Genetics-(higher level)

Genetic Crosses†††††††††††††††††† Dominant and Recessive Genes††††††††† Incomplete Dominance

Pedigree Studies†††††††††††††††† Sex Determination††††††††††††† Gregor Mendel

Mendelís 1st Law††††††††††††††††† Mendelís 2nd Law†††††††††††††††† Summary of Mendelís Laws

Gene Linkage††††††††††††† Sex Linkage

††††††††† Genetics is the science of heredity and how new life changes and varies in their characteristics. Sexual reproduction in humans involves 2 gametes; one male gamete, the sperm, and one female gamete, the egg.

As was discussed in the cell division webpage the gametes are haploid (n). When fertilisation occurs the resulting fertilised egg is diploid (2n) Genetic variations occur as a result of this union.

As was discussed in the heredity webpage the physical characteristics of organisms are developed as the protein builds up their bodies. These proteins are formed as a result of genes carried on chromosomes.

In genetics genes are represented by letters. There are usually 2 different types of the same gene; one is dominant and one is recessive. An example of this is: T is a gene for tall and t is a gene for short. The two versions of the same gene are called alleles. The two alleles are formed at the same position, or locus, on the chromosome.

REMEMBER: THE 2 ALLELES ARE GOTTEN FROM THE 2 GAMETES OF SEXUAL REPRODUCTION. THAT IS WHY THEY CAN DIFFER.

IN ASSEXUAL REPRODUCTION THE NEW LIFE IS ALWAYS IDENTICAL TO THE PARENT CELL BECAUSE THERE IS ONLY 1 PARENT.

Dominant genes will always prevent the recessive gene from working. If a person has 2 dominant alleles for tallness: T T then he will be tall. If a person has 1 dominant and I recessive allele for tallness: T t then he will be tall. The only way the recessive gene will be expressed is if he has 2 recessive alleles for short:††† t t then he will be short.

If the pair of genes controlling the characteristic has identical alleles, TT or tt, we call it a homozygous pair of genes. If the pair of genes controlling the characteristic are different alleles, Tt or Cr, or Cw we call it a heterozygous pair of genes.

When we express the genotype of a characteristic we state the pair of alleles. Genotype examples are TT, Tt, tt, BB, Bb, bb, etc.

When we state the physical characteristic expressed by the genotype we are stating is phenotype. The phenotype for TT is tall while the phenotype for tt is short. These phenotypes could vary because of environment effects. This is especially true in terms of genotypes and phenotypes for intelligence. The upbringing, environment, and education experiences greatly affect the phenotype.

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GENETIC CROSSES

REMEMBER: THE 2 ALLELES ARE GOTTEN FROM THE 2 GAMETES OF SEXUAL REPRODUCTION. THAT IS WHY THEY CAN DIFFER.

When working on genetic crosses you must state the capital letter which represents the dominant allele and the small letter that represents the recessive allele.

The following is an example of how to work out genetic crosses:

   

 

     A is a dominant characteristic.

 

 

     a is a recessive characteristic.

This bird has two genes for red feathers.

  Its genotype is AA.

Its phenotype is RED

This bird has two genes for blue feathers.

  Its genotype is aa.

Its phenotype is BLUE

This Punnett Square shows how we can diagram the genes.

The orange bird has two dominant A genes. We put two A s along the top of the square.

The blue bird has two recessive a genes. We put two a s down along the left side of the square.

All the offspring have the genes Aa.

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Dominant and Recessive Genes


They will all have orange feathers (phenotype), but will carry a recessive gene for blue feathers (genotype).The progeny are the offspring produced.
This is called the F1 generation progeny.

 

Now suppose that two individuals from the F1 generation become parents. Here they are!
The baby birds are called the F2 generation. You can see how their genes work out. The offspring are coded in the squares. One bird will be orange with two AA genes.
Two birds will be orange with genes coded Aa.
One bird will be blue and will have two recessive aa genes. Individual nests of birds may not turn out exactly like this, but if there are many baby birds, they will work out genetically with the ratios 1:2:1.

Genotype: 1 AA, 2 Aa, 1 aa

††††††††††††††††††††††††††††††††††††††††††††††† Phenotype: 3 orange feathers, 1 blue feathers

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Incomplete Dominance

Incomplete dominance is the situation where two different alleles are equally dominant. When this occurs the heterozygous genotype that is produced is an intermediate phenotype (blend) between the two respective homozygous genotypes. This is also called codominance.

In this example AA genotypes have red, Aa genotypes pink and aa genotypes whitish flowers. Note that the heterozygous genotype Aa is a blend of red and white. ††

Note: This is the F1 generation.

 

 

 

In the F2 generation two pink flowers will produce 2 pink phenotypes, 1 red phenotype, and 1 white phenotype.

 

 

 

 

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Pedigree Studies

††††††††† A pedigree is a diagram showing the genetic history of a group of related organisms.

A pedigree showing the occurrence of a recessive trait in three generations of a family

.

Circles in a pedigree represent females and squares represent males. A horizontal line between a circle and a square indicates a marriage or partnership. Vertical lines indicate the children from the marriage or partnership. In this activity, a filled-in circle or square shows that the individual has both alleles for the trait. A half-filled-in circle or square indicates that the individual has one recessive allele for the trait.

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Sex Determination

Our cells have 46 chromosomes. There are 23 pairs. Remember, we get 23 from our mother and 23 from our father. We have 22 pairs of chromosomes called autosomes. These are the chromosomes that control our body growth, enzymes, etc. We have 1 pair of sex chromosomes.

If the person is a male that pair is composed of an X and a Y chromosome. The maleís genotype is XY.

If the person is a female she will have a sex chromosome composed of 2 Y chromosomes. The femaleís genotype is XX.

The Punnett Square for the fertilisation of a sperm and an egg is:

If the child has a genotype of XX then it becomes a girl. If the child has a genotype of XY then it becomes a boy.

As you can see the ratio of males to females is 1:1.

†††††††††

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Gregor Mendel

Gregor Mendel studied 7 characteristics of pea plants. He studied:

 

As a result of Mendelís work the study of genetics began. He discovered that, although an organism may have genotypes for 2 different physical traits (phenotypes) the organism will only exhibit one of those traits.

The work he did for the height of the plants is a follows:

He discovered although the parents of the first generation had TT and tt genotypes all the progeny of the F1 generation were tall.

He then discovered that in the F2 generation there was a 3:1 ratio between tall and short.

The Punnett Square for Mendelís F1 generation is:


The Punnett Square for Mendelís F2 generation is:

 

 

 

Note that in the F1 generation the parents were homozygous. One was TT and one was tt. All the progeny were tall because all the progeny had Tt genotypes with T being the dominant characteristic.

In the F2 generation a Tt plant was used to self-pollinate itself. All of the phenotypes in this combination were Tt. As a result the F2 progeny were 3 Tt and1 tt.

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Mendelís Laws

First Law- The Law of Segregation

1.           This law states:

a.   In diploid organisms, chromosomes occur in matching pairs. These pairs are called homologous chromosomes. These are chromosomes that pair at meiosis, have the same length and banding pattern plus carrying the same genes at the same locus. They have the same sequence of genes.

 

Notice that these two chromosomes are homologous because they have alleles at the same position on the chromosome but one allele is for purple flowers and the other for white flowers.

b.   For each characteristic or trait organisms inherit two alternative forms of that gene, one from each parent. These alternative forms of a gene are called alleles.

c.   When gametes (sex cells) are produced, allele pairs separate or segregate leaving them with a single allele for each trait.

d.   When the two alleles of a pair are different, one is dominant and the other is recessive.

Example:

If your eyes are blue, green or grey you have two alleles for blue eyes (bb), then your gametes must have a blue allele (b); if your eyes are brown you might have two brown allele (BB), then your gametes have one allele for brown (B) or you might have one allele of each kind (Bb), in which case you make two kinds of gametes some contain the brown allele (B) and some contain the blue allele (b).

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Second Law- Law of Independent Assortment

Mendelís Second Law involves dihybrid crosses. Dihybrid crossing involves the study of 2 characteristics at the same time.The Law of Independent Assortment states that alleles for different traits are distributed to sex cells (& offspring) independently of one another.

Mendel noticed during all his work that the height of the plant and the shape of the seeds and the color of the pods had no impact on one another.  In other words, being tall didn't automatically mean the plants had to have green pods, nor did green pods have to be filled only with wrinkled seeds, the different traits seem to be inherited independently.

The genotypes of our parent pea plants will be:

RrGg x RrGg where
"R" = dominant allele for round seeds
"r" = recessive allele for wrinkled seeds
"G" = dominant allele for green pods
"g" = recessive allele for yellow pods

Notice that we are dealing with two different traits: (1) seed texture (round or wrinkled) & (2) pod color (green or yellow).  Notice also that each parent is hybrid for each trait (one dominant & one recessive allele for each trait).

 

RG

Rg

rG

rg

 

RG

RRGG
round

RRGg
round

RrGG
round

RrGg
round

 

Rg

RRGg
round

RRgg
round

RrGg
round

Rrgg
round

 

rG

RrGG
round

RrGg
round

rrGG
wrinkled

rrGr
wrinkled

 

Rg

RrGg
round

Rrgg
round

rrGg
wrinkled

rrgg
wrinkled

 

We need to "split" the genotype letters & come up with the possible gametes for each parent.  Keep in mind that a gamete (sex cell) should get half as many total letters (alleles) as the parent and only one of each letter. So each gamete should have one "R or r" and one "G or g" for a total of two letters. There are four possible letter combinations: RG, Rg, rG, and rg. So, when the two parentsí gametes form a new organism the punnett square will look like this:
 

The results from a dihybrid cross are always the same:

9/16 boxes (offspring) show dominant phenotype for both traits (round & green),
3/16 show dominant phenotype for first trait & recessive for second (round & yellow),
3/16 show recessive phenotype for first trait & dominant form for second (wrinkled & green), &
1/16 show recessive form of both traits (wrinkled & yellow).

So, as you can see from the results, a green pod can have round or wrinkled seeds, and the same is true of a yellow pod.  The different traits do not influence the inheritance of each other.  They are inherited INDEPENDENTLY.

Interesting to note is that if you consider one trait at a time, we get "the usual" 3:1 ratio of a single hybrid cross (like we did for the Law of Segregation). For example, just compare the color trait in the offspring; 12 green & 4 yellow (3:1 dominant: recessive).  The same deal with the seed texture; 12 round & 4 wrinkled (3:1 ratio).  The traits are inherited INDEPENDENTLY of each other.

Animation Explaining Independent Assortment of Genes

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SUMMARY OF MENDELíS LAWS

LAW

PARENT CROSS

OFFSPRING

DOMINANCE

TT x tt
tall x short

100% Tt
tall

SEGREGATION

Tt x Tt
tall x tall

75% tall
25% short

INDEPENDENT ASSORTMENT

RrGg x RrGg
round & green x round & green

9/16 round seeds & green pods
3/16 round seeds & yellow pods
3/16 wrinkled seeds & green pods
1/16 wrinkled seeds & yellow pods

 

Click here to go to practice questions using Punnett squares

 

Click here to go to practice questions about genetics and Mendelís Laws

 

Interactive animation with Mendelís peas

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Linkage (Gene Linkage)

Gene linkage occurs when traits for 2 separate characteristics occur on the same chromosome.


The characters Mendel examined happened to be on separate chromosomes. That is why he observed independent assortment. If, however, the genes are on the same chromosomes, they will be inherited together. For example, consider the following parental nuclei. Both father and mother have a pair of chromosomes with alleles for two different genes:


If we look at this with a punnet square what is going to happen in the next generation:

The phenotype ratios are still 3:1, but there are fewer genotype combinations than in the usual cross involving two alleles.

Remember: With independent assortment the phenotypes resulted in a 9:3:3:1 ratio.

 

View this animation of crossing over

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Sex-linked Genes or Sex linkage

†††††† Genes or traits whose controlling genes are on the X sex chromosome but not on the Y sex chromosome. As a result recessive phenotype occurs more often in males than in females.

There is yet another, unrelated, special case that means something totally different, yet has a similar-sounding name. This is sex-linked genes, genes located on one of the sex chromosomes (X or Y) but not the other. Since, typically the X chromosome is longer, it bears a lot of genes not found on the Y chromosome, and thus most sex-linked genes are X-linked genes. One example of a sex-linked gene is fruit fly eye colour. An X chromosome carrying a normal, dominant, red-eyed allele would be symbolized by a plain X, while the recessive, mutant, white-eyed allele would be symbolized by X' or Xw. A fly with genotype XX' would normally be a female with red eyes, yet be a carrier for the white-eyed allele. Because a male typically only has one X chromosome, he would normally be either XY and have normal, red eyes or X'Y and have white eyes. The only way a female with two X chromosomes could have white eyes is if she would get an X' allele from both parents making her X'X' genotype. The cross between a female carrier and a red-eyed male would look like this:

 

 X 

 Y 

 X 

XX

XY

XX'

X'Y

 X' 

Notice that while there is a ďtypicalĒ ratio of ĺ red-eyed to ľ white-eyed, all of the white-eyed flies are males.

Sex-linked traits act just like recessive ones except they also bow to the will of the sex of the child. Genes are carried on things called chromosomes. Most people already know that in humans the man has an X and Y chromosome and the female has two X chromosomes. This is the reason that only the man can determine the sex of the child. Women can only provide X chromosomes while a man can provide either. Now, the X chromosome is bigger and can carry more genetic information on it than the Y can.

This is where sex-linked traits come in. Because the X is bigger it means that some genes carried on it are not carried on the Y chromosome. These genes can be expressed even without a corresponding partner on an X chromosome. They also cannot be blocked out unless there is another X chromosome carrying a dominant partner.

In humans, two well-known X-linked traits are haemophilia and red-green colour-blindness. Haemophilia is the failure (lack of genetic code) to produce certain substance needed for proper blood-clotting, so a haemophiliacís blood doesnít clot, and (s)he could bleed to death from an injury that a normal person might not even notice.

One human sex-linked trait is Haemophilia. Haemophilia is a disease that keeps a person's blood from clotting when he is cut. Haemophiliacs can easily bleed to death and must be very careful not to injure themselves. Many also take daily injections to help the problem. Because haemophilia is a sex-linked disease most of the people who have it are men. Women can carry the gene for haemophilia but will not be affected by it because their second X chromosome will block it out with a healthy gene. They must have two copies of the defective gene to display the disease. Inheriting two copies is highly unlikely. Men carrying haemophilia do not have another X chromosome so they will have haemophilia with only one gene for it. Mothers carrying one gene for haemophilia take a great risk with having children because there is a 50% chance their sons will end up with the disease. Here's how it works:

How hemophilia is carried.

As you can see, at least half of her children (boxes 1 and 2) will inherit the defective gene. One, a daughter, will only carry the gene. The other, a son, will have the disease haemophilia. The last two children (boxes 3 and 4) will carry healthy genes. Of course these are only the possibilities of what her children could end up with. She could very well end up giving it to all her children or none at all. It's just a matter of chance. Now let's take a look at what will happen if this woman's haemophiliac son has children with a healthy woman:

How hemophilia is carried.

In this case half the children will still inherit the gene. All the sons will be safe but all the daughters will end up carrying haemophilia and could end up passing it on to their children. It is in this way that sex-linked genes can disappear and reappear from generation to generation.

 

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Review Questions

1. Which of the following is a possible abbreviation for a genotype?

A. BC
B. Pp
C. Ty
D. fg

2. What is the best way to determine the phenotype of the feathers on a bird?

A. analyze the bird's DNA (genes)
B. look at the bird's feathers
C. look at the bird's beak
d. examine the bird's droppings

3. Which of the following pairs is not correct?

A. kk = hybrid
B. hybrid = heterozygous
C. heterozygous = Hh
D. homozygous = RR

4. The genes present in an organism represent the organism's __________.

A. genotype
B. phenotype
C. physical traits

5. Which choice represents a possible pair of alleles?

A. k & t
B. K & T
C. K & k
D. K & t

6. How many alleles for one trait are normally found in the genotype of an organism?

A. 1
B. 2
C. 3
D. 4

7. Which statement is not true?

A. genotype determines phenotype
B. phenotype determines genotype
C. a phenotype is the physical appearance of a trait in an organism
D. alleles are different forms of the same gene

8. Which cross would best illustrate Mendel's Law of Segregation?

A. TT x tt
B. Hh x hh
C. Bb x Bb
D. rr x rr

9. In the cross Yy x Yy, what percent  of offspring would have the same phenotype as the parents?

A. 25%
B. 50%
C. 75%
D. 100%

10. In a certain plant, purple flowers are dominant to red flowers.  If the cross of two purple-flowered plants produces somepurple-flowered and some red-flowered plants, what is the genotype of the parent plants?

A. PP x Pp
B. Pp x Pp
C. pp x PP
D. pp x pp

 

Base questions #11-14 on the following information:

A white-flowered plant is crossed with a pink-flowered plant.  All of the F1 offspring from the cross are white.

11. Which phenotype is dominant?

12. What are the genotypes of the original parent plants?

13. What is the genotype of all the F1 offspring?

14. What would be the percentages of genotypes & phenotypes if one of the white F1 plants is crossed with a pink-flowered plant?

15. Which of Mendel's Laws is/are illustrated in this question?

16. Crossing two dihybrid organisms results in which phenotypic ratio?

A. 1:2:1
B. 9:3:3:1
C. 3:1
D. 1:1

17. The outward appearance (gene expression) of a trait in an organism is referred to as:

A. genotype
B. phenotype
C. an allele
D. independent assortment

 


18. In the homologous chromosomes shown in the diagram, which is a possible allelic pair?

A. cD
B. Ee
C. AB
D. ee
 

 


19. The phenotype of a pea plant can best be determined by:

A. analyzing its genes
B. looking at it
C. crossing it with a recessive plant
D. eating it

20. Mendel formulated his Law of Segregation after he had:

A. studied F1 offspring
B. studied F2 offspring
C. produced mutations
D. produced hybrids

21. Which cross would produce phenotypic ratios that would illustrate the Law of Dominance?

A. TT x tt
B. TT x Tt
C. Tt x Tt
D. tt x tt

22. The mating of two curly-haired brown guinea pigs results in some offspring with brown curly hair, some with brown straight hair, some with white curly hair, and even some with white straight hair.  This mating illustrates which of Mendel's Laws?

A. Dominance
B. Segregation
C. Independent Assortment
D. Sex-Linkage

 

Click here to go to the answers

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Answers


1. Which of the following is a possible abbreviation for a genotype?

A. BC
B. Pp - genotypes are made up of 2 of the same letter (either 2 capital, 2 lowercase, or one of each)
C. Ty
D. fg

2. What is the best way to determine the phenotype of the feathers on a bird?

A. analyze the bird's DNA (genes)
B. look at the bird's feathers - "phenotype of the feathers" means what the feathers look like, so look at them
C. look at the bird's beak
d. examine the bird's droppings

3. Which of the following pairs is not correct?

A. kk = hybrid - Kk would be hybrid (one capital, one lowercase of the same letter)
B. hybrid = heterozygous
C. heterozygous = Hh
D. homozygous = RR

4. The genes present in an organism represent the organism's __________.

A. genotype
B. phenotype
C. physical traits

5. Which choice represents a possible pair of alleles?

A. k & t
B. K & T
C. K & k - allele means 2 forms of the same gene. so this choice shows 2 forms of the same letter K or k
D. K & t

6. How many alleles for one trait are normally found in the genotype of an organism?

A. 1
B. 2 - one allele is inherited from each parent
C. 3
D. 4

7. Which statement is not true?

A. genotype determines phenotype - (note that the environment does play a role in influencing phenotype too)
B. phenotype determines genotype
C. a phenotype is the physical appearance of a trait in an organism
D. alleles are different forms of the same gene - (see question #5)

8. Which cross would best illustrate Mendel's Law of Segregation?

A. TT x tt
B. Hh x hh
C. Bb x Bb - both parent show dominant trait, but some recessive offspring will be produced (each parent carries a "b")
D. rr x rr

9. In the cross Yy x Yy, what percent  of offspring would have the same phenotype as the parents?

A. 25%
B. 50%
C. 75% - in the completed p-square, 3 of 4 boxes will have at least 1 "Y", producing the dominant phenotype (same as parents)
D. 100%

10. In a certain plant, purple flowers are dominant to red flowers.  If the cross of two purple-flowered plants produces somepurple-flowered and some red-flowered plants, what is the genotype of the parent plants?

A. PP x Pp
B. Pp x Pp - for any offspring to be recessive, each parent MUST have at least one "p"
C. pp x PP - only one parent is purple, this CAN'T be an answer
D. pp x pp - neither parent is purple, this CAN'T be an answer

 

 

Base questions #11-15 on the following information:

A white-flowered plant is crossed with a pink-flowered plant.  All of the F1 offspring from the cross are white.

11. Which phenotype is dominant? white

12. What are the genotypes of the original parent plants? WW (pure white) x ww (pink)

13. What is the genotype of all the F1 offspring? Ww (white)

14. What would be the percentages of genotypes & phenotypes if one of the white F1 plants is crossed with a pink-flowered plant?

50% heterozygous  white & 50% homozygous recessive pink.
 

The cross for this question would be "Ww (white F1) x ww (pink)".
The alleles of the  white parent are above the columns & those of the pink parent are in front of the rows. 2 of 4 boxes (50%) are "Ww", which is heterozygous & would have the dominant trait (white).  The other 2 of 4 boxes (50%) are "ww", which is homozygous recessive & would have the recessive trait (pink).
 
 

15. Which of Mendel's Laws is/are illustrated in this question? Dominance is illustrated by the original cross (WW x ww).

 

16. Crossing two dihybrid organisms results in which phenotypic ratio?

A. 1:2:1 - genotype ratio of a hybrid cross, ex: Tt x Tt
B. 9:3:3:1- dihybrid means hybrid for two different traits. An example could be GgYy x GgYy.
C. 3:1 - phenotype ratio of a hybrid cross
D. 1:1

17. The outward appearance (gene expression) of a trait in an organism is referred to as:

A. genotype
B. phenotype
C. an allele
D. independent assortment

 


18. In the homologous chromosomes shown in the diagram, which is a possible allelic pair?

A. cD
B. Ee- a possible allelic pair but NOT SHOWN IN THE DIAGRAM, so this CAN'T be an answer
C. AB
D. ee - an "allelic pair" is always two forms of the same letter.  In this example they are two lowercase "e's".
 

 


19. The phenotype of a pea plant can best be determined by:

A. analyzing its genes
B. looking at it
C. crossing it with a recessive plant
D. eating it

20. Mendel formulated his Law of Segregation after he had:

A. studied F1 offspring -
B. studied F2 offspring - he crossed two hybrids (F1's) and got a second generation --- the F2.
C. produced mutations - Mendel knew NOTHING about mutations so this CAN'T be an answer
D. produced hybrids

21. Which cross would produce phenotypic ratios that would illustrate the Law of Dominance?

A. TT x tt - one parent tall, the other short, all offspring would be tall
B. TT x Tt
C. Tt x Tt - illustrates Segregation
D. tt x tt

22. The mating of two curly-haired brown guinea pigs results in some offspring with brown curly hair, some with brown straight hair, some with white curly hair, and even some with white straight hair.  This mating illustrates which of Mendel's Laws?

A. Dominance
B. Segregation
C. Independent Assortment - the question involves two different traits (hair color & hair texture), this is the only law that deals with two different traits
D. Sex-Linkage - Mendel knew NOTHING about sex-linkage so this CAN'T be an answer

 

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The Punnett Square Practice Page

 

 

QUESTION #1

Let's say that in seals, the gene for the length of the whiskers has two alleles.  The dominant allele (W) codes long whiskers & the recessive allele (w) codes for short whiskers.

a)    What percentage of offspring would be expected to have short whiskers from the cross of two long-whiskered seals, one that is homozygous dominant and one that is heterozygous?

b)    b) If one parent seal is pure long-whiskered and the other is short-whiskered, what percent of offspring would have short whiskers?

ANSWER

 

QUESTION #2

In purple people eaters, one-horn is dominant and no horn is recessive. Draw a Punnett Square showing the cross of a purple people eater that is hybrid for horns with a purple people eater that does not have horns. Summarize the genotypes & phenotypes of the possible offspring.

ANSWER


 

 

 

QUESTION #3

A green-leafed luboplant (not a real plant) is crossed with a luboplant with yellow-striped leaves.  The cross produces 185 green-leafed luboplants. Summarize the genotypes & phenotypes of the offspring that would be produced by crossing two of the green-leafed luboplants obtained from the initial parent plants.

ANSWER

 

QUESTION #4

Mendel found that crossing wrinkle-seeded plants with pure round-seeded plants produced only round-seeded plants.  What genotypic & phenotypic ratios can be expected from a cross of a wrinkle-seeded plant & a plant heterozygous for this trait (seed appearance)?

ANSWER

 

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Punnett Square Solutions

 

Punnett Square Solutions Question #1

In seals, the gene for the length of the whiskers has two alleles.  The dominant allele (W) codes long whiskers & the recessive allele (w) codes for short whiskers.

a)  What percentage of offspring would be expected to have short whiskers from the cross of two long-whiskered seals, one that is homozygous dominant and one that is heterozygous? ANSWER: 0%.

I personally like to write down the info given in the question on my paper first.  So I start by writing:
W = allele for long whiskers
w = allele for short whiskers
A homozygous dominant seal would be "WW" (homozygous dominant = 2 CAPITAL letters).
A heterozygous seal would be "Ww" (heterozygous = 1 CAPITAL & 1 lowercase).
The cross is in the question therefore: WW x Ww.

The Punnett Square would look like this:

The possible gametes from the homozygous parent seal are on the left in front of the rows, & the possible gametes from the heterozygous parent are above the columns. We fill in the boxes by copying "one letter from the left, one letter from the top".

Analyzing our results, we find that 50% of our offspring (2 of 4 boxes) are "WW", and 50% (2 of 4 boxes) are "Ww". In terms of phenotype (what they would look like) 100% would have long whiskers (because all of the offspring have at least one "W", which codes for long whiskers).

So the answer to question 1a is: 0% would have short whiskers.  The only way to have short whiskers is to be "ww", and that combo is not possible from the parents in this cross.

b) If one parent seal is pure long-whiskered and the other is short-whiskered, what percent of offspring would have short whiskers? ANSWER: 0%.

Again, I suggest starting by defining symbols like so:
W = allele for long whiskers
w = allele for short whiskers
"Pure" is the same as homozygous, so "pure long-whiskered" would be "WW".
If you're a seal, the only way to have short whiskers is to have the homozygous recessive genotype, in other words be "ww".
So our cross is: WW x ww.

The Punnett Square would be:

The alleles from the long-whiskered parent (WW) are out in front of the rows (at the left), & the alleles of the short-whiskered parent are above the columns.  By the way, we could switch that around & it would not change our answer at all.  What I'm saying is: it doesn't matter where you put the parents (top or side).
Anyway, all our offspring (4 of 4 boxes) have the same genotype: "Ww" & would all end up with long whiskers. To summarize the offspring:
genotype = 100% heterozygous (Ww)
phenotype = 100% long-whiskered.

So our answer to Question 1b is also: 0% would be short-whiskered.
 
 

TIP
In any cross involving at least one parent that is homozygous dominant (2 CAPITAL letters), 100% of the offspring will have the dominant trait in their phenotype. 
This is illustrated by Questions 1a & 1b.

 

 

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Punnett Square Solutions Question #2

 

In purple people eaters, one-horn is dominant and no horn is recessive. Draw a Punnett Square showing the cross of a purple people eater that is hybrid for horns with a purple people eater that does not have horns. Summarize the genotypes & phenotypes of the possible offspring.
ANSWER:

Genotypes of Offspring

Phenotype(s) of Offspring

50% hybrid (Hh)
50% homozygous recessive (hh)

50% one-horn
50% no horns

No specific letter is given in the question to use as an abbreviation, so it's UP TO YOU! 
H = dominant allele for one horn
h = recessive allele for no (zero) horns

A purple people eater that is "hybrid" has one of each letters (the definition of hybrid), so that parent is "Hh". A purple people eater without horns has the recessive phenotype and the only way to have a recessive phenotype is to have a homozygous recessive genotype, which is 2 lowercase letters, "hh".
So our cross for this question is: Hh x hh.

The punnett square should be:

The alleles carried in the sex cells of the purple people eaters are split up & placed "outside" the p-square.  The alleles from the one-horn eater are on the left, and the alleles of the eater without horns are above each column. Copy one letter from the left & one from the top to fill-in the boxes.  The combinations inside the boxes are the possible genotypes (with respect to horns) of purple people eater offspring from these two parent purple people eaters.
Analyzing the data is simple count how many of each genotype & phenotype are found in each of the four boxes.  So, here we have 2 of 4 boxes "Hh" (50% hybrid, one horn), and 2 of 4 boxes "hh" (homozygous recessive, no horns).



 

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Punnett Square Solutions Question #3

 

A green-leafed luboplant is crossed with a luboplant with yellow-striped leaves.  The cross produces 185 green-leafed luboplants. Summarize the genotypes & phenotype of the offspring that would be produced by crossing two of the green-leafed luboplants obtained from the initial parent plants.
ANSWER:

Genotypes of the F2 Offspring

Phenotype(s) of F2 Offspring

25% homozygous dominant (GG)
50% hybrid (Gg)
25% homozygous recessive (gg)

75% green-leafed
25% yellow-striped leaves

First write down the letters & what they stand for. Since the parent luboplants have different leaf colors and 100% of the offspring resemble only one parent (i.e. they are all green), green is the dominant trait. It makes sense then to use:
G = dominant allele for green leaves
g = recessive allele for yellow-striped leaves
 

TIP: This is important to recognize

When two parents have opposite traits, 
and all their offspring look like only one of the parents, 
the trait the offspring have is the DOMINANT TRAIT.

The 185 "F1" offspring are all hybrids. How do I know?The yellow-striped parent MUST BE "gg". The 185 offspring had to have inherited a "g" from that parent plant because that parent plant has no "G's" to pass on. Since the 185 offspring are ALL green, they must have a dominant allele for green ("G"), so their entire genotype is "Gg".
That first cross must have been GG x gg, & its p-square would look like this:

 
 

Notice that 100% are hybrid (Gg) and 100% would look green.  IF that green parent had "Gg" for a genotype, then we would get half of the offspring with a homozygous recessive genotype (gg), which would give us 50% yellow-striped luboplants.  THIS IS NOT WHAT HAPPENED. The questions clearly states that all of the 185 plants are green, pretty good evidence that green-leafed parent luboplant is "GG" & not "Gg".
The offspring of this cross, by the way, are referred to as the "first filial" or "F1" generation.
 
 

Now, our question has to do with crossing two members of this F1 generation. That cross would be: Gg x Gg.
The punnett square showing this cross of two hybrids is:

Summary of results:

Genotypes of the F2 Offspring

Phenotype(s) of F2 Offspring

1 of 4 boxes (25%) homozygous dominant (GG)
2 of 4 boxes (50%) hybrid (Gg)
1 of 4 boxes (25%) homozygous recessive (gg)

 3 of 4 boxes (75%) green-leafed
1 of 4 boxes (25%) yellow-striped leaves


 
 

 

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Punnett Square Solutions Question #4

 

Mendel found that crossing wrinkle-seeded plants with pure round-seeded plants produced only round-seeded plants.  What genotypic & phenotypic ratios can be expected from a cross of a wrinkle-seeded plant & a plant heterozygous for this trait?
ANSWER: 50% HYBRID ROUND-SEEDED, & 50% HOMOZYGOUS RECESSIVE WRINKLE-SEEDED

The first thing to figure out is which trait is dominant & which is recessive. We get this from the 1st sentence.  If a wrinkled x round cross produces all round, then round is dominant & wrinkled is recessive.
Define our symbols:
R = dominant allele for round seeds
r = recessive allele for wrinkled seeds

Our wrinkle-seeded parent MUST be "rr", because the only way for a recessive trait to show up is if the genotype is homozygous recessive, which is 2 lowercase letters (rr).  Our parent that is "heterozygous for this trait" is "Rr", because heterozygous = hybrid= 1 CAPITAL & 1 lowercase.
So our cross for this problem is: rr x Rr.
The punnett square you drew should look something like this:

Again, you may have your "r's" on top & the "R" & "r" on the left, the combos inside the p-square will end up the same. No problem.  Remember, "one from the left & one from the top" when you are filling in the boxes. Of the offspring in this cross, 2 of 4 (50%) are hybrid (Rr) and would have round seeds, and 2 of 4 (50%) are homozygous recessive (rr) and would have wrinkled seeds.

 

 

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