In order to receive full credit, you must fill-in answers appropriately on the a
ID: 59239 • Letter: I
Question
In order to receive full credit, you must fill-in answers appropriately on the attached answer sheet and attach all your scratch-paper/work sheets used to solve the problems. In humans, normal skin pigmentation is due to a dominant gene, albinism to its recessive allele. A woman with normal skin pigmentation marries a man who is albino. If they have a child who is albino, what are the genotypes of the parents and their child? If these two individuals were to have more children, what would the ratio of phenotypes be? Assume that mermaids are real and their tails can have solid or striped fins. You conduct a series of crosses and obtain the results given in the table Define gene symbols and give all the possible genotypes of the parents of each cross. The autosomal gene for brachydactyly. short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that: The first child will have brachydactyly? The first two children will have brachydactyly? Their first child will be a brachydactyly girl?Explanation / Answer
Q1 (a) There are 2 alleles for skin pigmentation- A and a. Normal skin pigmentation is due to dominant gene and albinism is due to recessive allele.
Genotypes for normal skin pigmentation can be either AA or Aa
Genotype for albino is aa
A woman with normal skin pigmentation marries a man who is albino and they have a child who is also albino. This is feasible only if the woman and man have Aa and aa genotypes respectively.
b) - aa; - Aa. Children born to these parents will have normal skin pigmentation (Aa) and albino (aa) phenotypes in 1:1 ratio.
Q2 The mermaids can have solid or striped fins. Based on the given information, it can be understood that solid fin is a dominant character and striped fin is a recessive character.
Assuming, S- represents dominant character and s- recessive character. Hence, genotypes for solid fin is SS and Ss whereas mermaids with striped fins will possess ss genotype.
a)Solid fins and striped fins containing mermaids were mated, it resulted in 1:1 (60:55) of striped and solid fins mermaids. Therefore, genotypes of solid and striped fins mermaid parents are Ss and ss respectively.
b)Solid and solid fins containing mermaids were mated, it resulted in only solid fins mermaids. Therefore, genotypes of both the solid fins mermaid parents is SS.
c)Striped and striped fins containing mermaids were mated, it resulted in only striped fins mermaids. Therefore, genotypes of both the striped fins mermaid parents is ss.
d)Solid and solid fins containing mermaids were mated, it resulted in solid fins: striped fins mermaids in a ratio of 3:1 (92:30). Therefore, genotypes of solid fins mermaid parents are Ss.
e)Solid fins and striped fins containing mermaids were mated, it resulted in only solid fins mermaids. Therefore, genotypes of solid and striped fins mermaid parents are SS and ss respectively.
Q3 Finger length is controlled by a gene with 2 alleles i.e. B and b. Brachydactyly is a dominant over normal finger length.
Therefore, genotypes for brachydactyly are BB and Bb whereas normal finger length corresponds to bb genotype. According to the question, a female heterozygous for brachydactyly (Bb) marries a man with normal finger length (bb).
a)Children born to these parents will have Bb and bb genotypes (1:1 ratio) which corresponds brachydactyly and normal finger length. Therefore, there is 50% probability that their first child will have brachydactyly.
b)The probability that their first two children will be brachydactyly is 25%.
c)There are 50% chances that their first child has brachydactyly. Within this 50%, there equal probability that the child born is a girl or a boy. Therefore, the probability that their first child will be brachydactyly girl is 25%.