Question

ProblemAssume that long fingers are inherited as a recessive trait with 80% penetrance. Two people heterozygous for long fingers mate. What is the probability that their first child will have long fingers?L= short fingersl= long fingersL(short) >l (long)Ll X Ll short short gametes: ½ L ½ l ½ L ½ l fertilization: ½ L ½ l ½ L ¼ LL¼ Llshort short ½ l¼ Ll¼ ll short long

Answer 1

Answer:

The probability that their first child will have long fingers is 20%  

Explanation:

Available data:

• long fingers are inherited with 80% penetrance

• L represents the dominant allele for short fingers

• l represents the recessive allele for long fingers

Cross) Two people heterozygous for long fingers mate

                            Ll          x         Ll

Gametes)    1/2 L    1/2l         1/2L    1/2l

Punnet Square)  

                               L         l

                     L      LL      Ll

                       l     Ll       ll

F1) 25% or 1/4 LL

     50% or 1/2 Ll

     25% or 1/4 ll

Long fingers are inherited with 80% penetrance, which means that 80% is the proportion of the population that expresses the phenotype long fingers. Hence, the probability that the first child in this cross will have long fingers is:

100% ------- 80%

25% -------- X = 20%