Practice Questions: More Extensions of Mendelian Genetics

Practice questions: More Extensions of Mendelian Genetics

Complete the following problems. Do your work on a separate sheet of paper if more space is needed.

1. A rooster with a particular comb shape called walnut was crossed to a hen with a type of comb called single. The F1 progeny all had walnut combs. When F1 males and females were crossed to each other, 93 walnut and 11 single combs were found among the progeny, along with 29 chickens with a new kind of comb called rose and 32 birds with another new type of comb called pea.

a. Explain how comb shape is inherited – Conventional inheritance of two traits

b. What progeny would result from crossing a homozygous rose hen with a homozygous pea rooster? What phenotypes and ratios would be seen in the F2 progeny? – all walnut

c. A particular walnut rooster is crossed with a pea hen and the progeny consisted of: 12 walnut, 11 pea, 3 rose and 4 single chickens. What are the likely genotypes of the parents?A-B-; aaB-; A-bb; aabb

d. A different walnut rooster was crossed to a rose hen and all of the progeny were walnut. What are the possible genotypes of the parents?A-Bb walnut

a) AABB X aabb

walnut single

F1AaBb X AaBb

walnut walnut

F2 93/165 walnut A-B-

11/165 single aabb

29/165 rose A-bb

32/165 pea aa Bb

Therefore a 9:3:3:1 ratio of walnut: rose: pea: single

b) pea x rose

AAbb x aaBBAaBb all walnut

c) walnut x pea

AaBb x aaBb 12 walnut A-B-

11 pea aaB-

3 rose A-bb

4 single aabb

3:3:1:1 ratio

d) walnut x rose

AABB x A-bb A-Bb walnut

2. A black mare was crossed to a chestnut stallion and the mating produced a bay son and a bay daughter. These two offspring were mated to each other several times and they produced offspring with four coat colors: black, bay, chestnut and liver (sort of grayish brown). Crossing a liver grandson back to the black mare gave a black foal and crossing a liver granddaughter back to the chestnut stallion gave a chestnut foal. Explain how coat color is inherited in these horses and give the possible genotypes for each horse.

Black female x chestnut male

aaB- x A-bb

F1 AaBb x AaBb

bay bay

F2 BlackaaB-

Bay A-B-

Chestnut A-bb

Liver aabb

liver X black (grandmother)

aabb x aaB-  black aaBb

liver X chestnut (grandfather)

aabb X A-bb  chestnut Aabb

Has to be two genes because one gene cannot produce 4 phenotypes

A bay female and bay male is not possible if the trait is X-linked

So this is inherited by two genes

black phenotype = dominant B

chestnut phenotype = dominant A

bay phenotype = dominant A & B

A-B- = bay

aaB- = black

A-bb = chestnut

Aabb = liver

3. In a species of tropical fish, a colorful orange and black variety called Montezuma occurs. When two of these fish are crossed, 2/3 are Montezuma and 1/3 are the wild-type dark grayish green color. Montezuma is a single-trait gene and these fish are never true-breeding.

a. Explain the inheritance pattern and show how your explanation accounts for the phenotype ratios given

b. In this same species, the shape of the dorsal fin is altered from normal to ruffled by being homozygous for a recessive allele designated f. What progeny would you expect to obtain and in what proportions from the cross of a Montezuma fish homozygous for normal fins to a wild-type fish with ruffled fins?

c. What phenotypic ratios of progeny would be expected from crossing two of the Montezuma progeny you obtained in cross b?

a) Mm x Mm  2/3 Mm and 1/3 mm

this is because MM = lethal

b) MmFF Xmmff

Montezuma, normal x wild-type, ruffled MmFf Montezuma, normal & mmFfwild-type, normal

c) MmFf xMmFf

normally you would get

9 M-F- but 3 are MMF- (lethal) 6/12 Montezuma, normal

3 mmF- 3/12 wild-type normal

3 M-ff  but 1 is MMff (lethal)2/12 Montezuma, ruffled

1 mmff1/12 wild-type, ruffled

** butMMF- and MMff are lethal – so the ratio is 6:3:2:1

4) In the fruit fly, very dark (ebony) body color is determined by the e allele. The e+ allele produces a normal, wild-type, honey colored body. In heterozygotes for the two alleles, a dark marking called a trident can be seen on the thorax of the body but otherwise the body is honey-colored.

a. What kind of inheritance would produce this trident marking?Incomplete dominance

b. When female e+e flies are crossed to male e+e files, what is the probability that the progeny will have the trident marking?50% probability

c. Flies with the trident markings mate among themselves. Of the 300 progeny, how many would be expected to have a trident, how many would be ebony and how many honey-colored?150 trident; 75 wild-type; 75 black

a. e+e+ = wild type

ee = black/ebony

e+ e = trident

** Incomplete dominance

b. e+e X e+e  1 e+e wild type; 2 e+e trident; 1 ee black

c. 300 X 50% = 150 e+e trident

300 x 25% = 75 e+e+ wild type and 75% ee black

5. Duplication of genes is an important evolutionary mechanism. As a result, many, cases are known in which a species has two or more identical genes.

a. Suppose there are two genes - A and B - that specify production of the same enzyme. An abnormal phenotype results only if an individual does not make any of that enzyme. Only one of these two genes needs to be functional in order to make the enzyme.What ratio of normal to abnormal progeny would result from a mating between two parents of genotype AaBb where A and B represent alleles that specify production of the enzyme, while a and b do not? 15:1 ratio

AaBb XAaBb

9 A-B- = enzyme produced

3 aaB- = enzyme produced

3 A-bb = enzyme produced

1 aabb = no enzyme produced

Therefore a 15:1 ratio results

b. Suppose now there are three genes specifying production of this enzyme and again that a single functional allele is sufficient for a wild-type phenotype. What ratio of normal vs. abnormal progeny would result?

AaBbCc XAaBbCc

¼ AA¼ BB¼ CC

½ Aa½ Bb½ Cc

¼ aa¼ bb¼ cc

So ¾ of these progeny will be functional for each of these Punnett square

Single functional allele is sufficient

This means:

A-bbcc = normal

¾ x ¼ x ¼ = 3/64

aaB-cc = normal

¾ x ¼ x ¼ = 3/64

aabbC- = normal

¾ x ¼ x ¼ = 3/64

But there are other genotypes that will produce a functional enzyme

A-B-C- = normal

¾ x ¾ x ¾ = 27/64

aaB-C- = normal (one homozygous recessive, two dominant)

¼ x ¾ x ¾

There are a total of 3 combinations for this = 27/64

aabbC- (two homozygous recessive, one dominant) = normal

¼ x ¼ x ¾ = 3/64

There are a total of 3 combinations for this = 9/64

27/64 + 27/64 + 9/64 = 63/64 normal progeny

Want a short cut for this??

aabbcc = non-functional

¼ x ¼ x ¼ = 1/64 non-functional This means that 63/64 are functional

6. A woman with type O blood, whose father has type A and whose mother has type Bhas a child with type O. There is a dispute over the identity of the child's father. Two men are possible fathers. One is type AB and the other is type A.

a. What is the mother's genotype?

b. Which man could be the father?

c. If this man is the father, what is his genotype?

d. What are the genotypes of the woman's parents?

AO X BOOO mother X type A (A--) or type AB male OO child results

Type A father possible? Yes with a genotype of AO

Type AB father possible? Yes with a genotype of AB and the Bombay genotype of hh

OOhh X ABhhAOhh child (type O)

BOhh child (type O)

7. In the peppered moth, Bistonbetularia, there are three alleles that determine body color. These alleles are all at the same locus. The allele for pale color (m) is recessive. A second allele (M’), which is dominant to m, produces a mottled color called insularia. The third allele (M), which is dominant to both of the other two, produces a melanic moth (very dark colored).

A female moth having the typical pale color (i.e. mm) is mated to a male melanic. If half the progeny are melanic and half are insularia, what were the genotypes of the two parents? MM’ and mm

Dominance series: M > M’> m

Melanicinsularia > pale

MM’ x mm  MM’ melanic progeny & M’minsularia progeny in equal ratio

Melanic pale

In other words – to be melanic you need a single M allele (i.e. M--); to be insularia you need an M’ allele and no M allele (M’m is the only genotype possible for this phenotype); to be pale you need to be homozygous recessive mm

The equal ratio as a result of a cross with a homozygous recessive moth is the expected result of a test-cross and this means that the other parent is heterozygous MM’

8. In wheat kernel color is determined by a pair of genes in a quantitative way. Each of the two genes can have two alleles (A1, A2, B1, B2). Kernel color ranges from red, when four type 1 alleles are present, to white, when four type 2 alleles are present. Three intermediate colors (dark pink, medium pink, and light pink) can occur depending on the relative numbers type 1 and type 2 alleles. Calculate the expected proportions of the five phenotypes that would be produced by a cross between two wheat plants with medium pink kernels that are heterozygous for both genes: A1A2B1B2 x A1A2B1B2 .
A1A2B1B2 X A1A2B1B2

If you have 2 “1” and “2” alleles (e.g. A1A2B1B2) = medium pink

If you have 3 “1” alleles (e.g. A1A1B1B2, A1A2B1B1) = dark pink

If you have 3 “2” alleles (e.g. A1A2B2B2, A2A2B1B2) = light pink

If you have all “1” alleles (A1A1B1B1) = red

If you have all “2” alleles (A2A2B2B2) = white

Don’t panic – this is a typical dihybrid cross except instead of A and a, you use A1 and A2 and B and b you use B1 and B2

So do a Punnett square or a branching diagram

1/16 red:3/8 med pink: ¼ dark pink: ¼ light pink: 1/16 white

9. Coat colors of Labrador retrievers depend upon the action of at least two genes. At one locus is the gene for coat color with the dominant allele resulting in a black color and bb producing a brown color – known as a chocolate lab. However, at the second locus, a recessive epistatic inhibitor of coat color pigment (ee) prevents the expression of color alleles at another independently assorting locus, producing a yellow coat color. When the dominant condition exists at the inhibitor locus (E-), the alleles of the other locus may be expressed and observed, so the genotype E-B- produces black and E-bb producing chocolate. Two dihybrid black labs are mated together.

a. determine the phenotypic proportions expected in the progeny

B-- = black

B = brown/chocolate

E-- = coloration possible

E = prevents coloration - yellow

EeBb xEeBb

9 E-B- = black

3 eeB- = yellow due to ee genotype

3 E-bb = chocolate due to bb genotype

1 eebb = yellow due to ee genotype

Therefore 9 black: 4 yellow: 3 chocolate

This ratio is recessive epistasis

b. determine the chance of choosing, from among the black progeny, a genotype that is homozygous at both loci.

Chance of being EEBB = 1/9 from among the 9 E-B- progeny

10. In horses, the dominant allele B can produce a black coat color. However, if a dominant W allele is also present, the horse’s coat color will be white. If a heterozygous white (BbWw) mare is crossed with a heterozygous white (BbWw) stallion, what could the offspring's phenotype and genotype be?

You were missing some important information here – sorry about that!

The dominant W allele masks the black coat color

B allele with ww = black (BBww, Bbww)

W allele with any B allele = white (bbW- and B-W-)

bbww = chestnut

White phenotypes would be 9 B-W-

3 bbW-

Black phenotypes would be 3 B-ww

The chestnut phenotype would be 1 bbww

11. Horses can also be bay in color. This dominant gene allele (A) masks the dominant black color, but not the white gene or the recessive chestnut color. What is the phenotype and genotype of the offspring when a bay mare (AaBBww) is crossed with a bay stallion (AaBbww)?

Remember:

B = Black (BBww, Bbww)

W = White (B-W-, bbW-)

bbww = Chestnut

The A allele will mask only the B-ww genotype and produce a bay coat color

AaBBww X AaBbwwNormally these horses would be black but the Aa genotype masks this coat color

Bay Bay

Do a branching diagram for an AaBB x AaBb cross and then add the ww genotype to each result

1 AABBww = bay

1 AABbww= bay

2 AaBBww = bay6 Bay to 2 Black

2 AaBbww = bay

1 aaBBww = black

1 aaBbww = black

12. Horses also carry a dilution gene (D). If only one allele is the dominant D(i.e. D-), and a dominant A allele(i.e. A-)and the recessive chestnut color genes are present(i.e. bbww), then the horse will be a palamino. If there are two dominant dilution alleles (DD) and all else is the same as previously mentioned, the horse will be a pseudo albino. What is the phenotype and genotype of the offspring when a palomino mare (AAbbDdww) is crossed with a palomino stallion (AabbDdww)?

AAbbDdwwX AabbDdww – ignore the bb and ww for now and do a AADd x AaDd dihybrid cross

To be palamino, the horse needs the following:

A-

Dd

bb

ww

You need a dominant A and the Dd genotype to be a palomino horse – in other words, if you have a Dd genotype the horse is a palamino if there is an A allele in the genotype

If the DD genotype is present – the palamino horse becomes a pseudo-albino

If the dd genotype is present – it doesn’t matter what the A alleles are because the bbww makes the horse chestnut

So look at the D alleles first, then the A alleles and then consider the bbww

1 AAbbDDww = pseudo-albino

2 AAbbDdww = palamino

1 AAbbddww = chestnut (because of bbww)

1 AAbbDDww = pseudo-albino

2 AabbDdww = palamino

1 Aabbddww = chestnut (because of bbww)

13. In clover plants, the pattern on the leaves is determined by a single gene with seven alleles that are related in a simple dominance series - the allele that determines the absence of a pattern is recessive to the other six alleles, each of which produces a distinct pattern. All heterozygous combinations of alleles show complete dominance

a. How many different kinds of leaf patterns (including the absence of pattern) are possible in a population of clover plants where all seven alleles are present?

Assume the dominance series: A1>A2A3A4A5A6A7

Consider allele A1

What are the genotype patterns that will have A1 in them?

A1A1

A1A2

A1A3

A1A47 genotypes possible that contain the A1 allele

A1A5

A1A6

A1A7

Consider allele A2

What are the genotype patterns that will have A1 in them?

A2A2

A2A3

A2A46 genotypes possible

A2A5don’t count A2A1 because it’s already in the A1 list

A2A6

A2A7

A3 allele – 5 genotypes

A4 allele – 4 genotypes

A5 allele – 3 genotypes

A6 allele – 2 genotypes

A7 allele – 1 genotype (A7A7)

So the total # of genotypes for these 7 patterns = 7+6+5+4+3+2+1 = 28 genotypes

b. What is the largest number of different genotypes that could be associated with any one phenotype? Is there any phenotype that could be represented by only a single genotype?

pattern #1 has the largest # of genotypes

pattern #7 has the least # of genotypes

c. In a particular field, you find that the large majority of clover plants lack a pattern on their leaves even though you can identify a few plants representative of all possible pattern types. Explain this finding.

Only the homozygoteA7A7 reproduces the best out of all plants???

14. In fruit flies, the gene for white eyes is sex-linked recessive. (R) is red and (r) is white. Cross a white-eyed female with a normal red-eyed male.

a. What percent of the males will have red eyes? White eyes?No males will have red eyes

b. What percent of the females will have red eyes? White eyes?100% of females will have red eyes; no white eyed females possible

c. What total percent of the offspring will be white-eyed?50% of the offspring will be white eyed

d. What percent of the offspring will be carriers of the white eye trait?100% of females will carry the white eye trait

XrXr x XRYXRXrred-eyed females and XrY white-eyed males

15. Using the same information as for question #1, cross a heterozygous red-eyed female with a red-eyed male.

a. What are the genotypes of each parent?XRXr x XRY

b. What fraction of the children will have red eyes?¾ of offspring will have red eyes

c. What fraction of the children will have white eyes?¼ will be white eyed

d. What fraction of the female children will carry the white eyed trait?50% of females will carry white eye trait

XRXr x XRY  XRXR XRXrred-eyed females and XRYXrY white-eyed males

16. In humans, hemophilia is a sex-linked recessive trait. If a female who is a carrier for hemophilia (i.e. XHXh) marries a male with normal blood clotting, answer the following questions.

a. What fraction of the female children will have hemophilia?0% of females will be hemophiliac

b. What fraction of the female children will be carriers?50% of females are carriers

c. What fraction of the male children will have normal blood clotting? 50% of males are normal

d. What fraction of the male children will be carriers?0% of males are carriers

e. What fraction of the male children will have hemophilia?50% of males are hemophiliacs

Cross: XHXh x XHY  XHXH and XHXh females (all normal); XHY (normal) and XhY (hemophiliac) males

19. Two normal-visioned parents have a color-blind son. Give the genotype of both parents and the son.

XBXb x XBYXbY son

20. In cats, the allele (B) produces black color but (b) produces a yellow color. These alleles are incompletely dominant to each other. A heterozygote produces a tortoise shell color. The alleles (B) and (b) are sex-linked as well. Cross a tortoise shell female with a yellow male.

a. What percent of their offspring will be yellow?½ of offspring

b. What percent of their offspring will be black?¼ of offspring

c. What percent of their offspring will be tortoise shell?¼ of offspring (1/2 of females)

d. Why is it “impossible” to have a tortoise shell male offspring?Tortoiseshell requires 2 X chromosomes(although this could happen upon non-disjunction with the X chromosome producing a XXY male cat)

XbXb = tortoiseshell female

XBY = black male

XbY = yellow male

Cross: XbXb x XbY (do a Punnett square)

Progeny:

XBXb = tortoiseshell female

XbXb = yellow female

XBY = black male

XbY = yellow male