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Katena32 [7]
4 years ago
14

In order for a colorblind female to be born, ____________________. A- the mother must be a carrier B- the mother must be colorbl

ind C- both parents must be colorblind D- the father must be colorblind
Biology
1 answer:
Pani-rosa [81]4 years ago
7 0
The correct answer is D - the father must be colorblind.

Colorblindness is a recessive trait, so in order for it to appear, you would need either one heterozygous and one homozygous recessive parent, two heterozygous parents, or two homozygous recessive parents. 
Males only have one X chromosome to give to make a female. Their other chromosome is a Y. These sex-linked traits only appear on the X chromosome.
Since colorblindness is recessive, this X chromosome the father gives must be recessive, meaning <em>he must be colorblind.
</em>The mother can either be homozygous recessive (colorblind) or a carrier.
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7 0
4 years ago
In genetics, the dash symbol (-) is a "wild card" that stands for either the dominant allele or the recessive allele; for exampl
m_a_m_a [10]

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12 : 3 : 1

Explanation:

According to Mendel's principles, a dihybrid cross between two heterozygous summer squash parents (WwGg X WwGg) should yield offsprings, 12 with white, 3 with yellow and  1 with green color phenotype. The classic Mendelian phenotypic ratio for dihybrid cross is 9:3:3:1. In this case, genotypes W- G- and W- gg produce white color phenotype. This is because we have 12: 3: 1 phenotypic ratio in which the alleles of two different genes assort independently into gametes.

<em>Dihybrid cross for heterozygous summer squash:</em>

                    WG         Wg            wG            wg          

    WG      WWGG   WWGg     WwGG     WwGg

    Wg       WWGg   WWgg     WwGg      Wwgg

    wG        WwGG    WwGg     wwGG      wwGg

      wg       WwGg    Wwgg     wwGg        wwgg

   Genotypes:

   W-G-  and W- gg  = white = 12

    ww G- = yellow = 3

     ww gg = green = 1

3 0
4 years ago
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