Genetics Problems Review

2010

Monohybrid CrossIllustrated

True-breedinghomozygous recessiveparent plant

True-breedinghomozygous dominantparent plant

An F1 plantself-fertilizesand producesgametes:

F1 PHENOTYPES

F2 PHENOTYPES

aa

Aa

AA

aaAa

Aa

Aa Aa

Aa Aa

Aa Aa

Aa Aa

Aa

Aa

AA

aa

A

A

A

A

a a

a

a

AA

Figure 11.7Page 181

Monohybrid #1• In pea plants, spherical seeds (S) are dominant to dented seeds (s).

In a genetic cross of two plants that are heterozygous for the seed shape trait, what fraction of the offspring should have spherical seeds?

• We know: Ss x Ss• A. None • B. 1/4 • C. 1/2 • D. 3/4 • E. All

Monohybrid #1 Solution

• We know: Ss x Ss• One fourth of the offspring will be homozygous dominant

(SS), one half will be heterozygous (Ss), and one fourth will be homozygous recessive (ss).

• A. None • B. 1/4 • C. 1/2 • D. 3/4 • E. All

Monohybrid #2You try it!

• Long hair is dominant to short hair in guinea pigs. Show the results of a cross between a homozygous short haired male and a pure long haired female.

Monohybrid #2You try it! (Solution)

• Long hair is dominant to short hair in guinea pigs. Show the results of a cross between a homozygous short haired male and a pure long haired female.

• • L = long hair and dominant. l is recessive• • ll x LL gametes: l and L• • All offspring (F1’s) are Ll They are all long haired.

Monohybrid #3 (You try it!)

• An albino (lack of pigmentation) man marries a normally pigmented woman who had an albino mother. Show the types of children that this couple may have and the proportions of each.

Monhybrid #3 (Solution)

• An albino (lack of pigmentation) man marries a normally pigmented woman who had an albino mother. Show the types of children that this couple may have and the proportions of each. We know:

• aa = >albinism and AA or Aa => normal pigmentation• Albino father x carrier mother (inherited one a gene

from her mother)• aa x Aa => 1aa: 1Aa in offspring. So the changes of

having an albino child is 50% and normal pigmented child 50%

Incomplete Dominance

XHomozygous parent

Homozygous parent

All F1 are heterozygous

X

F2 shows three phenotypes in 1:2:1 ratio

Incomplete Dominance

Figure 11.10Page 184

Monohybrid #4 Incomplete Dominance

• Suppose a red flower and a white flower was cross-pollinated. The flowers from the resulting seeds were neither red nor white but pink. Explain.

• This is likely due to incomplete dominance. In some flowers, such as snapdragons, the heterozygous state is phenotypically an intermediate color.

Monohybrid #5 Incomplete Dominance (You try it!)

• Cross the offspring from the preceding problem. Give the phenotypic and genotypic ratios.

• Hint: Red x White = Pink• Red = RR and White rr => Pink R?

Monohybrid #5 Incomplete Dominance (Solution)

• Cross the offspring from the preceding problem. Give the phenotypic and genotypic ratios.

• Hint: Red x White = Pink• Red = RR and White rr => Pink R?• The cross: Rr x Rr => RR 2Rr rr

• Solution: The cross’ phenotypic ratio is 1:2:1 for red: pink: white. Genotypically 1 RR: 2 Rr : 1rr where RR = red and rr = white and pink = Rr

Monohybrid #6 Codominance

• Background:• Human blood type is determined by co-dominant

alleles. There are three different alleles, known as IA, IB, and i. The IA and IB alleles are co-dominant, and the i allele is recessive.

• The possible human phenotypes for blood group are type A, type B, type AB, and type O. Type A and B individuals can be either homozygous (IAIA or IBIB, respectively), or heterozygous (IAi or IBi, respectively).

ABO Blood Type:Allele Combinations

Range of genotypes:

Blood types:

IA IA

IA i IA IB IB i

IB IB

ii

A AB B O

or or

Figure 11.11Page 184

Monohybrid #6 Codominance

• A woman with type A blood and a man with type B blood could potentially have offspring with which of the following blood types?

• A. type A • B. type B • C. type AB • D. type O • E. all of the above

Monohybrid #6 Codominance (Solution)

• A woman with type A blood and a man with type B blood could potentially have offspring with which of the following blood types?

• A. type A • B. type B • C. type AB • D. type O • E. all of the above

• But if the man was type O rather than type B, offspring of type B and type AB would not be possible.

Monohybrid #7 (You try it!)

• A couple, both type O have a child who is type AB. Is this possible? Explain your answer.

Monhybrid #7 (Solution)

• A couple, both type O produce a child who is typed AB. Is this possible? Explain your answer.

• No this is not possible since the genotype of both parents is ii. Neither carry the A or B gene.

• It’s more likely that a mistake was made at the hospital such as assigning the wrong child to the couple, or there was a problem with the actual blood typing results for the child.

Dihybrid Cross: F1 Results

AABB aabbx

AaBb

AB AB ab ab

TRUE-BREEDING PARENTS:

GAMETES:

F1 HYBRID OFFSPRING:

purple flowers, tall

white flowers,dwarf

All purple-flowered, tall

Figure 11.9 (1)Page 183

1/16aaBB

1/16aaBb

1/16aaBb

1/16Aabb

1/16Aabb

1/16AAbb

1/16AABB

1/16AABb

1/16AaBB

1/16AaBb

1/16AABb

1/16AaBb

1/16AaBB

1/16AaBb

1/16AaBb

1/4 AB 1/4 Ab 1/4 aB 1/4 ab

1/16aabb

1/4 AB

1/4 Ab

1/4 aB

1/4 ab

AaBb AaBbX

1/16 white-flowered, dwarf

3/16 white-flowered, tall

3/16 purple-flowered, dwarf

9/16 purple-flowered, tall

Dihybrid Cross: F2 Results

Figure 11.9(2)Page 183

Dihybrid Problem #1

• Which of the following genetic crosses would be predicted to give a phenotypic ratio of 9:3:3:1?

• A. SSYY x ssyy • B. SsYY x SSYy• C. SsYy x SsYy• D. None of these

Dihybrid Problem #1 (Solution)

• Which of the following genetic crosses would be predicted to give a phenotypic ratio of 9:3:3:1?

• A. SSYY x ssyy (Wrong)• All offspring would have genotype of SsYy and a phenotype

dominant for both traits (spherical, yellow-seeded plants.) • B. SsYY x SSYy (Wrong)• All offspring would have genotype of either SSYY, SSYy,

SsYY, or SsYy, and a phenotype dominant for both traits (smooth, yellow-seeded plants.)

• C. SsYy x SsYy (Correct)

Dihybrid #2

• Background• In mice, the ability to run normally is a

dominant trait. Mice with this trait are called running mice (R). The recessive trait causes mice to run in circles only. Mice with this trait are called waltzing mice (r). Hair color is also inherited in mice. Black hair (B) is dominant over brown hair (b).

Dihybrid #2

• Cross a heterozygous running, heterozygous black mouse with a homozygous running, homozygous black mouse. What is the probable phenotype ratio?

Dihybrid #2 (Solution)

• Cross a heterozygous running, heterozygous black mouse with a homozygous running, homozygous black mouse. What is the probable phenotype ratio?

• RrBb x RRBB• • 100% Running black (All off spring will have

at least one dominant gene for the two traits)

Dihybrid #3 (You try it!)

• Cross a homozygous running, heterozygous black mouse with a waltzing brown mouse. What is the probable phenotype ratio?

Dihybrid #3 (Solution)

• Cross a homozygous running, heterozygous black mouse with a waltzing brown mouse. What is the probable phenotype ratio?

• RRBb x rrbb• • ½ running black and ½ running brown

Dihybrid #4 (You try it!)

• Cross a heterozygous running, heterozygous black mouse with a heterozygous running, heterozygous black mouse. What is the probable phenotype ratio?

Dihybrid #4 (Solution)

• Cross a heterozygous running, heterozygous black mouse with a heterozygous running, heterozygous black mouse. What is the probable phenotype ratio?

• RrBb x RrBb• • 9:3:3:1• • 9/16 running black; 3/16 running brown; 3/16 waltzing

black; 1/16 waltzing brown

X-Linked Recessive Inheritance (Sex Linkage)

• Males show disorder more than females

• Son cannot inherit disorder from his father

Sex Linkage #1

• In a cross between a pure bred, red-eyed female fruit fly and a white-eyed male, what percent of the male-offspring will have white eyes? (white eyes are X-linked, recessive)

• A. 100% • A. 75% • C. 50% • D. 25% • E. 0%

Sex Linkage #1

• In a cross between a pure bred, red-eyed female fruit fly and a white-eyed male, what percent of the male-offspring will have white eyes? (white eyes are X-linked, recessive)

• A. 100% • A. 75% • C. 50% • D. 25% • E. 0% Correct• All of the males and all of the females are red-eyed.

Sex Linkage #2Hemophilia in humans is due to an X-chromosome mutation. What will be the results

of mating between a normal (non-carrier) female and a hemophilic male? A. half of daughters are normal and half of sons are hemophilic. B. all sons are normal and all daughters are carriers. C. half of sons are normal and half are hemophilic; all daughters are carriers. D. all daughters are normal and all sons are carriers. E. half of daughters are hemophilic and half of daughters are carriers; all sons are

normal.

Sex Linkage #2• Hemophilia in humans is due to an X-chromosome mutation. What will be the results

of mating between a normal (non-carrier) female and a hemophilic male? • A. (Incorrect)• half of daughters are normal and half of sons are hemophilic. • B. • all sons are normal and all daughters are carriers. • Daughters inherit a normal allele from their mother and the hemophilia allele from

their father. Sons inherit the normal allele from their mother. Correct! • C. (Incorrect)• half of sons are normal and half are hemophilic; all daughters are carriers. • D. (Incorrect)• all daughters are normal and all sons are carriers. • E. (Incorrect)• half of daughters are hemophilic and half of daughters are carriers; all sons are normal.

Sex Linkage #3 (You try it!)

• A human female "carrier" who is heterozygous for the recessive, sex-linked trait causing red-green color blindness (or alternatively, hemophilia), marries a normal male. What proportion of their male progeny will have red-green color blindness (or alternatively, will be hemophiliac)?

• A. 100% • B. 75% • C. 50% • D. 25% • E. 0%

Sex Linkage #3 (Solution)• A human female "carrier" who is heterozygous for the recessive, sex-

linked trait causing red-green color blindness (or alternatively, hemophilia), marries a normal male. What proportion of their male progeny will have red-green color blindness (or alternatively, will be hemophiliac)?

• A. 100% • B. 75% • C. 50% Correct!• Half the sons would be expected to inherit the allele from their

mother and be afflicted because they are hemizygous. Half the daughters would be carriers like their mothers.

• D. 25% • E. 0%

Sex Linkage #4 (You try it!)

• A red-green colorblind male marries a normal female. What percentage of their female offspring will be red-green colorblind carries?

• A. 0 %• B. 25%• C. 50 %• D. 75%• E. 100%

Sex Linkage #4 (Solution)

• A red-green colorblind male marries a normal female. What percentage of their female offspring will be red-green colorblind carries?

• A. 0 %• B. 25%• C. 50 %• D. 75%• E. 100% All of his daughters would receive an Xb

chromosome from him and an X chromosome from their mother => X Xb (Carriers)