Genetics Practice Problems #7 - MARRIC



Genetics Practice Problems #7 Dihybrid crosses KEY

 

1.   In pepper plants, green (G) fruit color is dominant to red (g) and round (R) fruit shape is dominant to square (r) fruit shape.  These two genes are located on different chromosomes.

a.   What gamete types will be produced by a heterozygous green, round plant?

the green round plant will produce GR, Gr, gR, and gr gametes in equal proportion since the genes are unlinked

 

b.   If two such heterozygous plants are crossed, what genotypes and phenotypes will be seen in the offspring and in what proportions?

this will give 9/16 green round, 3/16 green square, 3/16 red round, and 1/16 red square phenotypes;

  

|  |GR |Gr |gR |gr |

|GR |GGRR |GGRr |GgRR |GgRr |

|Gr |GGRr |GGrr |GrRr |Ggrr |

|gR |GgRR |GgRr |ggRR |ggRr |

|gr |GgRr |Ggrr |ggRr |ggrr |

 

        G_R_ = green, round                  ggR_ =  red, round

        G_rr  = green, square ggrr =  red, square

Proportion Genotype Phenotype

|1 |GGRR |Green-round | |

|2 |GGRr |Green-round | |

|2 |GgRR |Green-round |9 |

|4 |GgRr |Green-round | |

|1 |GGrr |Green-square | |

|2 |Ggrr |Green-square |3 |

|1 |ggRR |Red-round | |

|2 |ggRr |Red-round |3 |

|1 |ggrr |Red-square |1 |

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1. About 70% of Americans perceive a bitter taste from the chemical phenylthiocarbamide (PTC). The ability to taste the chemical results from a dominant allele (T) and not being able to taste PTC is the result of having two recessive alleles (t). Albinism is also a single locus trait with normal pigment being dominant (A) and the lack of pigment being recessive (a).

A normally pigmented woman who cannot taste PTC has a father who is an albino taster. She marries a homozygous, normally pigmented man who is a taster but who has a mother that does not taste PTC.

a. What are the genotypes of the possible children? Hint: first determine the genotypes of the parents.

We know that the woman has normal pigment which means she must have at least one A. Her father is albino and because the albino allele is recessive, his genotype is aa. What does this make her skin genotype?

We also know that the woman cannot taste PTC. Because the ability to taste PTC is dominant, what does this make her genotype for the tasting trait?

Putting both traits together, we see that the woman's overall genotype is Aatt.

Now, what about her husband? You have been told that he is homozygous for normal pigment. What is his genotype for skin color?

He is a taster and so must have at least one T. However, we also know that his mother can not taste PTC so she must be homozygous recessive. With this information, what is his genotype for taste?

Putting both traits together, you see that his overall genotype is AATt.

| |At |At |at |at |

|AT |AATt |AATt |AaTt |AaTt |

|AT |AATt |AATt |AaTt |AaTt |

|At |AAtt |AAtt |Aatt |Aatt |

|At |AAtt |AAtt |Aatt |Aatt |

What percentage of the children will be albinos?

aa genotype necessary for the albino trait = 0%

What percentage of the children will be non-tasters of PTC?

T allele necessary to be a taster = 8/16 = 50%

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2. Wolves are sometimes observed to have black coats and blue eyes. Assume that these traits are controlled by single locus genes and are located on different chromosomes. Assume further that normal coat color (N) is dominant to black (n) and brown eyes (B) are dominant to blue (b).

Suppose the alpha male and alpha female of a pack (these are the dominant individuals who do most of the breeding) are black with blue eyes and normal colored with brown eyes, respectively. The female is also heterozygous for both traits.

How many of the offspring (assume 16) living in the pack will have each of the following genotypes?

(Dad) nnbb x NnBb (mom)

|Gametes X |nb |nb |nb |nb |

|NB |NnBb |NnBb |NnBb |NnBb |

|Nb |Nnbb |Nnbb |Nnbb |Nnbb |

|nB |nnBb |nnBb |nnBb |nnBb |

|nb |nnbb |nnbb |nnbb |nnbb |

Possible genotypes 4 – NnBb 4 – Nnbb 4 – nnBb 4 - nnbb

What percent of the offspring will be normal colored with blue eyes?

4/16 Nn (normal) bb (blue eyes) = 25%

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In the breeding season, male Anole lizards court females by bobbing their heads up and down while displaying a colorful throat patch. Assume for this question that both males and females bob their heads and have throat patches. Assume also, that both traits are controlled by single locus genes on separate chromosomes. Now, suppose that anoles prefer to mate with lizards who bob their heads fast (F) and have red throat patches (R) and that these two alleles are dominant to their counterparts, slow bobbing and yellow throats.

A male lizard heterozygous for head bobbing and homozygous dominant for the red throat patch mates with a female that is also heterozygous for head bobbing but is homozygous recessive for yellow throat patches.

(Dad) FfRR x Ffrr (mom)

|Gametes X |FR |FR |fR |fR |

|Fr |FFRr |FFRr |FfRr |FfRr |

|Fr |FFRr |FFRr |FfRr |FfRr |

|fr |FfRr |FfRr |ffRr |ffRr |

|fr |FfRr |FfRr |ffRr |ffRr |

Possible genotypes: 4 – FFRr 8 – FfRr 4 - ffRr

a. How many of the F1 offspring have the preferred fast bobbing / red throat phenotype (assume 16 young)? 12 = 4 FFRr + 8 FfRr

b. What percentage of the offspring will lack mates because they have both slow head bobbing and yellow throats? 0% will be both

c. What percentage of the offspring will have trouble finding mates because they lack one of the dominant traits? 4/16 = 25% will be slow bobbing

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4. Carrion beetles lay their eggs in dead animals and then bury them in the ground until they hatch. Assume that the preference for fresh meat (F) is dominant to the preference for rotted meat and that the tendency to bury the meat shallow (S) is dominant to the tendency to bury the meat deep. Suppose a female carrion beetle homozygous dominant for both traits mates with a male homozygous recessive for both traits. What will be the genotype of the F1 generation?

What will be the phenotype of the F1 generation?

What will be the genotypic ratio of the F2 generation (FFSS : FFSs : FFss : FfSS : FfSs : Ffss : ffSS : ffSs : ffss)?

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5. Suppose in a strain of soybeans, high oil (H) content in the seeds is dominant to low oil content and four seeds (E) in a pod is dominant to two seeds in a pod. A farmer crosses two soybean plants, both with high oil content and four seeds per pod. The resulting F1 offspring have a phenotypic ratio of 9:3:3:1 (High oil / four seeds : High oil / two seeds : Low oil / four seeds : Low oil / two seeds). What genotype were the parent plants?

Suppose the farmer chooses two of the high oil / four seed plants and crosses them. The F2 generation have all high oil / four seed phenotypes. What were the genotypes of the plants chosen by the farmer to cross?

Which known genotypes might the farmer cross her high oil / four seed plants with to determine their genotype?

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