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HOW TO SOLVE GENETICS PROBLEMS

 

1. Read the problem.

2. Determine what traits are dominant and which are recessive. Often you must marshal background knowledge to do this – which may not be explicitly mentioned in the problem.

3. Are any letters assigned to the genes? If not, make some up. We usually take the dominant characteristic and use the first letter of that word. For example, if polydactyly (extra fingers) is dominant over the normal five–fingered condition, we would pick P for the dominant gene, and small p for the recessive normal allele.

4. Determine, if possible, the genotypes of the parents. In 9 out of 10 problems this information is given, or at least implied. Sometimes you have to deduce it from other information given. Write it down so that you can remember what it is, e.g. Pp.

5. Determine all the possible kinds of gametes that can be made by each parent. Be careful, remember that a gamete can ordinarily receive only one gene of a pair of alleles. This is the part that most people have trouble with! e.g. P p.

6. Make a Punnett square, using each of the gametes for one parent across the top of each column, those of the other parent go vertically. If you have done step 5 properly you shouldn’t have any trouble with this step.

7. Work the cross carefully.

8. Now read the problem again. Find out exactly what it is asking for. Don’t assume too much. This is another place where many people get lost.

9. In most problems, these steps should get you through adequately. Some are slightly altered – for example, if the genotype of one of the parents is unknown, and that is what the problem wants you to discover. You may assign that parent something like A_ or __ genotype and see if that helps. Put the offspring genotypes in the square and work backward. Remember this won’t get all the problems – there is still nothing like real understanding – but it can help organize your attack on a genetic problem. And of course, unless you understand the terms, such as homozygous, heterozygous, dominant, recessive, allele, and so on, you cannot begin to think of working problems.

Finally, the actual genetic information you need to solve these problems often appears concealed rather than revealed by the wording of the problem. Learn to translate such a sentence, “Mary is normally pigmented but had an albino father”, into its logical consequence: “Mary is heterozygous for albinism” and then into “Mary is Cc”. Notice that, in this kind of a problem you may need to solve several subsidiary problems before you can proceed with the final solution.


PROBLEMS AND SOLUTIONS

Theoretical problems

1. In human families it is often observed that certain characteristics may “skip” a generation, then reappear. How would you explain this in the light of the facts expounded by Mendel?

If a parent who is homozygous dominant for a trait marries a homozygous recessive person, then all of their children will express the dominant allele’s phenotype. These children would be carriers and when marrying another carrier or a homozygous recessive person, the recessive phenotype could reappear. e.g. Aa x Aa produce AA, Aa and aa.



 

2. In certain Norwegian families there is an inherited condition known as “woolly hair”. Individuals showing this characteristics have hair which resembles sheep’s wool. A study of family pedigrees shows that a person never has woolly hair unless at least one parent also has woolly hair. How would this character most likely be inherited? Explain.

Dominant traits do not skip generations. Only one dominant allele is required for expression in the phenotype. Since woolly hair only occurs when one parent has it too, it is due to a dominant allele not a recessive one.

3. The hornless condition in cattle is dominant over horned. A cattleman has a herd of hornless cattle only, but some horned cattle occasionally appear. These are removed from the range before they can reproduce. Assuming that this man has good fences which can keep out stray bulls, how could this be explained?

One of this farmer’s original cattle was a hornless carrier animal for the gene for horns. When the original cattle reproduce, most of their offspring will be hornless carriers of the recessive allele for horns. Occasionally, during mating, two recessive alleles will meet to produce a horned animal. These homozygous animals are quickly removed by the farmer so that they won’t reproduce with other cattle. It is unlikely that the recessive allele will quickly be removed especially if the Hardy Weinberg conditions apply.

Determining of the first generation genotype and phenotype (I-st Mendel’s Law)

4. In Holstein cattle the spotting of the coat is due to a recessive gene while a solid–coloured coat is dominant. What types of offspring might be produced by a cross between two spotted animals? Show how you reach your conclusion. The gene P is responsible for coat pattern.

Let P be the dominant allele for a solid coloured coat.

Let p be the recessive allele for a spotted coat.

Parents: pp x pp (no P present)

Punnett table

sperm sperm

p p
p pp pp
p pp pp

egg

egg

All of the offspring of two spotted animals are also spotted.

5. In cats the gene for short hair is dominant over the gene for long hair (angora). A short–haired tom is mated with an Angora female. She bears eight kittens, six short–haired and two with long hair. How do these numbers compare with the expected ratio? If you mated these same cats four more times and obtained a total of forty offspring, would you expect the results to be a closed approximation of the expected ratios? Explain.

Let gene H control the length of the cat’s hair

Let H be the dominant allele for short hair.

Let h be the recessive allele for long hair.

Short–haired tom: H_ (we don’t know the second allele for sure)

Angora female: hh

Two of their offspring are long–haired with a genotype of hh. Therefore, each parent contributed an h allele and therefore the tom cat father has the genotype Hh.

Punnett table

eggs (all the same)

h h
H Hh Hh
h hh hh

sperm

sperm

Expected ratio is 1 Hh : 1 hh

Actual ratio is 3 Hh : 1 hh

Over a long range of matings, the actual ratio should approach the expected ratio.

6. In summer squash, white coloured fruit is dominant over yellow. If you place pollen from a yellow–fruited plant on the pistil of a hybrid white–fruited ( heterozygous ) plant, what type of seeds would you expect from the seed which come from this cross? Let the gene C control fruit colour in summer squash.

Let C be the dominant allele for a white–coloured fruit.

Let c be the recessive allele for a yellow–coloured fruit.

Punnett table

eggs eggs

C c
c Cc cc

Pollen

Cc : hybrid white–fruited plant

Cc : yellow–fruited plant

You can expect hybrid white–fruited plants and yellow–fruited plants in a 1 : 1 ratio

 

7. In Drosophila, vestigial wings and ebony colour are due to two separate recessive genes. The dominant alleles are normal (long) wings and normal (gray) body colour. What type of offspring would you expect from a cross between a homozygous vestigial ebony female and a normal double homozygous (long–winged, gray–bodied) male? If the F1 are allowed to breed among themselves what types of offspring would you expect in the F2? Show complete genotype and phenotype of both generations.

This is a dihybrid cross.

Gene L controls wing length and gene G controls body colour.

Let L be the dominant allele for normal long wings.

Let l be the recessive allele for vestigial wings.

Let G be the dominant allele for normal gray body colour.

Let g be the recessive allele for ebony body colour.

Homozygous vestigial ebony female : llgg

Normal (long–winged, gray–bodied) male : LLGG

Parents : llgg x LLGG

Punnett table

eggs

  lg
LG LlGg

 

sperms

 

Offspring from this mating are long–winged, gray–bodied Drosophila.

We raise an F2 by mating the F1 among themselves: LlGg x LlGg

 

Independent assortment of chromosomes

F1 gametes

F1 gamete

Types of offspring in the F2 generation of a dihybrid cross:

GENOTYPE PROPORTIONS PHENOTYPE
LLGG 1 / 16 normal (long, gray)
LLGg 2 / 16 Normal
LlGG 2 / 16 Normal
LlGg 4 / 16 Normal
LLgg 1 / 16 Long winged, gray body
Llgg 2 / 16 Long winged, ebony body
llGG 1 / 16 Vestigial wings, gray body
llGg 2 / 16 Vestigial wings, gray body
llgg 1 / 16 Vestigial wings, ebony body

 

8. Some dogs bark while trailing, others are silent. The barking trait is due to a dominant gene. Erect ears are dominant to drooping ears. What kind of pups would be expected from a double heterozygous erect–eared, barker mated to a drooped–eared, silent trailer? Gene B controls the barking ability; gene E controls ear shape.

Let B be the dominant allele for the barking trait.

Let b be the recessive allele for the silent trait.

Let E be the dominant for erect ears.

Let e be the recessive allele for drooping ears.

Heterozygous erect–eared barker: BbEe

Drooped–eared, silent trailer: bbee

 

Punnett table

  BE Be bE be
be BbEe Bbee bbEe bbee

Expected pups:

Erect – eared (heterozygous) barker: BbEe

Drooping – eared (homozygous) barker (heterozygous): Bbee

Erect – eared (heterozygous) silent (homozygous) trailer: bbEe

Drooping – eared silent trailer: bbee

 

First and second Mendel’s Laws

9. In Drosophila, vestigial wings and ebony colour are due to two separate recessive genes. The dominant alleles are normal (long) wings and normal (gray) body colour. What type of offspring would you expect from a cross between a homozygous vestigial ebony female and a normal double homozygous (long–winged, gray–bodied) male? If the F1 are allowed to breed among themselves what types of offspring would you expect in the F2? Show complete genotype and phenotype of both generations.

This is a dihybrid cross.

Gene L controls wing length and gene G controls body colour.

Let L be the dominant allele for normal long wings.

Let l be the recessive allele for vestigial wings.

Let G be the dominant allele for normal gray body colour.

Let g be the recessive allele for ebony body colour.

Homozygous vestigial ebony female : llgg

Normal (long–winged, gray–bodied) male : LLGG

Parents : llgg x LLGG

Punnett table

eggs

  lg
LG LlGg

 

sperms

 

Offspring from this mating are long–winged, gray–bodied Drosophila.

We raise an F2 by mating the F1 among themselves: LlGg x LlGg

 

Independent assortment of chromosomes

F1 gametes

F1 gamete

Types of offspring in the F2 generation of a dihybrid cross:

GENOTYPE PROPORTIONS PHENOTYPE
LLGG 1 / 16 normal (long, gray)
LLGg 2 / 16 Normal
LlGG 2 / 16 Normal
LlGg 4 / 16 Normal
LLgg 1 / 16 Long winged, gray body
Llgg 2 / 16 Long winged, ebony body
llGG 1 / 16 Vestigial wings, gray body
llGg 2 / 16 Vestigial wings, gray body
llgg 1 / 16 Vestigial wings, ebony body

10. If you made a test cross of the F1 males of the preceding problem what results would you expect to obtain? ( A test cross is a backcross .)

F1 male is LlGg x llgg female (homozygous recessive)

F1 sperms

  LG Lg lG lg
lg LlGg Llgg llGg llgg

eggs

Phenotypes are: 1 long, gray 1 vestigial, gray

1 long, ebony 1 vestigial, ebony.

11. In tomatoes, yellow fruit and dwarfed vine are due to recessive alleles of genes which produce the more common red fruit and tall vine. If pollen from the pure–line dwarf plant bearing red fruit is placed on the pistil of a pure–line tall plant bearing yellow fruit, what type of plant and fruit would be expected in the F1 ? If these are crossed among themselves, what results would be expected in the F2 ?

This is a dihybrid cross.

Gene R controls fruit color; gene T controls height.

Let R be the dominant allele for red fruit.

Let r be the recessive allele for yellow fruit.

Let T be the dominant allele for tall vine.

Let t be the recessive allele for dwarf vine.

Pureline (homozygous) dwarf plant bearing red fruit: ttRR

Pureline (homozygous) tall plant bearing yellow fruit: TTrr

Punnett table

  Tr
tR TtRr

 

F1 offspring are all red, tall–vined plants

TtRr x TtRr

TR Tr tR tr
TR TTRR TTRr TtRR TtRr
Tr TTRr TTrr TtRr Ttrr
tR TtRR TtRr ttRR ttRr
tr TtRr Ttrr ttRr ttrr

12. In pigeons, the checkered pattern is dependent on a dominant gene C and plain on the recessive allele c. red colour is controlled by a dominant gene B and brown by the recessive allele b. diagram completely a cross between homozygous checkered, red and plain, brown birds. Summarize the expected F2 results.

Let C be the dominant allele for checkered feathers and c for the recessive plain.

Let B be the dominant allele for red feathers and b for the recessive allele for brown. Genotype of homozygous checkered red birds: CCBB

Genotype of plain, brown birds: ccbb

Punnett table

  cb
CB CcBb

 

F1 offspring are heterozygous checkered, red birds

F2 offspring: 9 checkered red birds (express both dominant traits).

3 checkered brown birds (express one dominant trait).

3 plain brown bird (express both recessive traits).

1 plain brown bird (express both recessive traits).

 

 


Date: 2015-02-28; view: 2714


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