Question

1. Define each of the following terms and give an example.
   1. Simple (complete) dominance =

2. Incomplete dominance =

3. Codominance =

4. Monohybrid cross =

5. Dihybrid cross =

6. Sex-linked trait =

Simple Dominance with one gene:

2. A certain moth species can have either a long (L) or short (l) proboscis.
   1. If you cross a true-breeding long with a true-breeding short proboscis moth in the P generation, what will be the genotypes and phenotypes of the F1 generation?

1. 2. If you cross two of the F1 offspring together, what will be the genotypes and phenotypes of the F2 generation? Give the fraction of the offspring that will have each genotype and phenotype.
2. Roberts’ syndrome causes children to be born with misshapen bones in their faces, skulls, and limbs. Roberts’ syndrome is inherited on a recessive allele, while the dominant allele codes for normal bone shape.
   1. If a carrier for the Roberts’ allele marries someone who is homozygous dominant for the normal allele, what are the chances that their child will be born with Roberts’ syndrome?

2. If both parents were carriers, what would be the likelihood that their child would get Roberts’ syndrome?

4. The polled (meaning hornless) trait in cattle is dominant, the horned trait is recessive. A rancher has a polled bull (male) and mates the same bull to three cows (female). Cow A is horned and produces a polled calf. Cow B is horned and produces a horned calf. Cow C is polled and produces a horned calf. What are the genotypes of the four parents and their calves?

Simple Dominance with 2 genes

4. A gene in cats causes them to be black (dominant) or brown (recessive). A second gene causes cats to be agouti (have stripes on their hairs) or non-agouti (solid colored hairs).
   1. If you did a dihybrid cross between two black haired agouti cats, what fraction of the offspring would be expected to have each phenotype?

2. If a dihybrid black agouti cat mated with a brown cat that was heterozygous for the agouti gene, what fraction of the offspring would be expected to have each phenotype?

Test Cross

5. In mythical dragons, the ability to breath fire (F) is dominant over not being able to breathe fire (f). If you have a fire-breathing dragon and want to know whether it is homozygous or heterozygous, what could you do to find out? Explain how you would do the experiment.

6. In moths, having a long proboscis is dominant over having a short proboscis. You’re trying to figure out whether a particular female moth with a long proboscis is pure or a hybrid, so you cross it with a male moth that has a short proboscis. Half the offspring have long and half have short proboscii. What was the genotype of their mother?

7. In cats, black fur is dominant over brown fur. You have a black cat and you want to know whether it is true breeding for its black fur color, so you mate it with a brown cat. All 10 of their kittens are white. What was the genotype of your parent black cat? Can you be absolutely sure?

Incomplete Dominance

8. In rabbits, alleles for short fur (S) and long fur (L) are incompletely dominant. Hybrids have medium length fur. If you did a monohybrid cross with these rabbits, what ratio of short fur: medium fur: long fur would you see in the offspring?
9. In Andalusian fowl (similar to chickens), if you mate a true breeding white fowl with a true breeding black fowl, all the offspring will be blue. If you were to mate a blue fowl with a white fowl, what fraction of the offspring would be blue?

10. In cats, a particular gene determines whether the cat has white patches on top of its other color(s). Homozygous dominant cats are more than half white. Hybrid cats are less than half white, and homozygous recessive cats have no white patches at all. If a cat with white only on its feet mated with a cat that had no white at all, what fraction of the kittens would have each genotype and phenotype?

Codominance

11. In some horses, hair color is determined by one gene with two alleles. Homozygotes may be either red or white, and heterozygotes are roan (a mixture of red and white hairs). If you crossed two roan horses together, what fraction of the offspring would be roan?

12. Human blood type alleles are codominant. Lucy has type A blood. Her daughter Jenna has type O blood. Jenna’s father Ralph has never had his blood type tested.

1. What is Lucy’s genotype?

2. What are the possible genotypes that Ralph could have?

13. A woman with type 0 blood gives birth to a baby with type 0 blood. In a court case she claims that a man with type A blood is the father of the child. Could he be the father? Show why he can or cannot. Can it be proven on this evidence alone that he is the father?

Sex-linked genes

14. Hemophilia is a rare genetic disorder that results from a deficiency in blood clotting factors. When a person without hemophilia gets a cut, the bleeding lasts for a short time and then the blood clots to form a scab. People with hemophilia have only 1-5% of the blood clotting factors that other people have. Thus, when a hemophiliac gets a cut, it may continue to bleed for days or weeks, sometimes causing dangerous amounts of blood loss.  Hemophiliacs have to do special exercises to prevent their muscles from tearing and producing internal bleeding, and they can take supplemental blood clotting factors, but hemophilia has no known cure. Hemophilia is caused by a recessive allele that is located on the X chromosome.

If X^H is the allele for normal blood clotting, and X^h is the allele for hemophilia, draw a cross between a normal father and a mother who is a carrier for hemophilia. Then list the percentages of offspring that would get each genotype and phenotype.

15. Colorblindness is a sex-linked recessive trait. If a woman who was a carrier married a man who was colorblind, what percentage of their children (males and females) would be colorblind?

16. For a male to be colorblind, does he have to get the colorblindness allele from his mother, from his father, or both? For a female to be colorblind, does she have to get the colorblindness allele from his mother, from his father, or both?

Answer 1

Answer:

1. a) Simple (complete) dominance:

When one allele is completely dominant over another allele then it is known as simple (complete) dominance. In this condition the dominant allele completely masks or hides the effect of recessive allele in heterozygous condition.

Example: Height of pea plant is an example of simple (complete) dominance. The tall height allele (T) is completely dominant over dwarf height allele (t). The possible genotypes for two allele T and t are— TT (homozygous dominant), Tt (heterozygous), and tt (homozygous recessive) and the possible phenotype will be tall (for TT and Tt) and dwarf (for tt).

1. b) Incomplete dominance:

When neither allele is dominant over other then it is known as incomplete dominance. In this condition heterozygous genotype produce a different phenotype.

Example: In Four O’clock plants (Mirabilis jalapa) when the red flower plant (RR, homozygous dominant) is crossed with the white flower plant (rr, homozygous recessive) all plants in F1 generation pink flower plant (Rr, heterozygous). So, the red colour producing allele, R is not dominant over white colour producing allele, r.

1. c) Co-dominance:

When two alleles lack dominant and recessive relationship and both are observed phenotypically to same degree then it is known as co-dominance. In this condition heterozygous genotype produce characters of both phenotypes.

Example: MN blood group in human is an example of co-dominance. In heterozygous condition (LMLN) the characters of both allele LM and LN are produced.

1. d) Monohybrid cross:

Monohybrid cross is a cross between two organism on the basis of the two alleles of one gene.

Example: A cross of purple flowers plant (PP, homozygous dominant), with white flowers plant (pp, homozygous recessive) resulted in all purple flowers plant (Pp, heterozygous) in the F1 generation. If F1 purple flower plants are self-crossed the result come in F2 generation—

Gametes for Pp are P and p.

	P	p
P	PP (purple flower)	Pp (purple flower)
p	Pp (purple flower)	pp (white flower)

Genotypic ratio in F2 generation— PP:Pp:pp = 1:2:1

Phenotype ratio in F2 generation— purple:white = 3:1

1. e) Dihybrid cross:

Dihybrid cross is a cross between two organism on the basis of the four alleles of two genes.

Example: A cross of yellow and round plant (YYRR, homozygous dominant), with green and wrinkled plant (yyrr, homozygous recessive) resulted in all purple flowers plant (YyRr, heterozygous) in the F1 generation. If F1 purple flower plants are self-crossed the result come in F2 generation—

Gametes for YyRr—

	Y	y
R	YR	yR
r	Yr	yr

Self cross of YyRr:

	YR	Yr	yR	yr
YR	YYRR (yellow and round)	YYRr (yellow and round)	YyRR (yellow and round)	YyRr (yellow and round)
Yr	YYRr (yellow and round)	YYrr (yellow and wrinkled)	YyRr (yellow and round)	Yyrr (yellow and wrinkled)
yR	YyRR (yellow and round)	YyRr (yellow and round)	yyRR (green and round)	yyRr (green and round)
yr	YyRr (yellow and round)	Yyrr (yellow and wrinkled)	yyRr (green and round)	yyrr (green and wrinkled)

Genotypic ration = YYRR:YyRR:yyRR:YYRr:YyRr:yyRr:YYrr:Yyrr:yyrr = 1:2:1:2:4:2:1:2:1

Phenotypic ratio = yellow and round:yellow and wrinkled: green and round:green and wrinkled = 9:3:3:1

1. f) Sex-linked trait:

Genes which are present in the sex chromosome are called as sex linked genes and the inheritance of the sex-linked genes is known as Sex-linked trait.

Example: An example of sex-linked trait is hemophilia in human which is an X-linked trait.

2. A certain moth species can have either a long (L) or short (l) proboscis and one is completely dominant over another.

If I cross a true-breeding long (LL, homozygous dominant), with a true-breeding short (ll, homozygous recessive) proboscis moth in the P generation, the genotypes will be Ll (heterozygous) and phenotypes will be long proboscis in the F1 generation.

If you cross two of the F1 offspring together, the genotypes and phenotypes of the F2 generation will be—

Gametes for Ll are L and l.

	L	l
L	LL (long proboscis)	Ll (long proboscis)
l	Ll (long proboscis)	ll (short proboscis)

Genotypic ratio in F2 generation— LL:Ll:ll = 1:2:1

Phenotype ratio in F2 generation— long proboscis: short proboscis= 3:1

3. Roberts’ syndrome causes children to be born with misshapen bones in their faces, skulls, and limbs. Roberts’ syndrome is inherited on a recessive allele (r), while the dominant allele (R) codes for normal bone shape.

If a carrier for the Roberts’ allele (Rr, heterozygous) marries someone who is homozygous dominant (RR) for the normal allele, the chances that their child will be born with Roberts’ syndrome are—

Gamete of Rr is R and r; gamete of RR is R and R.

	R	r
R	RR	Rr
R	RR	Rr

Genotypic ratio in— RR:Rr = 2:2 = 1:1

Phenotype of the RR (homozygous dominant) and Rr (heterozygous) both normal bone shape. Thus the chance that their child will be born with Roberts’ syndrome is 0%.

If both parents were carriers (Rr, heterozygous), the likelihood that their child would get Roberts’ syndrome are—

Gametes for Rr are R and r.

	R	r
R	RR	Rr
r	Rr	rr

Genotypic ratio— RR:Rr:rr = 1:2:1

Phenotype ratio— normal bone shape:Roberts’ syndrome = 3:1

Thus, the chance that their child will be born with Roberts’ syndrome is 25%.

4. The polled (meaning hornless) trait in cattle is dominant (P), the horned trait is recessive (p). A rancher has a polled bull (male) and mates the same bull to three cows (female). Cow A is horned and produces a polled calf. Cow B is horned and produces a horned calf. Cow C is polled and produces a horned calf. The genotypes of the four parents and their calves are—

The male is heterozygous (Pp).

The Cow A female is homozygous recessive (pp) and her calf is heterozygous (Pp).

	P	p
p	Pp (polled)	pp (horned)
p	Pp (polled)	pp (horned)

The Cow B female is homozygous recessive (pp) and her calf is homozygous recessive (pp).

	P	p
p	Pp (polled)	pp (horned)
p	Pp (polled)	pp (horned)

The Cow C female is heterozygous (Pp) and her calf is homozygous recessive (pp).

	P	p
P	PP (polled)	Pp (polled)
p	Pp (polled)	pp (horned)