4. Joe, a genetics student at LMU, attended Jane and Paul\'s wedding. While chat

Question

4. Joe, a genetics student at LMU, attended Jane and Paul's wedding. While chatting idly with the mother of the bride, he noticed that she had very prominent mid-digital hair, a dominant trait. His interest was piqued and he added to his observations that her ear lobes were free, her little fingers crooked and she has a prominent widow's peak. He jotted down these observations on his cocktail napkin and sought out Jane. Jane it turms out has free ear lobes, mid-digital hair, straight little fingers and a widow's peak Her father has attached ear lobes, mid-digital hair, crooked little fingers and a widow's peak. His survey of Paul's family revealed that Paul had free ear lobes, mid-digital hair, cooked little fingers and a widow's peak. His mother had free ear lobes, mid-digital hair, straight little fingers and a straight hair line. His father had free ear lobes, no mid-digital hair, crooked little fingers and a widow's peak. Joe remembers that free ear lobes, mid-digital hair, crooked little fingers and widow's peaks are all dominant traits. He retires to a quiet table to figure out the genotypes of these 6 individuals. For any allele he could not be certain about, he put a? (e.g. M2 if he couldn't tell the individual was MM or Mm). What did he jot down on the napkin? 5. Joe was invited to the Christening of Jane and Paul's baby. During the ceremony he discovered the napkin from Jane and Paul's wedding in his pocket. Naturally, after the ceremony, he heads straight for the baby. He determines that the baby has attached ear lobes, no mid-digital hair, straight little fingers and a straight hair line. He uses this information to change some of the ?s on his last survey. What are the genotypes of Jane, Paul and also the baby? What are the chances that Jane and Paul's next baby will be phenotyically like the first baby for these four traits? What is the probability that Jane and Paul's next baby will be genotypically like the first baby for these four traits? What is the probability that the next baby will not be like the first baby for any of these three traits (i.e., will have free lobes, mid-digital lair, crooked little fingers and widow's peak?)

Explanation / Answer

Lets first assign the letters for each (Alleles) for each phenotype

Free ear lobes= F? (it can be FF or Ff); Attached ear lobes= ff (recessive trait)

Mid digital hair= M?( it can be MM or Mm); no mid-digial hair= mm (recessive trait)

Crooked little fingers= C? (it can be CC or Cc); Straight little fingers= cc ( recessive trait)

Widows peak= W? (it can be WW or Ww); Straight hair line= ww (recessive trait).

4. So the possible genotypes are

Jane's mother= F? M? C? W?

Jane= F? M? cc W?

Jane's father= ff M? C? W?

Paul= F? M? C? W?

Paul's mother=F? M? cc ww

Paul's father= F? mm C? W?

5. Baby'sgenotype= ff mm cc ww

Since the baby has all recessive traits then both Jane and Paul have to be heterozygous for the 'doubtful' traits

Therefore the genotypes are

Jane= Ff Mm cc Ww

Paul= Ff Mm Cc Ww

The possible genotypes for the next baby's earlobes when Jane and Paul unite is FF Ff fF (free ear lobes, 3/4) and ff (attached ear lobes 1/4)

The possible genotypes for the next baby's mid-digits when Jane and Paul unite is MM Mm mM (mid-digial hair, 3/4) and mm (no mid-digital hair 1/4)

The possible genotypes for the next baby's little fingers when Jane and Paul unite is cC ( crooked little fingers1/2) and cc (straight little fingers, 1/2)

The possible genotypes for the next baby's hair line when Jane and Paul unite is WW Ww wW ( widows peak, 3/4) and ww (straight hairline, 1/4).

What are the chances that their next baby will be phenotypically like the first baby for these four traits?

ans. 1/4 x 1/4 x 1/ 2 x 1/4= 1/128

What is the probability that their next baby will be genotypically like the first baby for these four traits?

ans. 1/4 x 1/4 x 1/ 2 x 1/4= 1/128

What is the probability that the next baby will not be like the first baby for any of these three traits?

ans. 3/4 x3/4 x1/2 x3/4=27/128

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.