If a homozygous tall pea plant and also a homozygous brief pea plant room crossed. I m sorry of the following is observed, assuming pea plant height exhibits finish dominance?


Explanation:

When a homozygous high pea plant and also a homozygous quick pea plant are crossed, the next generation space all heterozygous and also the leading allele is express in every the plants, and the recessive trait appears to disappear. Since we room not told whether the high or short allele is dominant, us cannot i think either.

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Explanation:

Dominant alleles are neither better, nor worse 보다 recessive alleles. Castle are simply expressed in the phenotype the a heterozygous pair. Several deleterious, and even deadly disorders prefer Huntington"s disease, are inherited via one autosomal leading pattern.

Note that the frequency of one allele or genotype in a populace must be determined experimentally, and also can be linked to environmental influences, autosomal versus sex-linked patterns, and numerous other factors.


Explanation:

The genotype describes the hereditary makeup or alleles of the individual, when the phenotype explains the physical characteristic that the individual.


Explanation:

The parental generation describes the generation in i m sorry the 2 organisms space crossed; the F1 generation is the very first filial generation, or the offspring developed from the cross; the F2 generation is the 2nd filial generation, or the offspring created from a cross between two F1 organisms.


Let us assume that flower color is either violet or white in a particular species. Shade is determined by finish dominance, and also purple is leading to white.


A purple flower sprouts in the garden one day, and the gardener would favor to recognize if it will certainly only produce purple flower if the is pollinated. If she wants to use only one generation to recognize its genotype, exactly how should she pollinate the flower?


Explanation:

When attempting to determine if an biology is heterozygous or homozygous for a dominant trait, the is ideal to use a check cross. A check cross entails crossing the flower in inquiry with a homozygous recessive flower. Because a white flower can only add a white allele, we can determine if the purple flower in question is heterozygous or homozygous. Any kind of white flower in the following generation will certainly confirm the the violet flower is heterozygous. If they space all purple, we deserve to confirm that the flower is homozygous for the trait.


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Example question #1 : expertise Punnett Squares and also Test the cross


Let us assume the purple flower are dominant to white flowers. A pure reproduction purple flower is crossed v a pure breeding white flower. The F1 generation is self pollinated.


What percentage of the F2 generation will certainly be homozygous? 


Possible Answers:

Further details is needed to identify the answer


25%


100%


50%


Correct answer:

50%


Explanation:

The F1 generation will be completely heterozygous, hence the F2 generation is the result of a heterozygous self-cross. A Punnet square reveals the 75% of the generation will certainly be purple (PP or Pp) and 25% will be white (pp). Of the 3 purple flower in the punnett square, two of them room heterozygous for shade (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 room two homozygous flowers, and also two heterozygous flowers in the punnett square, therefore 50% is the correct answer.


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Example concern #3 : knowledge Punnett Squares and Test the cross


A botanist find a pea plant cultivation in his backyard and wants to know its genotype for color, knowing that the allele for eco-friendly is leading to the allele for yellow. He breeds the unknown plant v a well-known homozygous dominant pea plant. If the F2 generation is three-fourths green and also one-fourth yellow, what to be the genotype that the unknown plant?


Possible Answers:

Impossible come tell


Either homozygous dominant or heterozygous


Heterozygous


Homozygous dominant


Homozygous recessive


Correct answer:

Homozygous recessive


Explanation:

The feasible genotypes that the unknown plant are GG, Gg, or gg. To develop the F1 generation, the is crossed v a plant with genotype GG.

Scenario 1: GG x GG, an outcome is every GG in F1; F2 cannot perhaps contain a yellow (gg) plant

Scenario 2: Gg x GG, an outcome is fifty percent Gg and fifty percent GG; F2 will certainly contain both green and yellow tree of every genotypes

Scenario 3: gg x GG, an outcome is every Gg in F1; F2 will certainly be 75% green and 25% yellow

In order to have actually the ratios defined in the question, the unknown plant must be homozygous recessive.


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Example concern #1 : Mendelian And populace Genetics


In a certain species of bird, yellow beaks are dominant to orange beaks, and also blue feather are leading to black color feathers.


Two heterozygous birds room crossed. What fraction of the offspring would be meant to have actually yellow beaks and blue feathers?


Possible Answers:

*


*


*



Correct answer:


Explanation:

This concern requires us to perform a dihybrid cross. We can represent the gene for beak color with the prize "A" for leading yellow and "a" because that the recessive orange. Likewise, for feather color, we can use "B" because that blue feathers and "b" because that black.

The problem states that both birds space heterozygous for each trait, implying the our overcome is between two birds v the genotype AaBb.

Now us look in ~ the gametes that have the right to be produced by these parents: AB, Ab, aB, and also ab. This gametes deserve to then be used to make a punnet square.

Offspring: 1 AABB, 3 Aabb, 8 AaBa, 3 aaBb, 1 aabb

There space 16 complete offspring. 12 the them bring the dominant A allele, offering them the yellow beak phenotype.

Yellow beaks: 1 AABB, 3Aabb, 8 AaBb

Finally, of these 12, 9 bring the leading B allele because that blue feathers.

Yellow beaks and blue feathers: 1 AABB, 8 AaBb

This offers a complete of 9 the end of the 16 offspring that will certainly express both the yellow beak and also blue feather phenotypes.

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You need to be acquainted with the 9:3:3:1 phenotypic proportion resulting native dihybrid crosses.