Question
Tall (T) is dominant over short ( t ) and black hair (B) is dominant to white hair (b). multicolumn(2)(|c|)( TB ) & Tb & tB & tb TB & TTBB & TTBb & TtBB & TtBb Tb & TTBb & TTbb & TtBb & Ttbb tB & TtBB & TtBb & ttBB & ttBb tb & TtBb & Ttbb & ttBb & ttbb What fraction of the offspring are heterozygous tall and heterozygous for black hair? 1 / 16 4 / 16 2 / 16 8 / 16
Answer
4.7
(152 Votes)
Poppy
Master · Tutor for 5 years
Answer
The fraction of the offspring that are heterozygous tall and heterozygous for black hair is 1/4.
Explanation
## Step 1The problem is about Mendelian genetics, specifically about the inheritance of two traits: height and hair color. According to the information provided, tall (T) is dominant over short (t), and black hair (B) is dominant over white hair (b). ## Step 2The problem asks for the fraction of offspring that are heterozygous tall and heterozygous for black hair. Heterozygous means having two different alleles for a particular trait. In this case, heterozygous tall (Tt) means having one allele for tall (T) and one for short (t). Similarly, heterozygous for black hair (Bb) means having one allele for black hair (B) and one for white hair (b). ## Step 3Looking at the provided Punnett square, we need to find the cells that represent the genotype TtBb. This genotype represents the offspring that are heterozygous tall and heterozygous for black hair.## Step 4Counting the number of TtBb cells in the Punnett square, we find that there are 4 such cells. ## Step 5The total number of cells in the Punnett square is 16 (since it's a 4x4 grid). Therefore, the fraction of offspring that are heterozygous tall and heterozygous for black hair is 4/16.## Step 6However, the question asks for the fraction in its simplest form. The fraction 4/16 can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 4. This gives us the fraction 1/4.