Problems on patterns of inheritance

1. Suppose you are designing a new Jurassic Park and you would like to have the crowd drawing presence of

Velociraptor without the danger of having your crowds eaten. You have managed to come up with two varieties,

very aggressive (A), which is apparently a dominant trait in Velociraptors, and non-aggressive. They also come in

two colors, red (R), which is dominant, and green. At the moment you have a green female that is homozygous

dominant for aggression and a non-aggressive male that is homozygous dominant for color. What gametes will be

produced by the male and female?

Ans. Aggression is a dominant trait and is indicated by A. The color red is also a dominant trait in Velociraptors and is

indicated by R. The female is homozygous dominant for aggression and the male is a wimp. What are their

aggressive genotypes? Female AA, Male aa. Now for color, the female is green and the male is homozygous

dominant for color. What are their genotypes for color? Female rr, Male RR. Putting the two together we find that

the female has a diploid genotype of AArr and the male, aaRR.

Female: AArr Male: aaRR. Thus, gametes produced from male: aR and female: Ar

2. Wildcats in Kansas come in three colors, blue, red, and purple. This trait is controlled by a single locus gene with

incomplete dominance. A homozygous (CBCB) individual is blue, a homozygous (CbCb) individual is red, and a

heterozygous (CbCB) individual is purple.

a. What would be the genotypes and phenotypes of the offspring if a blue wildcat were crossed with a red one?

Ans. Genotype of parent 1: CBCB. Genotype of parent 2: CbCb. In all progeny, genotype would be CbCB. The phenotype

would be purple.

b. What are the genotypic and phenotypic ratios of the F2 generation?

Ans. The Genotypic Ratio would be 1:2:1. The phenotypic ratio would be 1:2:1

3. The lubber grasshopper is a very large grasshopper, and is black with red and yellow stripes. Assume that red stripes

are expressed from the homozygous RR genotype, yellow stripes from the homozygous rr genotype, and both from

the heterozygous genotype.

a. What will be the phenotypic ratio of the F1 generation resulting from a cross of two grasshoppers, both with red

and yellow stripes (red : both : yellow)?

C

B

C

B

C

b

C

b

C

B

C

b

C

B

C

b

C

b

C

B

C

b

C

B

C

B

C

b

C

B

C

B

C

B

C

b

C

B

C

b

C

b

C

B

C

b

C

b

Ans. 1:2:1

b. What genotypes would be produced by crossing a grasshopper with both color stripes and one with yellow stripes?

Ans. Rr, rr

c. What phenotypes would be produced by crossing a grasshopper with both color stripes and one with yellow stripes?

Ans. Both colors, yellow stripes

4. Suppose you have two rose plants, both with pink flowers. You cross the two plants and are surprised to find that,

while most of the offspring are pink, some are red and some are white.

a. You decide that you like the red flowers and would like to make more. What cross would you perform to produce the

most red flowered plants?

Ans. Red x Red

b. Your mother decides she would like some of the pink flowered roses. Which cross would give you the most pink

flowered plants?

Ans. Red x White

5. A naturalist visiting an island in the middle of a large lake observes a species of small bird with three distinct types of

beaks. Those with short, crushing beaks (BB) consume hard shelled nuts, those with long, delicate beaks (bb) pick the

seeds from pine cones, and those with intermediate beaks (Bb), consume both types of seeds though they are not as

good at either. Assume that this difference in beak morphology is the result of incomplete dominance in a single locus

gene.

a. Which of the mated pairs below will have the best adapted offspring in a year in which most of the food available is in

the form of hard shelled nuts?

Ans. Options:

Short x short (BB x BB) Short x intermediate (BB x Bb)

Short x long (BB x bb) Intermediate x intermediate (Bb x Bb)

Intermediate x long (Bb x Bb) Long x long (bb x bb)

b. What would be the phenotypic (Short:Intermediate:Long) ratio of the F1 generation resulting from a cross of Bb x

bb?

Ans. RATIO = 0:2:2

c. How many offspring of an intermediate x short beak cross will have long beaks (assume 4)?

Ans. Count up how many of the four possible offspring in the square have the long beak phenotype = 0

6. Racoons have rings around their tails and a habit of washing their food in water before eating it. Suppose that both

of these traits are controlled via incomplete dominance so that wide bands on the tail are BB, medium sized bands are

Bb, and narrow bands are bb and that washing all their food is WW, washing some of their food is Ww, and washing

no food is ww.

a. How many of each genotype will be in the F1 generation resulting from a cross of two racoons, both with medium

sized tail bands and that wash some of their food (assume 16)?

Ans.

Count up how many of the 16 possible offspring in the square have the different genotypes and phenotypes.

genotype number phenotype phenotype

BBWW 1 Wide bands Washing all food

BBWw 2 Wide bands Washing some food

BBww 1 Wide band No washing food

BbWW 2 Intermediate band

Washing all food

BbWw 4 Intermediate band

Washing some food

Bbww 2 Intermediate band

Washing no food

bbWW 1 Narrow band Washing all food

bbWw 2 Narrow band Washing some food

bbww 1 Narrow band Washing no food

BW Bw bW bw

BW BBWW BBWw BbWW BbWw

Bw BBWw BBww BbWw Bbww

bW BbWW BbWw bbWW bbWw

bw BbWw Bbww bbWw bbww

b. How many of the F1 generation will have wide tail bands and won't wash any of their food?

Ans. One individual will have wide bands on tail and will not wash food at all.

7. About 70% of Americans perceive a bitter taste from the chemical phenylthiocarbamide (PTC). The ability to taste

this chemical results from a dominant allele (T) and not being able to taste PTC is the result of having two

recessive alleles (t). Albinism is also a single locus trait with normal pigment being dominant (A) and the lack of

pigment being recessive (a). A normally pigmented woman who cannot taste PTC has a father who is an albino

taster. She marries a homozygous, normally pigmented man who is a

taster but who has a mother that does not taste PTC.

a. What are the genotypes of the possible children?

Ans. Genotype of offspring = AATt, AaTt, AAtt, Aatt 4:4:4:4

b. What percentage of the children will be albinos?

Ans. Albino children = 0%

C. What percentage of the children will be non-tasters of PTC?

Ans. Non-tasters of PTC = 50%

Solution:

First determine the genotypes of the parents. We know that the woman has normal pigment which

means she must have at least one A. Her father is albino and because the albino allele is recessive, his

genotype is aa. What does this make her skin genotype? Aa

We also know that the woman cannot taste PTC. Because the ability to taste PTC is dominant, what does

this make her genotype for the tasting trait? tt

Putting both traits together, what is the woman's overall genotype? Aatt.

Now, what about her husband? You have been told that he is homozygous for normal pigment. What is

his genotype for skin color? AA

He is a taster and so must have at least one T. However, we also know that his mother cannot taste PTC

so she must be homozygous recessive. With this information, what is his genotype for taste? Tt

Putting both traits together, what is his overall genotype? AATt.

Now that you have the genotypes of the parents, perform a Punnett square to determine what the F1 offspring

would be.

At At at at

AT AATt AATt AaTt AaTt

AT AATt AATt AaTt AaTt

At AAtt AAtt Aatt Aatt

At AAtt AAtt Aatt Aatt

8. Wolves are sometimes observed to have black coats and blue eyes. Assume that these traits are controlled by

single locus genes and are located on different chromosomes. Assume further that normal coat color (N) is

dominant to black (n) and brown eyes (B) are dominant to blue (b). Suppose the alpha male and alpha female

of a pack (these are the dominant individuals who do most of the breeding) are black with blue eyes and

normal colored with brown eyes, respectively.

a. The female is also heterozygous for both traits. How many of the offspring (assume 16) living in the pack will

have each of the following genotypes? NNBB, NNBb, NNbb, NnBB, NnBb, Nnbb, nnBB, nnBb, nnbb

Ans. NNBB:0, NNBb:0, NNbb:0, NnBB:0, NnBb:4, Nnbb:4, nnBB:0, nnBb:4, nnbb:4

b. What percent of the offspring will be normal colored with blue eyes?

Ans. 25%

Solution

First determine the genotypes of the parents. We know that the female is heterozygous for both traits so

her genotype must be NnBb

The male is black with blue eyes. These are both recessive traits and must be homozygous to be

expressed. What, then, is the male's genotype? nnbb

Now that you have the genotypes of the parents, perform a Punnett square to determine what the F1

offspring would be

Now simply count up the number of offspring of each genotype and phenotype produced.

9. In the breeding season, male Anole lizards court females by bobbing their heads up and down while displaying

a colorful throat patch. Assume that both males & females show this behavior. Assume also, that both traits

are controlled by single locus genes on separate chromosomes. Now, suppose that anoles prefer to mate with

lizards who bob their heads fast (F) and have red throat patches (R) and that these two alleles are dominant to

their counterparts, slow bobbing and yellow throats. A male lizard heterozygous for head bobbing and

homozygous dominant for the red throat patch mates with a female that is also heterozygous for head

bobbing but is homozygous recessive for yellow throat patches.

a. How many of the F1 offspring have the preferred fast bobbing / red throat phenotype (assume 16

young)?

NB Nb nB nb

nb NnBb Nnbb nnBb nnbb

nb NnBb Nnbb nnBb nnbb

nb NnBb Nnbb nnBb nnbb

nb NnBb Nnbb nnBb nnbb

Ans. 12 of the F1 offspring have the preferred fast bobbing / red throat phenotype (assume 16 young).

b. What percentage of the offspring will lack mates because they have both slow head bobbing and yellow

throats?

Ans. None.

c. What percentage of the offspring will have trouble finding mates because because they lack one of the

dominant traits?

Ans. 25%

Solution:

Once again, first determine the genotypes of the parents. We know that the female is heterozygous for the rate

of head bobbing but homozygous recessive for the color of her throat patch. So, what must her genotype be? Ffrr

The male is also heterozygous for head bobbing but is homozygous dominant for throat patch color. What, then,

is the male's genotype? FfRR

Now that you have the genotypes of the parents, perform a punnett square to determine what the F1 offspring

would be

10. Carrion beetles lay their eggs in dead animals and then bury them in the ground until they hatch. Assume that the

preference for fresh meat (F) is dominant to the preference for rotted meat and that the tendency to bury the

meat shallow (S) is dominant to the tendency to bury the meat deep. Suppose a female carrion beetle homozygous

dominant for both traits mates with a male homozygous recessive for both traits.

a. What will be the genotype of the F1 generation?

Ans. FfSs

b. What will be the phenotype of the F1 generation?

Fr Fr fr fr

FR FFRr FFRr FfRr FfRr

FR FFRr FFRr FfRr FfRr

fR FfRr FfRr ffRr ffRr

fR FfRr FfRr ffRr ffRr

Ans. Fresh meat / shallow

c. What will be the genotypic ratio of the F2 generation (FFSS: FFSs: FFss: FfSS: FfSs: Ffss: ffSS: ffSs: ffss)?

Ans. 1:2:1:2:4:2:1:2:1

Solution:

Determine the genotypes of the parents. We know that the female is homozygous dominant for both

traits. What is her genotype?

The male is also homozygous for both traits but he is recessive. So, what does this make the male's

genotype?

Now that you have the genotypes of the parents, perform a Punnett square to determine what the F1

offspring would be

F1 F2

11. Suppose in a strain of soybeans, high oil (H) content in the seeds is dominant to low oil content and four

seeds (E) in a pod is dominant to two seeds in a pod. A farmer crosses two soybean plants, both with high oil

content and four seeds per pod. The resulting F1 offspring have a phenotypic ratio of 9:3:3:1 (High oil / four

seeds : High oil / two seeds : Low oil / four seeds : Low oil / two seeds).

a. What genotype were the parent plants?

Ans. HhEe x HhEe

Solution:

FS FS FS

FS

fs FfSs FfSs FfSs FfSs

fs FfSs FfSs FfSs FfSs

fs FfSs FfSs FfSs FfSs

fs FfSs FfSs FfSs FfSs

FS Fs fS

fs

FS FFSS FFSs FfSS Ffss

Fs FFSs FFss FfSs Ffss

fS FfSS FfSs ffSS ffSs

fs FfSs Ffss ffSs ffss

The easiest way, is to have remembered what type of cross will produce a 9:3:3:1 phenotypic ratio. HhEe x HhEe

12. In a condition called Turner Syndrome individuals are born with genotype XO due to nondisjunction of sex

chromosome. One such individual was found to be color blind. Upon analyzing the parents it was found that

the father had normal vision. List the possibilities of non-dysjunction that has happened in the parental

gametes that could have resulted in such a child. Also, clearly explain which of the two, is the affected parent.

Ans. Turner Syndrome has genotype XO; the individual is color blind; color blindness is an X-linked disorder; so

the individual has a faulty X chromosome; Father is reported to be normal; so the faulty X chromosome has

come from the mother. As mother has contributed one X chromosome to the foetus, her meiotic division is

normal and the maternal gamete had one X chromosome No nondisjunction. It was the paternal gamete

which was without any sex chromosome. So, non dysjunction has happened in the father in Meiosis I to

produce 50% gametes without any sex chromosome and 50% gametes with both XY sex chromosomes.

Fertilization of the normal maternal gamete with the paternal gamete without any sex chromosome results in

an Turner individual. The inherited X chromosome from mother is faulty for color blindness hence

individual is born with the disorder.

13. A sex-linked recessive gene c produces red-green color blindness in humans. A normal woman whose father

was color-blind marries a color-blind man. i) What genotypes are possible for the mother of the color-blind

man? ii) What are the chances that the first child from this marriage will be a color-blind boy? iii) Of all the

girls produced by these parents, what percentage is expected to be color-blind? iv) of all the children (sex not

specified) from these parents, what proportion is expected to be normal?

Ans. i) Cc/cc as the woman is Cc & the man cY ii) iii) 50% iv)

13. a) A normal woman whose father had haemophilia mates with a normal man and produces a Klinefelters son (XXY) who is also a haemophilic. Explain how this is feasible and what kind of non-disjunction can explain this result? Ans Since the mother is normal but has a haemophylic father, her genotype is XHXh where H = normal gene and h = haemophilic gene. When she mates with a normal male (XHY) then in the 1st meiotic division, she produces two gametes with the constitution XH and Xh whereas, the male produces XH and Y gametes. But in the 2nd meiotic division, due to non-disjunction in the mother, the XH gamete will produce two more of its type but the Xh gamete will produce one gamete with Xh Xh constitution and another with zero chromosomes. Now when this Xh Xh gamete fuses with the Y gamete of the normal male then a klinefelters haemophilic son will be produced. b) White eye colour in a fly is due to a sex-linked recessive gene w and wild type (red) eye colour, due to its dominant allele w+. A laboratory population of these flies was found to contain 150 red eyed males and 50 white eyed males (where XY represents the males and XX the females). Estimate the frequency of w+ allele and w allele in the gene pool. Ans. The genotype of 50 white eyed males is wY and that of 150 red eyed males is w+Y.

So, 50 of the 200 X chromosomes in this sample carry the recessive allele w.

q = 50/200 = 0.25 or 25% w alleles and p = 1- 0.25 = 0.75 or 75% w+ alleles

c) On chromosome 3 of corn there is a dominant gene (A1), which, together with the dominant gene (B1) on chromosome 9, produces coloured seeds. All other genetic combinations of the alleles of these two genes produce colourless seeds. Two pure colourless strains are crossed to produce an all coloured F1. i) What were the genotypes of the parental strains and F1?

Ans. P: A1 A1b1b1 x a1 a1B1B1; F1: A1 a1B1b1

ii) Using a punnett- square estimate the phenotypic proportions expected among the F2? Ans. 9/16 coloured : 7/16 colourless where the gametes are A1 B1, A1b1, B1a1, a1b1.

Gametes male/female

A1 B1 A1b1 B1a1 a1b1

A1 B1 A1A1B1B1 (coloured) A1 A1B1 b1 (coloured)

A1 a1 B1 B1 (coloured)

A1 a1 B1 b1 (coloured)

A1b1 A1A1B1b1 (coloured) A1 A1b1b1 (colourless)

A1 a1 B1 b1 (coloured)

A1 a1 b1 b1 (colourless)

B1a1 A1a1B1B1 (coloured) A1 a1 B1 b1 (coloured)

a1 a1 B1 B1 (colourless)

a1a1B1 b1 (colourless)

a1b1 A1a1B1b1 (coloured) A1 A1 b1 b1 (colourless)

a1 a1 B1 b1 (colourless)

a1b1 a1b1 (colourless)