In monohybrid inheritance experiments involving guinea pig coat color, where the allele fo… — JAMB UTME

In monohybrid inheritance experiments involving guinea pig coat color, where the allele for black fur ( $B$ ) is completely dominant over the allele for white fur ( $b$ ), match each specified parental cross or test cross scenario on the left with its corresponding phenotypic or genotypic outcome in the offspring on the right.

Cross between a heterozygous black guinea pig ( $Bb$ ) and a white guinea pig ( $bb$ )A $1:1$ phenotypic and genotypic ratio ( $50\%$ heterozygous black, $50\%$ homozygous white)
Cross between two heterozygous black guinea pigs ( $Bb \times Bb$ ) yielding a total of 160 offspringExpected phenotypic distribution of approximately 120 black fur and 40 white fur offspring
A test cross of a dominant black guinea pig that produces 100% black offspring across multiple littersIndicates the tested parent is homozygous dominant ( $BB$ ), as no recessive allele was contributed by that parent
Cross between pure-breeding black ( $BB$ ) and pure-breeding white ( $bb$ ) parents to produce the $F_1$ generation100% uniform heterozygous genotype ( $Bb$ ) exhibiting the black fur phenotype

Cevap

The correct pairings match each monohybrid inheritance scenario to its precise Mendelian ratio or outcome: (1) Heterozygous black crossed with white matches the 1:1 genotypic and phenotypic ratio; (2) Two heterozygous black parents producing 160 offspring matches 120 black and 40 white offspring (3:1 ratio); (3) A test cross producing 100% dominant offspring confirms a homozygous dominant parent (BB); and (4) Crossing pure-breeding parents yields 100% uniform heterozygous F1 offspring.

Each cross directly demonstrates a fundamental aspect of Mendel's First Law. Heterozygote

\times

homozygous recessive gives a

1:1

ratio; two heterozygotes produce a

3:1

phenotypic ratio (120:40 out of 160); a test cross yielding zero recessive offspring confirms homozygous dominance; and contrasting pure lines produce uniform

F_1

heterozygotes.

Adım Adım Çözüm

Analyze the cross between

Bb

and

bb

Gametes: Parent 1 produces

B

and

b

in equal proportion (

1:1

), Parent 2 produces only

b

. Offspring genotypes:

\frac{1}{2} Bb

(black fur) and

\frac{1}{2} bb

(white fur).

Mendel's Law of Segregation dictates that alleles segregate during gamete formation so that each gamete carries only one allele for each gene.

Calculate expected offspring numbers for

Bb \times Bb

with total

N = 160

Monohybrid phenotypic ratio is

3

dominant :

1

recessive. Dominant count

= \frac{3}{4} \times 160 = 120

; Recessive count

= \frac{1}{4} \times 160 = 40

A monohybrid cross of two heterozygotes generates a genotypic ratio of

1 BB : 2 Bb : 1 bb

, which simplifies to a

3:1

phenotypic ratio under complete dominance.

Evaluate the test cross of an unknown black parent with a recessive

bb

parent

If the parent were

Bb

, white offspring would appear in approximately

50\%

of cases. Since

100\%

of offspring are black, the unknown parent must be homozygous dominant (

BB

A test cross uses a known homozygous recessive individual to reveal whether an organism expressing a dominant trait is homozygous or heterozygous.

Determine the outcome of crossing homozygous contrasting parents (

BB \times bb

)

All

F_1

offspring receive

B

from the black parent and

b

from the white parent, making

100\%

of the

F_1

generation heterozygous (

Bb

This illustrates Mendel's principle of uniformity in the

F_1

generation when crossing pure lines.

Anahtar Kavram

Mendel's Law of Segregation and Monohybrid Inheritance Ratios

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru