Practical Review Problems

  1. Using the letter A, explain a simple monohybrid cross (always beginning with homozygous parents) and following 2 generations.

  2. Using the letters A and B, explain a dihybrid cross (always beginning with homozygous parents) and following 2 generations.

  3. Repeat the exercise above using the branching system.

  4. Explain the principle of independent assortment or second Mendel’s law.

  5.  Be familiar with these terms.

Alleles

chi-square test

cross

cross-fertilization

dihybrid cross

dominant

F1

F2 generation

first filial generation

 

genes

gene segregation

genotype

hereditary traits

heterozygous

homozygous

locus

monohybrid crosses

pedigree analysis

 

P generation

phenotype

principle of independent assortment

principle of segregation

Punnett square

pure-breeding

 

reciprocal crosses

recessive

second law

self-fertilization

selfing

trihybrid crosses

test cross

true-breeding

wild-type allele

 

Some help for problem solving.
Using probabilities to easily solve some problems. 
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1A couple in Indiana are both heterozygous for the autosomal recessive allele for albinism. They have two children. What is the probability that both children will be phenotypically identical with regard to skin color? 

A)        3/4      B)        1/16     C)        1/4       D)        9/16     E)        5/8

Solution:  Every time that a couple has a child is an independent event.  To make it easy to understand think about it!  Having a child in 2004 is completely independent from having a second child in 2006.  One event does not interfere with the second event, meiosis and fertilization for one event are independent from meiosis and fertilization for the second event.  In math independent events multiply.  This is explained in your Mendelian Genetics lab.

In problem 1 albinism is recessive so:  N=normal   n=albino

Parents: heterozygous: Nn

First Child

P            Nn x Nn

F          ¼  NN; ½ Nn and ¼ nn  and the phenotypes are  ¾ normal and ¼ albino

Second Child:

P            Nn x Nn

F          ¼  NN; ½ Nn and ¼ nn  and the phenotypes are  ¾ normal and ¼ albino

What is the probability that both children are normal?  Since having these two children are independent events and independent events multiply, then ¾ x ¾  = 9/16

What is the probability that both children will be albino, same rationale ¼ x ¼ = 1/16

But the probability that they both will be phenotypically identical (both normal or both albinos) is the combination (addition) of both independent events

9/16 + 1/16 = 10/16 or 5/8!

 

 

Another kind of problems

2.  In the cross between a female AaBbccDdee  and male AabbCcDdee, what proportion of the progeny will be phenotypically identical to the female parent? (Assume independent assortment of all genes and complete dominance.)

A)        1/16    B)        3/16     C)        3/8       D)        9/16     E)        9/64

Because the inheritance of each gene is independent, to solve this problem you can analyze each gene independently and determine the phenotypic proportion that look like the female parent AaBbccDdee.

1.      Aa x Aa = ¼ AA; ½ Aa and ¼ aa… ..it means that ¾ look like the mother Aa

2.      Bb x bb = ½ Bb ; ½ bb…………… .. .it means that ½  look like the mother Bb

3.      cc x Cc = ½ Cc ; ½ cc………………..it means that ½  look like the mother cc

4.      Dd x Dd= ¼ DD½ Dd and ¼ dd…….it means that ¾ look like the mother Dd

5.      ee x ee = all ee = 1

Phenotypic proportion that look like the mother

¾ x ½ x ½ x ¾ x 1 = 9/64

3.       If a plant of genotype Aa;Bb;Cc;Dd is selfed and the genes assort independently, how many different genotypes will be found among the progeny?

A)        4         B)        16        C)        24        D)        64        E)        81

 Using the same method that we used above

  1. Aa x Aa = ¼ AA; ½ Aa and ¼ aa = 3 different genotypes
  2. Bb x Bb = ¼ BB; ½ Bb and ¼ bb = 3 different genotypes
  3. Cc x Cc = ¼ CC, ½ Cc ; ¼  cc = 3 different genotypes
  4. Dd x Dd= ¼ DD½ Dd and ¼ dd = 3 different genotypes

 Therefore all independent events combined  = 3x3x3x3 = 81

4.       A couple are both heterozygous for two autosomal recessive diseases: cystic fibrosis (CF) and phenylketonuria (PKU). What is the probability that their first child will have either CF or PKU?

A)        0         B)        1/4       C)        1/2       D)        1/16     E)        9/16

PKU and cystic fibrosis are two independent diseases!

Parents are heterozygous (carriers) for CF and for PKU

CF = normal                                  P = PKU normal (no PKU)

cf = cystic fibrosis                         p = disease with PKU

Parents   Cf cf  x Cf cf

F1  ¼  Cf Cf    ½  Cf cf and ¼ cfcf    phenotypes ¾ normal and ¼ disease with CF

Parents Pp x Pp

F1  ¼ PP   ½ Pp    ¼ pp    phenotypes are ¾ normal and ¼ disease with PKU 

Since you are combining CF or PKU then the probably of one or the other is the addition of the separate probabilities  ¼ for CF + ¼ for PKU = ½

 

 

 Selected book problems. 

12.21   (modified from book).  Depict each of the crosses that follow, first using Mendelian or using Drosophila notation, give the genotype and phenotype of the F1 progeny that can be produced.

a.  In humans, a mating between two individuals, each heterozygous for the recessive trait phenylketonuria, whose locus is on chromosome 12.

            b. In humans, a mating between a female heterozygous for both phenylketonuria and X-linked color blindness and a male with normal color vision
                 who is heterozygous for phenylketonuria.

 Answer:

a.                                  Allele symbols:            P= normal       P= phenylketonuria                

Cross: Pp ×Pp
 
F1 Genotypes :1 PP:2 Pp:1 pp
 
F1 phenotypes: 3⁄4 normal, 1⁄4 phenylketonuria

b.
Allele symbols: C =normal c = color blind Cross:

Cc Pp ×CY Pp

F1 Genotypes:
1⁄16 CCPP                  1⁄8 CC Pp                  1⁄16 CC pp                  1⁄16 CcPP       1⁄8 CcPp

1⁄16 Cc pp                   1⁄16 CY PP                1⁄8 CY Pp                  1⁄16 CY pp      1⁄16 cY PP

1⁄8 cY Pp                    1⁄16 cY pp

 

F1 phenotypes:

3⁄8 normal females

1⁄16 phenylketonuria males

 

1⁄8 phenylketonuria females

3⁄16 color-blind males

 

3⁄16 normal males

1⁄16 color-blind phenylketonuria males

 

12.22   In Drosophila, white eyes are a sex-linked character. The mutant allele for white eyes (w) is recessive to the wild-type allele for brick-red eye color (w+). A white-eyed female is crossed with a red-eyed male. An F1 female from this cross is mated with her father, and an F1 male is mated with his mother. What will be the eye color of the offspring of these last two crosses?

Answer: The initial cross is ww´w+Y, so that the F1 females are ww+ and the F1 males are wY. The second set of crosses are therefore w+w´w+Y and www. The former will give all brick-red females (w+) and half white (wY) and half brick-red (w+Y) males. The latter will give only white-eyed males and females (wY and ww).

12.23   One form of color blindness in humans is caused by a sex-linked recessive mutant gene (c). A woman with normal color vision (c+) and whose father was color-blind marries a man of normal vision whose father was also color-blind. What proportion of their offspring will be color-blind? (Give your answer separately for males and females.)

Answer: Since fathers always give their X chromosome to their daughters, the woman must be heterozygous for the color-blind trait and is c+c. As her husband received his X chromosome from his mother and has normal color vision, he is c+Y. The cross is therefore c+c´c+Y. All daughters will receive the paternal X bearing the c+ allele and have normal color vision. As sons will receive the maternal X half will be cY and be color-blind and half will be c+Y and have normal color vision.

12.24  

In humans, red-green color blindness is recessive and X-linked, whereas albinism is recessive and autosomal. What types of children can be produced as the result of marriages between two homozygous parents—a normal-visioned albino woman and a color-blind, normally pigmented man?

Answer: Let c and c+ be the color-blind and normal vision alleles, respectively, and let a and a+ be the albino and normal pigmentation alleles, respectively. Then the cross can be represented as c+c+ aa´cY a+a+. As all the offspring will be a+a, all will have normal pigmentation. The offspring will be either c+c or c+Y and have normal color vision. The daughters will, however, be carriers for the color-blind trait.

12.25   In Drosophila, vestigial (partially formed) wings (vg) are recessive to normal long wings (vg+), and the gene for this trait is autosomal. The gene for the white-eye trait is on the X chromosome. Suppose a homozygous white-eyed, long-winged female fly is crossed with a homozygous red-eyed, vestigial-winged male.

a.  What will be the genotypes and phenotypes of the F1 flies?

b.  What will be the genotypes and phenotypes of the F2 flies?

c.  What will be the genotypes and phenotypes of the offspring of a cross of the F1 flies back to each parent?

Answer:

a.  The initial cross is ww vg+vg+´w+Y vgvg. The F1 consists of wY vg+vg (white, normal-winged) males and ww+ vg+vg (red, normal-winged) females.

b.  The F2 would be produced by crossing wY vg+vg males and w+w vg+vg females. In both the male and the female progeny, 18 will be white and vestigial, 18 will be red and vestigial, 38 will be white and normal winged, and 38 will be red and normal winged.

c.  If the F1 males are crossed back to the female parent, the cross is wY vg+vg´ww vg+vg+. All the progeny would be white and normal winged. If the F1 females are crossed back to the male parent, the cross is ww+ vg+vg´w+Y vgvg. Among the male progeny, there would be 1⁄4 white, vestigial; 1⁄4 red, vestigial; 1⁄4 white, normal winged; and 1⁄4 red, normal winged. Among the female progeny, half would be red and normal winged and half would be red and vestigial.

 

12.26   In Drosophila, two red-eyed, long-winged flies are bred together and produce the offspring listed in the following table:

                                                                                            Females                       Males

                      red-eyed, long-winged                                      3⁄4                              3⁄8

                      red-eyed, vestigial-winged                                1⁄4                              1⁄8

                      white-eyed, long-winged                                  —                               3⁄8

                      white-eyed, vestigial-winged                            —                               1⁄8

            What are the genotypes of the parents?

Answer: From problem 12.25, we know that w is X-linked while vg is autosomal. This can also be determined by considering just one trait at a time and examining the frequency of progeny phenotypes. The ratio of long-winged to vestigial-winged progeny is 3:1 (3⁄4 to 1⁄4) in both sexes, while the ratio of red-eyed to white-eyed progeny is all to none in females and 1:1 in males. This is consistent with vg being autosomal and w being X‑linked. The 3:1 ratio of long-winged to vestigial-winged progeny indicates that each parent was heterozygous at the vg locus. Since both parents had red eyes, both had (at least) one w allele. Since half of the sons are white eyed, the mother must have been heterozygous. Therefore, the parents were ww vgvg and wY vgvg.

12.27   In chickens, a dominant sex-linked gene (B) produces barred feathers, and the recessive allele (b), when homozygous, produces nonbarred (solid-color) feathers. Suppose a nonbarred cock is crossed with a barred hen.

a.  What will be the appearance of the F1 birds?

b.  If an F1 female is mated with her father, what will be the appearance of the offspring?

c.  If an F1 male is mated with his mother, what will be the appearance of the offspring?

Answer:

a.  In poultry, sex type is determined by a ZZ (male) and ZW (female) system. The cross can be depicted as bb (nonbarred cock) x BW (barred hen). The F1 progeny will be bW (nonbarred) hens and Bb (barred) cocks.

b.  The cross can be represented as bW x bb. All the progeny will be nonbarred.

c.  The cross can be represented as Bb x BW. The progeny will be 1⁄2 barred cocks (1⁄4 BB, 1⁄4 Bb), 1⁄4 barred hens (BW), and 1⁄4 nonbarred hens (bW).

12.28   A man (A) suffering from defective tooth enamel, which results in brown-colored teeth, marries a normal woman. All their daughters have brown teeth, but the sons are normal. The sons of man A marry normal women, and all their children are normal. The daughters of man A marry normal men, and 50 percent of their children have brown teeth. Explain these facts genetically.

Answer: Notice that the trait is transmitted from the father to his daughters, indicating crisscross inheritance. This is typical of an X-linked trait. Since the man marries a normal woman and all of their daughters have the trait, the trait must be dominant. The man’s X chromosome bearing the defective tooth enamel allele is inherited by all of his daughters and none of his sons. All of his daughters would therefore have defective tooth enamel and be heterozygous for the defective enamel allele. These daughters would transmit the defective enamel allele half of the time, giving rise to 50 percent of their children being affected.

12.29   In humans, differences in the ability to taste phenylthiourea are due to a pair of autosomal alleles. Inability to taste is recessive to ability to taste. A child who is a nontaster is born to a couple who can both taste the substance. What is the probability that their next child will be a taster?

Answer: Since the inability to taste the substance is recessive, the nontaster child must be homozygous for the recessive allele, and each of his parents must have given the child a recessive allele. Since both parents can taste, they must also bear a dominant allele. Let T represent the dominant (taster) allele, and t represent the recessive (nontaster) allele [Note: Mendelian notation is used here for convenience, but also because there is no value in assigning a normal (+) and abnormal allele.] Then the cross can be written as Tt ´Tt. The chance that their next child will be a taster is the chance that the child will be TT or Tt, or 3⁄4. 

 

12.35   An individual with Turner syndrome would be expected to have how many Barr bodies in the majority of cells?

Answer: None. Turner syndrome individuals are XO. They have only one X, so no X is inactivated.

12.36   An XXY individual with Klinefelter syndrome would be expected to have how many Barr bodies in the majority of cells?

Answer: All but one X chromosome is inactivated. An XXY individual, having two X chromosomes, has one inactivated.

12.44   Which of the following statements is not true for a disease that is inherited as a rare X‑linked dominant trait?

a.  All daughters of an affected male will inherit the disease.

b.  Sons will inherit the disease only if their mothers have it.

c.  Both affected males and affected females will pass the trait to half the children.

d.  Daughters will inherit the disease only if their fathers have it.

Answer: The only untrue statement is (d). Since daughters receive an X chromosome from each of their parents, they can inherit an X-linked dominant disease from either their mother or father.

12.45   Women who were known to be carriers of the X-linked recessive hemophilia gene were studied to determine the amount of time required for blood to clot. It was found that the time required for clotting was extremely variable from individual to individual. The values obtained ranged from normal clotting time at one extreme to clinical hemophilia at the other. What is the most probable explanation for these findings?

Answer: Since hemophilia is an X-linked trait, the most likely explanation is that random inactivation of X chromosomes (lyonization) produces individuals with different proportions of cells with a functioning allele. Normal clotting times would be expected in females with a functional h+ allele, i.e., females whose h-bearing X chromosome was very frequently inactivated. Clinical hemophilia would be expected in females without a functional h+ allele, i.e., females whose h+-bearing X chromosome was very frequently inactivated. Intermediate clotting times would be expected to be proportional to the amount of h+ function, which is related to the frequency of inactivation of the h+-bearing X chromosome.

12.46   Hurler syndrome is a genetically transmitted disorder of mucopolysaccharide metabolism resulting in short stature, mental retardation, and various bony malformations. Two specific types are described with extensive pedigrees in the medical genetics literature:

Type I: recessive autosomal

Type II: recessive X linked

You are a consultant in a hospital ward with several patients with Hurler syndrome who have asked you for advice about their relatives’ offspring. Being aware that both types are extremely rare and that afflicted individuals almost never reproduce, what counsel would you give to a woman with Type I Hurler syndrome, whose normal brother’s daughter is planning marriage, about the offspring of that proposed marriage? In your answer, state the probabilities that the offspring will be affected and whether male and female offspring have an equal probability of being affected.

Answer: Draw out the pedigree of the patient and try to assign genotypes to the relevant individuals. Let h represent the Hurler syndrome Type I allele. Then the patient is hh. Since we are told that Hurler syndrome patients virtually never reproduce, neither of her parents is expected to be hh. Still, in order for the patient to be hh, her parents must have both been Hh. (The cross was Hh´Hh.) Since her brother is normal, he is H–, with a 2⁄3 chance of being Hh. The brother’s daughter has a half chance of receiving either of the brother’s (H or h) alleles. Thus, the chance that the brother’s daughter has an h allele is 2⁄3 ´1⁄2=1⁄3. Since the trait is extremely rare, it is likely she will marry an HH individual, and have H– children. Therefore there is no chance the brother’s daughter will have affected children, as her husband most likely will provide a dominant, normal H allele. Since Type I is autosomal, there will be no sex differences.