Two-character genetic inheritance (problems solved). (Page 9)
Now we will see how you are able to solve genetic problems based on Mendel's third law.
Now we will see how you are able to solve genetic problems based on Mendel's third law.
In peas, the gene for skin color has two alleles: yellow (A) and green (a). The gene that determines skin texture has two others: smooth skin (B) and rough skin (b). Smooth yellow pea plants (AA, BB) are crossed with rough green pea plants (aa, bb), obtaining 1000 peas from these crosses.
What results are to be expected?
The aniridia (impaired vision) in man is due to a dominant factor (A). The migraine is due to another gene also dominant (J). A man who suffered from aniridia and whose mother did not, married a woman who suffered from migraine, but whose father did not suffer from it. Neither the man had a headache, nor the woman aniridia. What proportion of his children will suffer from both?
In peas, the gene that determines skin color has two alleles: yellow (A) and green (a). The gene that determines skin texture has two others: smooth skin (B) and rough skin (b). Smooth yellow pea plants (Aa, Bb) are crossed with smooth yellow pea plants (Aa, Bb). From these crosses plants are obtained that yield 220 kg of peas.
How many kilograms of each type will be obtained?
In peas, the gene for skin color has two alleles: yellow (A) and green (a). The gene that determines skin texture has two others: smooth skin (B) and rough skin (b). Smooth yellow pea plants (Aa, Bb) are crossed with smooth green pea plants (aa, Bb). 884 Kg of peas are obtained from these crosses.
What results are expected?
Fat mice can be produced by two independent genes: the "oo" genotype produces a fat mouse called obese and the "aa" genotype gives rise to a fat mouse called adipose. Dominant alleles produce normal growth. What phenotypic ratio of fat to normal mice is expected in the cross between two mice of the OoAa genotype?
A plant that has compound and serrated leaves is crossed with another plant that has simple, lobed leaves. Each parent is homozygous for one of the dominant characteristics and one of the recessive characteristics. What is the genotype of the F1 generation? What is its phenotype? If F1 individuals are crossed, what phenotypes will the F2 generation have and in what proportion? (Use the symbols C: compound, c: simple, A: lobed; a: serrated). Reason for the answers.
Suppose that in humans, the inheritance of hair and eye color is simple and is determined by two autosomal genes with the following relationships: brown eye color (A) dominant over blue (a) and dark hair (B ) dominant on blonde hair (b).
a) A man with brown eyes and dark hair marries a woman with blue eyes and dark hair and they have 2 children, one with brown eyes and blond hair and the other with blue eyes and dark hair. Reasonably indicate the genotypes of the parents and children.
b) If the man from the previous section with brown eyes and dark hair married a woman with blue eyes and blond hair. What genotypes and phenotypes could the couple's children have?
In Drosophila melanogaster, a recessive mutant allele, black, gives rise to a very dark body in homozygosity. The normal wild-type color is gray. Another sepia mutant allele, also recessive, gives rise to brown eyes. The normal color is red. By crossing a homozygous ♂ with red eyes and black body with a ♀ with sepia eyes and gray body, an F1 was obtained in which all the flies were red-eyed and gray-bodied. Subsequently, the ♂ and ♀ of F1 were crossed with each other to obtain F2.
a) What are the genotypes of the F1 parents and descendants?
b) Indicate the genotypic and phenotypic proportions of the F2 offspring.
Gray body (G)> black (g)
Red eye (R)> sepia (r)
a)
Male: RRgg
Female: rrG-. Since all the descendants of F1 are RrGg (red eyes and gray body), it has to be rrGG.
Parental: RRgg x rrGG
Gametes: Rg rg
F1 RrGg
b)
F1 crossover: RrGg x RrGg
Gametes: RG Rg rG rg RG Rg rG rg
Possible crosses to obtain the possible genotypes in F2.
|
RG |
Rg |
rG |
rg |
RG |
RRGG |
RRGg |
RrGG |
RrGg |
Rg |
RRGg |
RRgg |
RrGg |
Rrgg |
rG |
RrGG |
RrGg |
rrGG |
rrGg |
rg |
RrGg |
Rrgg |
rrGg |
rrgg |
Genotypes:
Possible phenotypes:
In certain animal species, gray hair (G) is dominant over white hair (g) and curly hair (R) over straight hair (r). An individual with curly gray hair, who has a white-haired father and a straight-haired mother, is crossed with another with straight white hair.
a) Can they have children with gray and straight hair? If so, in what percentage?
b) Can they have children with white and curly hair? If yes, in what percentage?
c) What Mendel's law does this question refer to? State it.
An individual with gray and curly hair (GR-) is crossed, which has a father with white hair (gg--) and a mother with straight hair (--rr), for which the individual is GgRr
The other individual in the crossbreed has smooth white hair (ggrr).
Crossing: GgRr x ggrr
Gametes: GR G r gR gr gr
F1: 25% GgRr (curly gray) 25% Ggrr (smooth gray) 25% ggRr (curly white) 25% ggrr (smooth white)
This exercise refers to Mendel's third Law .
In pea, the characters long stem and red flower dominate over dwarf stem and white flower. What will be the proportion of double homozygous plants that can be expected in the F2 obtained from a cross between two pure lines, one with a long stem and a white flower with another with a dwarf stem and a red flower? Indicate the genotype of all homozygous plants that can appear in F2. Reason the answer
The long stem (L) dominates over the dwarf stem (l).
The flower color red (A) dominates over the white (a).
Cross 1: LLaa (long stem, white flower) x llAA (dwarf stem, red flower)
Gametes: La lA
F1: LlAa (long stem, red flower)
Junction 2: LlAa x LlAa
Gametes: LA La lA la LA La lA la
F2: In the Punnett square we will observe all possible crosses
LA | La | lA | la | |
LA | LLAA | LLAa | LlAA | LlAa |
La | LLAa | LLaa | LlAa | Llaa |
lA | LlAA | LlAa | llAA | llAa |
la | LlAa | Llaa | llAa | llaa |
The genotypes of dihomozygous plants are as follows:
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