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In a maternity ward, four babies become accidentally mixed up. The ABO types of the four babies are known to be \(\mathrm{O}, \mathrm{A}, \mathrm{B},\) and \(\mathrm{AB}\). The \(\mathrm{ABO}\) types of the four sets of parents are determined. Indicate which baby belongs to each set of parents: (a) \(A B \times O,(b) A \times O,(c) A x\) \(\mathrm{AB},(\mathrm{d}) \mathrm{O} \times \mathrm{O}\).

Short Answer

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(a) Baby B, (b) Baby A, (c) Baby AB, (d) Baby O.

Step by step solution

01

Matching Parents and Baby Types

To assign babies to their parents, we'll assess blood type compatibility. A person's blood type is determined by one allele from each parent. Possible blood type alleles are A, B, and O. For case (d), O × O can only produce O since O is recessive. Therefore, baby O must belong to this pair.
02

Identifying Baby A's Parents

Next, consider the A blood type baby. Baby A cannot belong to parents (d) since that pair is already matched with baby O, nor can it belong to (c) A × AB, which cannot produce a pure A type (possible combinations AO, AB, BO). The combination A × O (case b) is the only pair that could produce an A baby (since AA or AO are possible, but AA is impossible with an O parent). Therefore, baby A belongs to these parents.
03

Determining Baby B's Parents

Now, look at the baby with blood type B. Baby B cannot belong to parents (d) or (b). The pair (a) AB × O can produce a B baby, with allele combinations being BO or AO (AO produces A type). Thus, baby B belongs to the AB × O parents.
04

Assigning Baby AB's Parents

The remaining baby, AB, must belong to parents (c) A × AB, as this combination can result in an AB type child (potential combinations AB, AA, AO, BO), fulfilling the condition without discrepancy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Blood Type Inheritance
The ABO blood group system is central to understanding how blood types are inherited. In this system, three main alleles determine a person's blood type: A, B, and O. Every individual carries two alleles, one inherited from each parent.
  • Alleles A and B are dominant over O.
  • This dominance means a person with at least one A allele will have an A-type phenotype, and the same goes for B.
  • The combination of alleles results in four potential blood types: A, B, AB, and O.
A mix-up in a maternity ward can be clarified by understanding these combinations and probabilities:
  • The O blood type is recessive, requiring two O alleles to be expressed.
  • Type AB can only appear if one parent contributes an A allele and the other a B allele.
  • Knowing these inheritance patterns allows us to predict which parents could possibly have a child with a particular blood type.
Genetics in Maternity
In maternity cases, especially those involving mix-ups, the genetic background helps determine parentage. Blood types act as genetic markers that aid in identifying parent and child compatibility. Each parent passes one of their two alleles to their offspring, making blood type determination a genetic puzzle that can be solved with the right keys. Consider the pairings and their possibilities:
  • An O × O pairing can only produce an O child due to the recessive nature of O alleles.
  • An A × O pairing can result in either an A or O child, depending on which alleles are passed on.
  • An AB parent can contribute either an A or B allele.
This genetic understanding is crucial for accurately assigning parentage and ensuring each baby is matched with the correct parents, especially in scenarios where documentation may be unclear or missing.
Allele Combinations
Allele combinations epitomize genetic diversity and the inheritance possibilities of blood types. Each parent has two alleles, but they can only pass one per trait to their child. For the ABO system:
  • A combination of AA or AO results in blood type A.
  • BB or BO results in blood type B.
  • AB produces blood type AB, as both alleles are co-dominant.
  • OO results exclusively in blood type O.
When evaluating potential parent-child blood type pairings, considering these combinations helps in solving puzzles related to inheritance and can confirm or refute potential biological relationships. This knowledge helps both in genetic consultations and resolving issues that may arise, such as in the case study from the maternity ward.

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Most popular questions from this chapter

A dominant allele \(H\) reduces the number of body bristles that Drosophila flies have, giving rise to a "hairless" phenotype. In the homozygous condition, \(H\) is lethal. An independently assorting dominant allele \(S\) has no effect on bristle number except in the presence of \(H\) in which case a single dose of \(S\) suppresses the hairless phenotype, thus restoring the hairy phenotype. However, \(S\) also is lethal in the homozygous (S/S) condition. a. What ratio of hairy to hairless flies would you find in the live progeny of a cross between two hairy flies both carrying \(H\) in the suppressed condition? b. When the hairless progeny are backcrossed with a parental hairy fly, what phenotypic ratio would you expect to find among their live progeny?

Many kinds of wild animals have the agouti coloring pattern, in which each hair has a yellow band around it. a. Black mice and other black animals do not have the yellow band; each of their hairs is all black. This absence of wild agouti pattern is called nonagouti. When mice of a true-breeding agouti line are crossed with nonagoutis, the \(F_{1}\) is all agouti and the \(F_{2}\) has a 3: 1 ratio of agoutis to nonagoutis. Diagram this cross, letting \(A\) represent the allele responsible for the agouti phenotype and \(a\) nonagouti. Show the phenotypes and genotypes of the parents, their gametes, the \(\mathrm{F}_{1}\), their gametes, and the \(\mathrm{F}_{2}\). b. Another inherited color deviation in mice substitutes brown for the black color in the wild-type hair. Such brown-agouti mice are called cinnamons. When wild-type mice are crossed with cinnamons, all of the \(\mathrm{F}_{1}\) are wild type and the \(\mathrm{F}_{2}\) has a 3: 1 ratio of wild type to cinnamon. Diagram this cross as in part \(a\), letting \(B\) stand for the wild-type black allele and \(b\) stand for the cinnamon brown allele. c. When mice of a true-breeding cinnamon line are crossed with mice of a true- breeding nonagouti (black) line, all of the \(\mathrm{F}_{1}\) are wild type. Use a genetic diagram to explain this result. d. In the \(F_{2}\) of the cross in part \(c,\) a fourth color called chocolate appears in addition to the parental cinnamon and nonagouti and the wild type of the \(\mathrm{F}_{1}\). Chocolate mice have a solid, rich brown color. What is the genetic constitution of the chocolates? e. Assuming that the \(A / a\) and \(B / b\) allelic pairs assort independently of each other, what do you expect to be the relative frequencies of the four color types in the \(\mathrm{F}_{2}\) described in part \(d ?\) Diagram the cross of parts \(c\) and \(d,\) showing phenotypes and genotypes (including gametes). f. What phenotypes would be observed in what proportions in the progeny of a backcross of \(\mathrm{F}_{1}\) mice from part \(c\) with the cinnamon parental stock? With the nonagouti (black) parental stock? Diagram these backcrosses. g. Diagram a testcross for the \(\mathrm{F}_{1}\) of part \(c .\) What colors would result and in what proportions? h. Albino (pink-eyed white) mice are homozygous for the recessive member of an allelic pair \(C / c,\) which assorts independently of the \(A / a\) and \(B / b\) pairs. Suppose that you have four different highly inbred (and therefore presumably homozygous) albino lines. You cross each of these lines with a true-breeding wild-type line, and you raise a large \(\mathrm{F}_{2}\) progeny from each cross. What genotypes for the albino lines can you deduce from the following \(\mathrm{F}_{2}\) phenotypes?

In the multiple-allele series that determines coat color in rabbits, \(c^{+}\) encodes agouti, \(c^{\mathrm{ch}}\) encodes chinchilla (a beige coat color), and \(c^{\mathrm{h}}\) encodes Himalayan. Dominance is in the order \(c^{+}>c^{c h}>c^{h} .\) In a cross of \(c^{+} / c^{c h} \times\) \(c^{\mathrm{ch}} / c^{\mathrm{h}},\) what proportion of progeny will be chinchilla?

A plant thought to be heterozygous for two independently assorting genes \((P / p ; Q / q)\) was selfed, and the progeny were \\[ 88 \quad P /-; Q /- \\] \(25 p / p ; Q /-\) \\[ 32 P /-; q / q \\] \(14 \mathrm{p} / \mathrm{p} ; \mathrm{q} / \mathrm{q}\) Do these results support the hypothesis that the original plant was \(P / p ; Q / q ?\)

Mice normally have one yellow band on each hair, but variants with two or three bands are known. A female mouse having one band was crossed with a male having three bands. (Neither animal was from a pure line.) The progeny were \(\begin{array}{lll}\text { Females } & \frac{1}{2} \text { one band } & \text { Males } & \frac{1}{2} \text { one band } \\ & \frac{1}{2} \text { three bands } & & \frac{1}{2} \text { two bands }\end{array}\) a. Provide a clear explanation of the inheritance of these phenotypes. b. In accord with your model, what would be the outcome of a cross between a three-banded daughter and a one-banded son?

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