Understanding Punnett Squares and Test Crosses - AP Biology
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Let us assume that flower color is either purple or white in a particular species. Color is determined by complete dominance, and purple is dominant to white.
A purple flower sprouts in the garden one day, and the gardener would like to know if it will only create purple flowers if it is pollinated. If she wants to use only one generation to determine its genotype, how should she pollinate the flower?
Let us assume that flower color is either purple or white in a particular species. Color is determined by complete dominance, and purple is dominant to white.
A purple flower sprouts in the garden one day, and the gardener would like to know if it will only create purple flowers if it is pollinated. If she wants to use only one generation to determine its genotype, how should she pollinate the flower?
When attempting to determine if an organism is heterozygous or homozygous for a dominant trait, it is best to use a test cross. A test cross involves crossing the flower in question with a homozygous recessive flower. Since a white flower can only contribute a white allele, we can determine if the purple flower in question is heterozygous or homozygous. Any white flowers in the next generation will confirm that the purple flower is heterozygous. If they are all purple, we can confirm that the flower is homozygous for the trait.
When attempting to determine if an organism is heterozygous or homozygous for a dominant trait, it is best to use a test cross. A test cross involves crossing the flower in question with a homozygous recessive flower. Since a white flower can only contribute a white allele, we can determine if the purple flower in question is heterozygous or homozygous. Any white flowers in the next generation will confirm that the purple flower is heterozygous. If they are all purple, we can confirm that the flower is homozygous for the trait.
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Let us assume that purple flowers are dominant to white flowers. A pure breeding purple flower is crossed with a pure breeding white flower. The F1 generation is self pollinated.
What percentage of the F2 generation will be homozygous?
Let us assume that purple flowers are dominant to white flowers. A pure breeding purple flower is crossed with a pure breeding white flower. The F1 generation is self pollinated.
What percentage of the F2 generation will be homozygous?
The F1 generation will be entirely heterozygous, thus the F2 generation is the result of a heterozygous self-cross. A Punnet square reveals that 75% of the generation will be purple (PP or Pp) and 25% will be white (pp). Of the three purple flowers in the punnett square, two of them are heterozygous for color (Pp). The other flower is homozygous for the purple allele (PP). In addition, the white flower is homozygous for the recessive white allele (pp).
There are two homozygous flowers, and two heterozygous flowers in the punnett square, so 50% is the correct answer.
The F1 generation will be entirely heterozygous, thus the F2 generation is the result of a heterozygous self-cross. A Punnet square reveals that 75% of the generation will be purple (PP or Pp) and 25% will be white (pp). Of the three purple flowers in the punnett square, two of them are heterozygous for color (Pp). The other flower is homozygous for the purple allele (PP). In addition, the white flower is homozygous for the recessive white allele (pp).
There are two homozygous flowers, and two heterozygous flowers in the punnett square, so 50% is the correct answer.
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If a male is heterozygous for the dimples allele and marries a female who does not possess the allele needed for dimples, what is the chance their child will have dimples? Dimples are an autosomal dominant trait.
If a male is heterozygous for the dimples allele and marries a female who does not possess the allele needed for dimples, what is the chance their child will have dimples? Dimples are an autosomal dominant trait.
By creating a Punnett square in which the male carries the label Dd and the female has the label dd, one can see the possible combinations of alleles that the child may have. D will symbolize the allele for dimples, and d will symbolize the allele for no dimples. The cross will be Dd x dd.
There is a 50% chance that the child does not obtain the allele needed for dimples (dd), and a 50% chance that the child is heterozygous (Dd). Because dimples is an autosomal dominant trait, heterozygosity will express dimples, leading to a 50% chance that the child will have dimples.
By creating a Punnett square in which the male carries the label Dd and the female has the label dd, one can see the possible combinations of alleles that the child may have. D will symbolize the allele for dimples, and d will symbolize the allele for no dimples. The cross will be Dd x dd.
There is a 50% chance that the child does not obtain the allele needed for dimples (dd), and a 50% chance that the child is heterozygous (Dd). Because dimples is an autosomal dominant trait, heterozygosity will express dimples, leading to a 50% chance that the child will have dimples.
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A botanist finds a pea plant growing in his backyard and wants to know its genotype for color, knowing that the allele for green is dominant to the allele for yellow. He breeds the unknown plant with a known homozygous dominant pea plant. If the F2 generation is three-fourths green and one-fourth yellow, what was the genotype of the unknown plant?
A botanist finds a pea plant growing in his backyard and wants to know its genotype for color, knowing that the allele for green is dominant to the allele for yellow. He breeds the unknown plant with a known homozygous dominant pea plant. If the F2 generation is three-fourths green and one-fourth yellow, what was the genotype of the unknown plant?
The possible genotypes of the unknown plant are GG, Gg, or gg. To create the F1 generation, it is crossed with a plant with genotype GG.
Scenario 1: GG x GG, result is all GG in F1; F2 cannot possibly contain a yellow (gg) plant
Scenario 2: Gg x GG, result is half Gg and half GG; F2 will contain both green and yellow plants of all genotypes
Scenario 3: gg x GG, result is all Gg in F1; F2 will be 75% green and 25% yellow
In order to have the ratios described in the question, the unknown plant must be homozygous recessive.
The possible genotypes of the unknown plant are GG, Gg, or gg. To create the F1 generation, it is crossed with a plant with genotype GG.
Scenario 1: GG x GG, result is all GG in F1; F2 cannot possibly contain a yellow (gg) plant
Scenario 2: Gg x GG, result is half Gg and half GG; F2 will contain both green and yellow plants of all genotypes
Scenario 3: gg x GG, result is all Gg in F1; F2 will be 75% green and 25% yellow
In order to have the ratios described in the question, the unknown plant must be homozygous recessive.
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A breeder wants to know her dog’s genotype. The breed she works with comes in two varieties: black and yellow. Black is dominant to yellow. She breeds her black dog with a yellow dog and gets a litter of three black dogs and three yellow dogs. What is the genotype of the parental black dog?
A breeder wants to know her dog’s genotype. The breed she works with comes in two varieties: black and yellow. Black is dominant to yellow. She breeds her black dog with a yellow dog and gets a litter of three black dogs and three yellow dogs. What is the genotype of the parental black dog?
We know that the yellow dog must be homozygous recessive and that the black dog must be either heterozygous or homozygous dominant. When crossed, their offspring can show either the dominant (black) phenotype or the recessive (yellow) phenotype. In order for offspring to show the recessive phenotype, they must inherit a recessive allele from each parent. To have this outcome, the black dog must carry a recessive allele even though it expresses the dominant trait; this makes the black dog heterozygous.
We can look at a punnett square to verify the result. We will use B as the dominant allele and b as the recessive allele.
Bb (black dog) x bb (yellow dog)
Offspring: Half Bb (black) and half bb (yellow)
We know that the yellow dog must be homozygous recessive and that the black dog must be either heterozygous or homozygous dominant. When crossed, their offspring can show either the dominant (black) phenotype or the recessive (yellow) phenotype. In order for offspring to show the recessive phenotype, they must inherit a recessive allele from each parent. To have this outcome, the black dog must carry a recessive allele even though it expresses the dominant trait; this makes the black dog heterozygous.
We can look at a punnett square to verify the result. We will use B as the dominant allele and b as the recessive allele.
Bb (black dog) x bb (yellow dog)
Offspring: Half Bb (black) and half bb (yellow)
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In a certain species of bird, yellow beaks are dominant to orange beaks, and blue feathers are dominant to black feathers.
Two heterozygous birds are crossed. What fraction of the offspring would be expected to have yellow beaks and blue feathers?
In a certain species of bird, yellow beaks are dominant to orange beaks, and blue feathers are dominant to black feathers.
Two heterozygous birds are crossed. What fraction of the offspring would be expected to have yellow beaks and blue feathers?
This question requires us to do a dihybrid cross. We can represent the gene for beak color with the symbol "A" for dominant yellow and "a" for the recessive orange. Likewise, for feather color, we can use "B" for blue feathers and "b" for black.
The problem states that both birds are heterozygous for each trait, implying that our cross is between two birds with the genotype AaBb.
Now we look at the gametes that can be produced by these parents: AB, Ab, aB, and ab. These gametes can then be used to make a punnet square.
Offspring: 1 AABB, 3 Aabb, 8 AaBa, 3 aaBb, 1 aabb
There are 16 total offspring. 12 of them carry the dominant A allele, giving them the yellow beak phenotype.
Yellow beaks: 1 AABB, 3Aabb, 8 AaBb
Finally, of these 12, 9 carry the dominant B allele for blue feathers.
Yellow beaks and blue feathers: 1 AABB, 8 AaBb
This gives a total of 9 out of the 16 offspring that will express both the yellow beak and blue feather phenotypes.
You should be familiar with the 9:3:3:1 phenotypic ratio resulting from dihybrid crosses.
This question requires us to do a dihybrid cross. We can represent the gene for beak color with the symbol "A" for dominant yellow and "a" for the recessive orange. Likewise, for feather color, we can use "B" for blue feathers and "b" for black.
The problem states that both birds are heterozygous for each trait, implying that our cross is between two birds with the genotype AaBb.
Now we look at the gametes that can be produced by these parents: AB, Ab, aB, and ab. These gametes can then be used to make a punnet square.
Offspring: 1 AABB, 3 Aabb, 8 AaBa, 3 aaBb, 1 aabb
There are 16 total offspring. 12 of them carry the dominant A allele, giving them the yellow beak phenotype.
Yellow beaks: 1 AABB, 3Aabb, 8 AaBb
Finally, of these 12, 9 carry the dominant B allele for blue feathers.
Yellow beaks and blue feathers: 1 AABB, 8 AaBb
This gives a total of 9 out of the 16 offspring that will express both the yellow beak and blue feather phenotypes.
You should be familiar with the 9:3:3:1 phenotypic ratio resulting from dihybrid crosses.
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In a newly discovered species, albinism is recessive to natural coloring. If an albino parent were crossed with a naturally colored parent, what would be the expected phenotypic ratios of the offspring?
In a newly discovered species, albinism is recessive to natural coloring. If an albino parent were crossed with a naturally colored parent, what would be the expected phenotypic ratios of the offspring?
Because we are only told the phenotype of the naturally colored parent, we do not know if it is homozygous dominant or heterozygous. To account for this, we must anticipate the phenotypic ratios of both dominant genotypes. Two crosses must be performed: one between a homozygous dominant parent and a homozygous recessive parent, and one between a heterozygous parent and a homozygous recessive parent. The resulting ratios would be 100% natural and 50% albino/50% natural respectively.
AA x aa: all offspring Aa (natural)
Aa x aa: half offspring Aa (natural), half offspring aa (albino)
Because we are only told the phenotype of the naturally colored parent, we do not know if it is homozygous dominant or heterozygous. To account for this, we must anticipate the phenotypic ratios of both dominant genotypes. Two crosses must be performed: one between a homozygous dominant parent and a homozygous recessive parent, and one between a heterozygous parent and a homozygous recessive parent. The resulting ratios would be 100% natural and 50% albino/50% natural respectively.
AA x aa: all offspring Aa (natural)
Aa x aa: half offspring Aa (natural), half offspring aa (albino)
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A new type of plant is shown to have two distinct traits for their seeds: color and shape. Green color is dominant to white, and a long shape is dominant to round. If two plants heterozygous for both traits are crossed, what would be the expected phenotypic ratios of the offspring?
A new type of plant is shown to have two distinct traits for their seeds: color and shape. Green color is dominant to white, and a long shape is dominant to round. If two plants heterozygous for both traits are crossed, what would be the expected phenotypic ratios of the offspring?
This question requires that we do a dihybrid cross. The cross in question is AaBb x AaBb, using A to represent dominant green color and B to represent dominant long shape. The parents are heterozygous for both traits, meaning they will carry one dominant color allele and one dominant shape allele.
The result of a punnet square for a dihybrid cross is: 1 AABB, 3 Aabb, 8 AaBb, 3 aaBb, 1 aabb.
This gives a total of sixteen different offspring. Two different genotypes carry dominant alleles for both traits: AABB and AaBb. There are a total of nine offspring between these two genotypes that will be green and long. The other genotypes all represent different phenotypes. Aabb will be green and round. aaBb will be white and long. aabb will be white and round. Together, this gives a final phenotypic ratio of 9:3:3:1.
This question requires that we do a dihybrid cross. The cross in question is AaBb x AaBb, using A to represent dominant green color and B to represent dominant long shape. The parents are heterozygous for both traits, meaning they will carry one dominant color allele and one dominant shape allele.
The result of a punnet square for a dihybrid cross is: 1 AABB, 3 Aabb, 8 AaBb, 3 aaBb, 1 aabb.
This gives a total of sixteen different offspring. Two different genotypes carry dominant alleles for both traits: AABB and AaBb. There are a total of nine offspring between these two genotypes that will be green and long. The other genotypes all represent different phenotypes. Aabb will be green and round. aaBb will be white and long. aabb will be white and round. Together, this gives a final phenotypic ratio of 9:3:3:1.
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Two individuals with the following genotypes are crossed:
AABbccDdEeFF x AaBBCCDdEeff
What is the probability that their offspring will have the genotype AaBbCcddEEFf?
Two individuals with the following genotypes are crossed:
AABbccDdEeFF x AaBBCCDdEeff
What is the probability that their offspring will have the genotype AaBbCcddEEFf?
To obtain the overall probability, multiply the individual probabilities for each locus.
Parent cross: AABbccDdEeFF x AaBBCCDdEeff
Offspring: AaBbCcddEEFf
A locus cross is AA x Aa. Half of the offspring will be AA and half will be Aa. The probability of Aa is one half.
B locus cross is Bb x BB. Half of the offspring will be BB and half will be Bb. The probability of Bb is one half.
C locus cross is cc x CC. All offspring will be Cc. The probability of Cc is one.
D locus cross is Dd x Dd. One-fourth of the offspring will be DD, half will be Dd, and one-fourth will be dd. The probability of dd is one-fourth.
E locus cross is Ee x Ee. One-fourth of the offspring will be EE, half will be Ee, and one-fourth will be ee. The probability of EE is one-fourth.
F locus cross is FF x ff. All offspring will be Ff. The probability of Ff is one.
Multiply all the probability values.
To obtain the overall probability, multiply the individual probabilities for each locus.
Parent cross: AABbccDdEeFF x AaBBCCDdEeff
Offspring: AaBbCcddEEFf
A locus cross is AA x Aa. Half of the offspring will be AA and half will be Aa. The probability of Aa is one half.
B locus cross is Bb x BB. Half of the offspring will be BB and half will be Bb. The probability of Bb is one half.
C locus cross is cc x CC. All offspring will be Cc. The probability of Cc is one.
D locus cross is Dd x Dd. One-fourth of the offspring will be DD, half will be Dd, and one-fourth will be dd. The probability of dd is one-fourth.
E locus cross is Ee x Ee. One-fourth of the offspring will be EE, half will be Ee, and one-fourth will be ee. The probability of EE is one-fourth.
F locus cross is FF x ff. All offspring will be Ff. The probability of Ff is one.
Multiply all the probability values.
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In a new species of beetle, black coloration is recessive to blue coloration. If a homozygous dominant blue beetle is mated to a black beetle, what are the expected phenotypic ratios?
In a new species of beetle, black coloration is recessive to blue coloration. If a homozygous dominant blue beetle is mated to a black beetle, what are the expected phenotypic ratios?
Because we are told that the blue parent is homozygous dominant, we can set up a simple cross. We know that the black beetle must be homozygous recessive to present the black phenotype. Using B as the dominant blue allele and b as the recessive black allele, we can see that the parent beetles have genotypes of BB and bb.
BB x bb
Offspring: All offspring will be Bb and present the dominant (blue) phenotype.
All resulting offspring will have at least one dominant allele, giving us a 100% blue ratio.
Because we are told that the blue parent is homozygous dominant, we can set up a simple cross. We know that the black beetle must be homozygous recessive to present the black phenotype. Using B as the dominant blue allele and b as the recessive black allele, we can see that the parent beetles have genotypes of BB and bb.
BB x bb
Offspring: All offspring will be Bb and present the dominant (blue) phenotype.
All resulting offspring will have at least one dominant allele, giving us a 100% blue ratio.
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A new species of insect was recently discovered. Scientists have found that orange coloration is recessive to yellow coloration. If a heterozygous yellow insect is mated to an orange insect, what are the expected phenotypic percentages of the offspring?
A new species of insect was recently discovered. Scientists have found that orange coloration is recessive to yellow coloration. If a heterozygous yellow insect is mated to an orange insect, what are the expected phenotypic percentages of the offspring?
We are told that the yellow parent is heterozygous, so we can set up a simple cross. We know that the orange insect must be homozygous recessive due to the fact that it displays the orange phenotype.
In our cross, we will use A as the dominant allele and a as the recessive allele. This gives the yellow parent a genotype of Aa and the orange parent a genotype of aa.
Aa x aa
Possible offspring: Aa, Aa, aa, aa
We can see that there are only two possible genotypic results. Half of the offspring will be Aa and half will be aa. The Aa offspring will be yellow and the aa offspring will be orange, giving a 1:1 phenotypic ratio.
We are told that the yellow parent is heterozygous, so we can set up a simple cross. We know that the orange insect must be homozygous recessive due to the fact that it displays the orange phenotype.
In our cross, we will use A as the dominant allele and a as the recessive allele. This gives the yellow parent a genotype of Aa and the orange parent a genotype of aa.
Aa x aa
Possible offspring: Aa, Aa, aa, aa
We can see that there are only two possible genotypic results. Half of the offspring will be Aa and half will be aa. The Aa offspring will be yellow and the aa offspring will be orange, giving a 1:1 phenotypic ratio.
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A new type of plant is shown to have two distinct traits for its seeds: seed color and seed shape. Green color is dominant to white, and a long shape is dominant to round. If two plants heterozygous for both traits were crossed, what is the probability that an offspring would show the dominant phenotype for both traits?
A new type of plant is shown to have two distinct traits for its seeds: seed color and seed shape. Green color is dominant to white, and a long shape is dominant to round. If two plants heterozygous for both traits were crossed, what is the probability that an offspring would show the dominant phenotype for both traits?
This question requires that we do a dihybrid cross. The cross in question is AaBb x AaBb, using A to represent dominant green color and B to represent dominant long shape. The parents are heterozygous for both traits, meaning they will carry one dominant color allele and one dominant shape allele.
The result of a punnet square for a dihybrid cross is: 1 AABB, 3 Aabb, 8 AaBb, 3 aaBb, 1 aabb.
This gives a total of sixteen different offspring. Two different genotypes carry dominant alleles for both traits: AABB and AaBb. There are a total of nine offspring between these two genotypes. The probability of an offspring being dominant for both traits is thus nine out of sixteen.
This question requires that we do a dihybrid cross. The cross in question is AaBb x AaBb, using A to represent dominant green color and B to represent dominant long shape. The parents are heterozygous for both traits, meaning they will carry one dominant color allele and one dominant shape allele.
The result of a punnet square for a dihybrid cross is: 1 AABB, 3 Aabb, 8 AaBb, 3 aaBb, 1 aabb.
This gives a total of sixteen different offspring. Two different genotypes carry dominant alleles for both traits: AABB and AaBb. There are a total of nine offspring between these two genotypes. The probability of an offspring being dominant for both traits is thus nine out of sixteen.
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A new type of plant is shown to have two distinct traits for its seeds: color and shape. Yellow color is dominant to gray, and a long shape is dominant to round. If two plants heterozygous for both traits were crossed, what is the probability that an offspring would show the recessive phenotype for both traits?
A new type of plant is shown to have two distinct traits for its seeds: color and shape. Yellow color is dominant to gray, and a long shape is dominant to round. If two plants heterozygous for both traits were crossed, what is the probability that an offspring would show the recessive phenotype for both traits?
This question requires that we do a dihybrid cross. If A represents dominant color and B represents dominant shape, then both parent plants have the genotype AaBb.
Parent cross: AaBb x AaBb
Possible offspring: AABB, AABb, AAbb, Aabb, aaBB, aaBb, AaBB, aabb, AaBb
The phenotypic ratios resulting from a dihybrid cross are always 9:3:3:1, where 9 represents both dominant phenotypes, each 3 represents one dominant and one recessive phenotype, and 1 represents both recessive phenotypes. In this cross, there would be 9 long yellow seeds, 3 round yellow seeds, 3 long gray seeds, and 1 round gray seed.
Only one out of the potential sixteen offspring from the cross will carry both recessive phenotypes.
This question requires that we do a dihybrid cross. If A represents dominant color and B represents dominant shape, then both parent plants have the genotype AaBb.
Parent cross: AaBb x AaBb
Possible offspring: AABB, AABb, AAbb, Aabb, aaBB, aaBb, AaBB, aabb, AaBb
The phenotypic ratios resulting from a dihybrid cross are always 9:3:3:1, where 9 represents both dominant phenotypes, each 3 represents one dominant and one recessive phenotype, and 1 represents both recessive phenotypes. In this cross, there would be 9 long yellow seeds, 3 round yellow seeds, 3 long gray seeds, and 1 round gray seed.
Only one out of the potential sixteen offspring from the cross will carry both recessive phenotypes.
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A breed of dog can be either of two colors. In this breed, brown is dominant to yellow. A brown dog mates with a yellow dog and produces a litter of six brown dogs. What is the genotype of the parental brown dog?
A breed of dog can be either of two colors. In this breed, brown is dominant to yellow. A brown dog mates with a yellow dog and produces a litter of six brown dogs. What is the genotype of the parental brown dog?
We know that the yellow dog must be homozygous recessive and that the brown dog must be either heterozygous or homozygous dominant. If B is used to represent the dominant brown allele and b is used to represent the recessive yellow allele, this means that the yellow parent must be bb and the brown parent could be either BB or Bb.
We know that all of the puppies must carry a dominant allele, since they all express the dominant phenotype. The yellow parent only carries the recessive allele. This indicates that every puppy must have inherited a dominant allele from the brown parent. The most likely genotype of the brown parent would be homozygous dominant, which would lead to all puppies being heterozygous and brown.
BB x bb
All offspring will be Bb and express the dominant brown phenotype.
Note, however, that we cannot determine the genotype of the brown parent for certain. Due to independent assortment, it is possible that a heterozygous brown parent gave the recessive allele to all offspring by chance.
Bb x bb
Half offspring will be Bb and half will be bb.
The probability of getting six brown puppies from this cross would be equal to one-half to the sixth power, or 1.56%. While this is a very small chance, it is not impossible and we cannot rule out a heterozygous genotype for the brown dog.
We know that the yellow dog must be homozygous recessive and that the brown dog must be either heterozygous or homozygous dominant. If B is used to represent the dominant brown allele and b is used to represent the recessive yellow allele, this means that the yellow parent must be bb and the brown parent could be either BB or Bb.
We know that all of the puppies must carry a dominant allele, since they all express the dominant phenotype. The yellow parent only carries the recessive allele. This indicates that every puppy must have inherited a dominant allele from the brown parent. The most likely genotype of the brown parent would be homozygous dominant, which would lead to all puppies being heterozygous and brown.
BB x bb
All offspring will be Bb and express the dominant brown phenotype.
Note, however, that we cannot determine the genotype of the brown parent for certain. Due to independent assortment, it is possible that a heterozygous brown parent gave the recessive allele to all offspring by chance.
Bb x bb
Half offspring will be Bb and half will be bb.
The probability of getting six brown puppies from this cross would be equal to one-half to the sixth power, or 1.56%. While this is a very small chance, it is not impossible and we cannot rule out a heterozygous genotype for the brown dog.
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If two parents are heterozygous for a trait and they have children, what is the percentage of the children that are heterozygous for the trait?
If two parents are heterozygous for a trait and they have children, what is the percentage of the children that are heterozygous for the trait?
For simplicity, we will assign the letter "a" for the gene of interest. Thus, the heterozygous genotype is Aa. You may sketch a punnet square of the cross: Aa x Aa to help illustrate that the combinations result in 50% chance of heterozygous offspring.
For simplicity, we will assign the letter "a" for the gene of interest. Thus, the heterozygous genotype is Aa. You may sketch a punnet square of the cross: Aa x Aa to help illustrate that the combinations result in 50% chance of heterozygous offspring.
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In a species of lizard, green coloration is dominant to blue coloration. A breeder wants to produce as many blue lizards as possible from one cross. Which of the following crosses will produce the highest percentage of blue lizards?
In a species of lizard, green coloration is dominant to blue coloration. A breeder wants to produce as many blue lizards as possible from one cross. Which of the following crosses will produce the highest percentage of blue lizards?
The question tells us that green is dominant to blue. This means that the C allele must correspond to green and the c allele must correspond to blue. We are looking for the cross that will produce the most blue offspring. Since blue is recessive, all blue offspring will be homozygous recessive; thus, we are essentially looking for the cross that carries the most recessive alleles.
Of the given crosses, cc x cc contains the greatest number of recessive alleles. Every offspring from this cross will be homozygous recessive and display the blue phenotype. This cross thus gives the highest percentage of blue offspring, with that percentage being 100%.
The question tells us that green is dominant to blue. This means that the C allele must correspond to green and the c allele must correspond to blue. We are looking for the cross that will produce the most blue offspring. Since blue is recessive, all blue offspring will be homozygous recessive; thus, we are essentially looking for the cross that carries the most recessive alleles.
Of the given crosses, cc x cc contains the greatest number of recessive alleles. Every offspring from this cross will be homozygous recessive and display the blue phenotype. This cross thus gives the highest percentage of blue offspring, with that percentage being 100%.
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If a homozygous brown rat mates with a heterozygous brown rat, what percent of the offspring will have white fur?
If a homozygous brown rat mates with a heterozygous brown rat, what percent of the offspring will have white fur?
Heterozygous organisms carry one dominant allele and one recessive allele. The dominant allele is expressed over the recessive allele, giving the organism the dominant phenotype. If the heterozygous rat in the question is brown, then we can conclude that brown is dominant to white.
The cross of these two rats would be:
Parents: BB (brown) x Bb (brown)
Offspring: half BB (brown), half Bb (brown)
Though half of the offspring will be homozygous and half will be heterozygous, all offspring will be brown. None of the offspring from this cross will show the white phenotype.
Heterozygous organisms carry one dominant allele and one recessive allele. The dominant allele is expressed over the recessive allele, giving the organism the dominant phenotype. If the heterozygous rat in the question is brown, then we can conclude that brown is dominant to white.
The cross of these two rats would be:
Parents: BB (brown) x Bb (brown)
Offspring: half BB (brown), half Bb (brown)
Though half of the offspring will be homozygous and half will be heterozygous, all offspring will be brown. None of the offspring from this cross will show the white phenotype.
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In pea plants, tall is dominant for height, green is dominant for color and round is dominant for pea shape.
A tall green plant is crossed with a short yellow plant. 50% of the offspring are tall and green, and 50% are tall and yellow.
What are the genotypes of the parent plants?
In pea plants, tall is dominant for height, green is dominant for color and round is dominant for pea shape.
A tall green plant is crossed with a short yellow plant. 50% of the offspring are tall and green, and 50% are tall and yellow.
What are the genotypes of the parent plants?
Let's look at each trait individually. We can see that all of the offspring from this cross will be tall; the tall allele must be dominant over the recessive allele. We also know that one parent must be homozygous dominant for the tall allele, since it is passed to every single offspring. The parent genotypes for height are TT (tall) and tt (short). The short plant must be homozygous recessive to show the recessive phenotype.
Now, let's look at color. Half of the offspring are green and half are yellow. This tells us that whichever parent displays the dominant phenotype is also heterozygous, allowing the recessive phenotype to appear in the offspring. Since we know that green is dominant, we know that the green plant must be heterozygous. In order to display the recessive yellow phenotype, the yellow plant must be homozygous recessive. The parent genotypes for color must be Gg (green) and gg (yellow).
Together, we can see that the tall-green plant will be TTGg and the short-yellow plant will be ttgg.
Let's look at each trait individually. We can see that all of the offspring from this cross will be tall; the tall allele must be dominant over the recessive allele. We also know that one parent must be homozygous dominant for the tall allele, since it is passed to every single offspring. The parent genotypes for height are TT (tall) and tt (short). The short plant must be homozygous recessive to show the recessive phenotype.
Now, let's look at color. Half of the offspring are green and half are yellow. This tells us that whichever parent displays the dominant phenotype is also heterozygous, allowing the recessive phenotype to appear in the offspring. Since we know that green is dominant, we know that the green plant must be heterozygous. In order to display the recessive yellow phenotype, the yellow plant must be homozygous recessive. The parent genotypes for color must be Gg (green) and gg (yellow).
Together, we can see that the tall-green plant will be TTGg and the short-yellow plant will be ttgg.
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A biologist genetically crosses two cats, both with black fur, which is dominant over white fur. The pair have 12 kittens, 9 with black fur and 3 with white fur. What is the parental cross?
A biologist genetically crosses two cats, both with black fur, which is dominant over white fur. The pair have 12 kittens, 9 with black fur and 3 with white fur. What is the parental cross?
The ratio of kittens with black fur to kittens to white fur is 3:1. Since the recessive phenotype is shown in the offspring, we know that both of the parents are carriers of the recessive genotype and are thus both heterozygous, Bb.
The ratio of kittens with black fur to kittens to white fur is 3:1. Since the recessive phenotype is shown in the offspring, we know that both of the parents are carriers of the recessive genotype and are thus both heterozygous, Bb.
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Assume that in the organism in question, the allele for brown fur (B) is dominant to the allele for white fur (b). Also assume that the allele for curly fur (C) is dominant to the allele for straight fur (c). Finally, assume these genes are independent.
Given parents with the genotypes BBCc and Bbcc, what fraction of offspring will display straight, brown fur?
Assume that in the organism in question, the allele for brown fur (B) is dominant to the allele for white fur (b). Also assume that the allele for curly fur (C) is dominant to the allele for straight fur (c). Finally, assume these genes are independent.
Given parents with the genotypes BBCc and Bbcc, what fraction of offspring will display straight, brown fur?
Each parent will contribute one allele at random from each of the two genes. The odds of having brown fur, given the parental genotypes, is 100% because all children will receive a dominant allele for brown fur (B) from the first parent. This means that the odds of having brown straight fur will depend only on the odds of having straight fur. Having straight fur requires one recessive allele (c) from each parent. The second parent will always contribute a recessive allele while the first will contribute a recessive allele half the time. Thus, the odds of straight fur is 50%, as are the odds of straight, brown fur. Alternatively, a Punnett square may be used. The square below shows the 50% offspring combinations with straight, brown fur highlighted in red.
Each parent will contribute one allele at random from each of the two genes. The odds of having brown fur, given the parental genotypes, is 100% because all children will receive a dominant allele for brown fur (B) from the first parent. This means that the odds of having brown straight fur will depend only on the odds of having straight fur. Having straight fur requires one recessive allele (c) from each parent. The second parent will always contribute a recessive allele while the first will contribute a recessive allele half the time. Thus, the odds of straight fur is 50%, as are the odds of straight, brown fur. Alternatively, a Punnett square may be used. The square below shows the 50% offspring combinations with straight, brown fur highlighted in red.
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