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Solutions for genetics problems bil 151


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SOLUTIONS FOR GENETICS PROBLEMS - BIL 151

1. Most humans (and vertebrates, in general) are genetically "programmed" to produce a brown skin pigment known as melanin. Albinism--the inability to produce melanin--is inherited in humans as an autosomal recessive trait. An individual with genotype AA or Aa will have normally pigmented skin, whereas an individual with genotype aa will produce no melanin. The albino's skin is very pale white, and the (highly vascularized) irises of the eyes appear red because no melanin obscures the red reflectance of blood. Predict the frequencies of all genotypes and phenotypes expected to result from the cross of an albino male with a heterozygous female.


Parental genotypes: male: aa female: Aa
Possible gametes of the male: a only

Possible gametes of the female: A or a
Punnett Square will yield 50% Aa (heterozygous; pigmented) and 50% aa (homozygous; albino)






A

a

a

Aa

Aa

2. Do the same for a mating of two heterozygotes.


Possible gametes of either parent: A or a
Punnett Square will yield 25% AA (homozygous pigmented) 50% Aa (heterozygous; pigmented) and 50% aa (homozygous; albino)





A

a

A

AA

Aa

a

Aa

aa

3. What can you say about the genetic constitution of the parents of an albino child?


Both parents must carry at least one albino allele for a child to be albino.
4. Given what you know about human A-B-o blood types, what parental cross would yield an F1 consisting of 50% type A, 25% type AB, and 25% type B?
These are the typical ratios expected from a cross between an individual with type AB blood and an individual with type A (Ao) blood





A

B

A

AA

AB

o

Ao

Bo

5. As a genetic advisor to a hospital you are confronted with the following problem: two women had babies in the hospital at about the same time on the same day. Mrs. Cartman took home a boy (named Happy), and Mrs. Simpson took home a girl (named Joy). However, Mrs. Simpson thought she recalled the delivery room nurses commenting on her child as being a boy. In any case, she requests immediate clarification of the situation.


The blood types of all concerned parents were determined as follows:
Mr. Cartman = B Mrs. Cartman = AB Happy = A

Mr. Simpson = B Mrs. Simpson = o Joy = o


Has there been an interchange of babies? Support your conclusion or be fired.
Mrs. Cartman can have ONLY the genotype AB. Her gametes will be either A or B.
Mr. Cartman could be either BB or Bo. So there are two possible outcomes:





A

B

B

AB

BB

B

AB

BB


Or…





A

B

B

AB

BB

o

Ao

Bo


The Cartmans could not have produced a type o baby, as Joy is. There has been no exchange of babies.
Just to be sure, let's look at the Simpsons' genotypes and possible outcomes.
Mrs. Simpson can have ONLY the genotype oo. Her gametes can carry ONLY the o allele.
Mr. Simpson could be either BB or Bo. Again, there are two possible outcomes:





o

B

Bo


Or…





o

B

Bo

o

oo


Joy is their baby.
(Actually, you could have avoided all this work if you'd just been astute enough to notice that Mrs. Cartman is the only one in the bunch with an A allele. She is the only one of the four parents who could have provided the A allele that her son Happy
6. The polled (hornless) condition in cattle (H) is dominant over the horned condition (h) and is autosomal. A polled bull (Spike) is bred to three different cows with the following results:

Faye is horned (hh). She produces a polled calf.

Holly is horned (hh) She produces a horned calf.

Clover is polled (Hh). She produces a horned calf.


What are the genotypes of each animal?

Spike: Hh Faye: hh


Holly: hh Clover: Hh
What offspring would you expect from each of these matings? Punnet squares:
Spike x Faye…





H

h

h

Hh

hh

h

Hh

hh


Spike x Holly…





H

h

h

Hh

hh

h

Hh

hh


Spike x Clover…





H

h

H

HH

Hh

h

Hh

hh

7. In cats, yellow fur (B) is dominant to black fur (b). The heterozygous "tortoise-shell" or "calico" condition is exhibited only by the female, and is a result of mosaic expression. The alleles B and b are sex-linked. What offspring (genotypes, phenotypes and frequencies) would you expect from the cross of a black male with a yellow female?


Remember: the allele occurs only on the X chromosome.





XB

XB

Xb

XBXb

XBXb

Y

XBY

XBY


Offspring are expected to be 100% yellow if male, and 100% calico if female.
8. What kinds of offspring would you expect to result from the cross of a black male with a tortoise-shell female?





XB

Xb

Xb

XBXb

XbXb

Y

XBY

XbY


Females will be 50% calico, 50% black.

Males will be 50% yellow, 50% black.
9. How about the cross of two tortoise-shell animals?
This is highly unlikely, as the only male tortoise-shell animal would have Klinefelter's Syndrome, and have a sex chromosome genotype of XXY (or even more X chromosomes than two).
If such a mating were to occur, however, the colors of the offspring would be extremely difficult to predict, as the aneuploid sex chromosomes would migrate in various unpredictable ways during meiosis.
Trust us. Just get all those cats neutered or spayed and don't even try this.
10. In guinea pigs rough hair (R) is dominant with respect to smooth hair (r), and black hair (B) is dominant with respect to white (b). These genes are not linked (i.e., the locus for hair color is not on the same chromosome as the locus for hair texture), nor are they sex-linked. Cross a guinea pig heterozygous for hair texture and color, with a white-smooth-haired guinea pig. Describe the predicted F1 in terms of phenotype and genotype frequencies.
Heterozygous guinea pig: RrBb

White, smooth haired guinea pig: rrbb





RB

Rb

rB

rb

rb

RrBb

Rrbb

rrBb

rrbb



Expected phenotypes:

25% rough black

25% rough white

25% smooth black

25% smooth white
11. Over several years, a pair of guinea pigs produced the following offspring
25% Rough-black 25% Rough-White

25% Smooth-black 25% Smooth-White


Describe the phenotypes and genotypes of each parent.
They are the same guinea pigs in the first part of the question. RrBb and rrbb.
12. Alas! Your prize potato patch is infested with greedy, potato-munching leperchauns! Not to

be confused with the more attractive and popular "leprechaun" (Homunculus patrickii), the "leperchaun" (Homunculus odoratus) has a strongly disagreeable odor, and extremities that fall off messily when the organism is handled. This is an effective defense mechanism.

At your friendly neighborhood BillyBob's Bargain Barn of Biological Control, you purchase a pair of special, mutant ferrets--each of which is true breeding. This strain of ferret is highly prized by potato farmers because of two simple, autosomal traits affecting (1) coat color and (2) appetite for leperchauns.

Coat color has two alleles. Black fur (B) is dominant to the more desirable brown fur (b), which camouflages the ferrets as they lie in ambush in the potato patch. Diet preference in these voracious carnivores also has two alleles. Unfortunately, the allele which drives the ferrets to chase and devour leperchauns (p) is recessive to the gene which drives them to chase and devour your ankles (P).


Your pair consists of a true-breeding black, ankle-biting male and a true-breeding brown, leperchaun-eating female. What type of offspring do you expect to obtain in the F1?
Recall that "true breeding" for a particular trait means that the organism in question is homozygous for that trait.
A true-breeding black ankle biter will be BBPP

A true-breeding brown leprechaun-eater will be bbpp
The F1 generation of this crossing should be 100% BbPp
What about the F2?
The F2 generation is the result of a dihybrid cross of two individuals in the F1 generation:





BP

Bp

bP

bp

BP

BBPP

BBPp

BbPP

BbPp

Bp

BBPp

BBpp

BbPp

Bbpp

bP

BbPP

BbPp

bbPP

bbPp

bp

BbPp

Bbpp

bbPp

bbpp


Black ankle biter is coded in BLACK

Black leperchaun-eater is coded in BLUE

Brown ankle biter is coded in PINK

Brown leperchaun-eater is coded in GREEN
It's the typical 9:3:3:1 ratio you'd expect!
13. If a black, ankle-biting ferret is selected from the F2 progeny above, what is the probability that it will breed true in succeeding generations?
Of the nine black, ankle-biting weasels in the above cohort, only ONE (BBPP) will breed true in succeeding generations. There's a one in nine chance.
14. Over several months, a black, ankle-biting ferret and a brown, ankle-biting ferret yield the following phenotypic ratios in their offspring:
64 black, leperchaun-eating 59 brown, ankle-biting

18 black, leperchaun-eating 23 brown, leperchaun eating


What are the genotypes of the parents?
Since there are ANY brown, leperchaun eating offspring, you know that the black anklebiter must be heterozygous for both traits, and that the brown ankle-biter is heterozygous at the P locus. Hence, the two parents have the genotypes:
BbPp and bbPp, respectively. Doing a Punnett square for these genotypes, you get…





BP

Bp

bP

bp

bP

BBPP

BbPp

bbPP

bbPp

bp

BbPp

Bbpp

bbPp

bbpp


As you can see, this does reflect the percentages (out of 164 total babies) of the offspring these two weasels produced over their production times:
37.5% black ankle-biters (3/8)

12.5% black leperchaun-eaters (1/8)

37.5% brown ankle-biters (3/8)

12.5% brown leperchaun-eaters (1/8)
15. When investigators study human genetics, they often must examine "family trees,"

or pedigrees. The following page shows two different pedigrees, each following the expression of a single genetic trait. To figure out the type of inheritance involved, consider each generation separately, and determine phenotypic ratios in each set of offspring. If you can "work" the following human pedigrees, you will have mastered several fundamental aspects of Mendelian genetics!


Let us make the assumption that only four types of inheritance are involved in the pedigrees described.
autosomal dominant X-linked dominant

autosomal recessive X-linked recessive


The symbols used are:
o = male m = female
When the symbol is filled in, the individual expresses the trait. When the symbol is empty, the individual does not express the trait.
For example, the figure below represents a marriage in which the female exhibits the trait, but the male does not. The symbols on below the "parental" symbols represent the children of this marriage. In this case, both sons exhibit the trait, but the daughter does not.


Pedigrees are constructed by a process of elimination. Begin by writing down all possible genotypes for each individual. Rule out each type of inheritance as it becomes evident that, from the information given, a certain type of inheritance is not possible. If a certain type of inheritance "fits" the pedigree, it means only that that type of inheritance is possible in the given situation--not that you are absolutely correct.
On the next page you will find two pedigrees, each typical of a particular type of inheritance (autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive, Y-linked dominant, or Y-linked recessive). Examine the pedigree, and fill in the most likely genotypes for each individual on the family tree. Finally, given what you have hypothesized about the family, write the type of inheritance in the space provided below each pedigree.
Pedigree analysis: The family trees
Pedigree I.


Type of inheritance: Autosomal dominant
If this were autosomal recessive, you would not expect individual #7 to not express the trait. Note that matings between non-expressing individuals always produce non-expressing offspring. This tells you that the expressed trait is dominant.
Pedigree II


Type of inheritance: X-linked recessive
Note that only the sons of the affected mother #1 inherit the trait, as they receive the X chromosome only from their mother.
Individual #4 is a carrier; one of her sons has inherited the trait. The mating between #7 and #8 reveals that this is an X-linked recessive trait, since all the X chromosomes in this family carry the recessive allele, and all individuals, either homozygous (females) or hemizygous (males) will express it.



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