your diagram for the preceding question to include the information that both of the mice crossed are heterozygous for brown (Bb).

 

16.   Suppose that after the crosses indicated in the following Drosophilia; the following were obtained in F2:  

 

            wild                  292

            curled                    9

            curled, spineless    92

            spineless                            7

                                          -------

                         TOTAL            400

Curled wings (cu) and spineless bristles (ss) are autosomal characters in Drosophilia. The genes giving rise to these characters are both in chromosome #3.

(a)    Starting with a wild‑type female and a curled, spineless male; prepare a diagram showing parents and progenies through an F2 generation.

(b)    Remembering again the special circumstance of no crossing‑over in male Drosophilia, is there some reasonably straight‑forward way to arrive at an estimate of the map distance between cu and ss?  Explain your method and calculate the distance.

 

17.   In cats, black is due to a pair of genes BB. The recessive is bb; a yellow cat. The coat color, tortoise‑shell is Bb. This is a sex-linked trait in cats.

(a)    A female yellow cat is bred to a black male. What will be the ratio of kittens?

(b)    What is the expected ratio of a cross between a tortoise-shell female and a black male?

(c)    Occasionally a sterile tortoise‑shell male cat is found. Explain.

 

18.   Purebred Holstein‑Friesian cattle are black and white. A recessive allele that, when homozygous, results in red and white is present but rare in this breed. Red and white calves are barred from registration, and therefore it is economically important to avoid using for breeding purposes black and white individuals that carry the undesirable recessive allele hidden in the heterozygous condition. How might you detect such heterozygosity in a bull to be used extensively in artificial insemination?

 

19.   For some reason (not yet fully understood) more male babies are conceived than female. By childhood, the number of boys and girls is the same. Can you think of a possible explanation for the higher mortality (both before birth and after) of male babies?

 

Thalassemia is a type of human anemia rather common in Mediterranean populations, but relatively rare in other peoples. The disease occurs in two form, called minor and major, the latter is much more severe. Severely affected individuals are homozygous for an aberrant gene; mildly affected persons are heterozygous. Persons free of this disease are homozygous for the normal allele.

 

20.   A man with Thalassemia minor marries a normal woman. With respect to Thalassemia, what types of children and in what proportions, may they expect?

21.   Both father and mother in a particular family have Thalassemia minor. What is the chance that their baby will be:  (a)  severely affected?

(b) mildly affected?

(c) normal?

 

22.   An infant has Thalassemia major.

(a)    From this information, what possibilities might you expect to find  if you checked the infant’s parents for anemia?

(b)    Thalassemia major is usually fatal in childhood. How does this fact modify your answer to the question above?

 

23.   The following questions refer to general genetic situations, in which the particular phenotypes are not specified.

(a)    A particular cross gives in F2 a modified dihybrid ration of 9:7. What phenotypic ratio would your expect in a testcross of the F1’s?

(b)    What phenotypic ratio would you expect from the testcross of an F1 giving a 13:3 ratio in F2 ?

(c)    A 9:3:3:1 ratio?

(d)    Indicate the basis of each of the two dihybrid ratios listed in questions (a), (b) and (c) above.

 

24.   In cattle, RR individuals have a red coat; rr a white coat; and the heterozygous are roan. The polled characteristic is determined by a dominant gene (P), the horned alternative by genotype (pp).

(a)    What will be the phenotype of F1 individuals after mating of a white, horned animal with a red that is homozygous for polled? What phenotypes will be found in F2, and in what proportions?

(b)    If the F1 individuals in the question above (a) are backcrossed with white, horned animals; what phenotypes will be produced and in what proportions?

 

25.   What kind of gametes does an individual of genotype Ww produce?

 

26.   What gametes does an individual of genotype WwXx produce?

 

27.   In four-o’clocks red is incompletely dominant to white, and bush is dominant to lazy (a creeping vine characteristic). What is the genotype ratio of an F1 X F1 cross?

 

28.   What is the genotypic ratio of a cross between AabbCcDd X AabbccDD?

 

29.   What is the genotypic ratio of a cross between TtSsRr X ttssrr?  What is such a cross called?

 

30.   What is the genotype of the progeny from the following cross: AaBbCc X AaBbCc?

 

31.   Assume that the frequencies of the blood group genes for O, A, and B are 0.65, 0.25, and 0.10 respectively in computing the frequency of individuals with the various types of blood.

(a)    Calculate the frequency of the A, B, AB, and O blood types in the human population.

(b)    Henry is blood type O, and so is his mother. What blood types may his father have?

(c)    What are the possible blood types of children resulting from the marriage of a type O father with a type AB mother?

 

32.   If the average proportion of heterozygotes in a population is ¼, what would it be after 3 generations of inbreeding?

 

33.   Give the gene content of the different eggs produced by a woman whose genotype is JjKkLl.

 

 

34.   In the fruit fly, sepia-eye is recessive to red-eye, and curved-wing is recessive to straight-wing.

(a)    If a true breeding sepia-eyed, straight-winged fly is mated with a true breeding red-eyed, curved-winged fly; what phenotypes will appear in the F1 generation?

(b)    If the two F1 flies are allowed to mate, what phenotypes will occur in the F2 generation and in what ratio?

(c)    How many different genotypes will occur in the F2 generation described in question (a) above?

 

35.   When plant A (a pure breeding tall) is crossed with plant B (a pure breeding short), the offspring are all intermediate in height. When two of the offspring are crossed, 1/16th of the next generation is as tall as the tall grandparent. How do you account for this phenomenon?

 

36.   Neglecting crossing-over, how many different kinds of ascospores can Neurospora crassa form by random assortment of its chromosomes, if seven is the haploid number?

 

37.   What sequence of bases in messenger RNA will be coded by the following triplets in the DNA molecule: GCTCATCCAAAAAGT?

 

38.   It is possible to buy sorghum seeds ¾ of which germinate into green seedlings and ¼ of which germinate into albino seedlings. The albino seedlings soon die.

(a)    Why?

(b)    If A = green and a = albino, what is the genotype of the seeds?

 

39.   Nondisjunction may lead to zygotes containing XXY.

(a)    What sex would this produce in a fruit fly?

(b)    In a human?

 

In garden peas the following genetic conditions prevail: The gene for tallness is dominant over the gene for dwarfism. Smooth is dominant to wrinkled. Yellow pod color is dominant to green.

 

40.   Cross a heterozygous smooth pea with a heterozygous wrinkled pea.

 

41.   A cross between a tall pea and a dwarf pea produced 86 tall plants and 81 dwarf plants.

(a)    What is the probable genotype of the tall plant?

(b)    What is the name for this type of cross?

 

42.   What will be the phenotypes of the offspring of:

(a)    homozygous yellow X green?

(b)    heterozygous yellow X green?

(c)    heterozygous yellow X homozygous yellow?

(d)    heterozygous yellow X heterozygous yellow?

 

43.   If two animals heterozygous for a single pair of genes are mated and have 200 offspring, about how many will have the dominant phenotype?

 

44.   Assume brown-eye dominant in humans over blue-eye. Remember that color-blindness is a sex-linked trait in humans. A blue-eyed woman whose mother was color-blind marries a brown-eyed man whose mother was blue-eyed. What are the expected ratios of their children?

 

45.   The Rh factor is carried by a pair of genes designated as DD for Rh+ and dd for Rh-. A woman who is Rh- marries a man who is Rh+ and whose parents were both Rh+. What is the possible genotypes of their children and their Rh blood types?

 

46.   Albinism in humans is inherited as a simple recessive trait. For the following families, determine the genotypes of the parents and offspring. When two alternative genotypes are possible, list both.

a.      Two non-albino (normal) parents have five children, four normal and one albino.

b.     A normal male and an albino female have six children, all albino.

 

47.   Determine the genotypes of the F2 plants given here by analyzing the phenotypes of the offspring of these crosses.

 

F2 Plants

Offspring

(a) round, yellow X round, yellow

¾ round, yellow

 

¼ wrinkled, yellow

 

 

(b) wrinkled, yellow X round, yellow

6/16 wrinkled, yellow

 

2/16 wrinkled, green

 

6/16 round, yellow

 

2/16 round, green

Is either of the crosses above (a or b) a test cross?

 

48.   In Drosophila melanogaster, gray body color is dominant to ebony body color, while long wings are dominant to vestigial wings. Work the following crosses through the F2 generation and determine the genotypic and phenotypic ratios for each generation. Assume the P1 individuals are homozygous.

a.      gray, long X ebony, vestigial

b.     gray, vestigial X ebony, long

c.      gray, long X gray, vestigial

 

49.   In one of Mendel’s dihybrid crosses, he observed 315 smooth, yellow; 108 smooth, green; 101 wrinkled, yellow; and 32 wrinkled, green F2 plants. Analyze these data using the chi-square test to see if:

a.      they fit a 9:3:3:1 ratio

b.     the smooth:wrinkled traits fit a 3:1 ratio

c.      the yellow:green traits fit a 3:1 ratio

 

50.   A geneticist, in assessing data that fell into two phenotypic classes, observed values of 250:150. She decided to perform chi-square analysis using two different null hypothesis: (a) the data fit a 3:1 ratio; and (b) the data fit a 1:1 ratio.

a.      Calculate the chi-square values for each hypothesis.

b.     What can be concluded about each hypothesis?

 

 

 

SOURCES FOR PROBLEMS:

Elementary Genetics, Singleton                                                  Genetics, Winchester

Essential Genetics, Klug & Cummings                                       Molecular Genetics, Horowitz

Human Genetics, Lewis                                                            Population Genetics, C.C. Li

General Genetics, Srb & Owen