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An autosomal allele \(N\) in humans causes abnormalities in nails and patellae (kneecaps) called the nail-patella syndrome. Consider marriages in which one partner has the nail-patella syndrome and blood type A and the other partner has normal nails and patellae and blood type O. These marriages produce some children who have both the nail-patella syndrome and blood type A. Assume that unrelated children from this phenotypic group mature, intermarry, and have children. Four phenotypes are observed in the following percentages in this second generation: nail-patella syndrome, blood type A \(66 \%\) normal nails and patellae, blood type 0 if \(16 \%\) normal nails and patellae, blood type A \(9 \%\) nail-patella syndrome, blood type \(\mathrm{O}\), \(9 \%\) Fully analyze these data, explaining the relative frequencies of the four phenotypes. (See pages \(214-215\) for the genetic basis of these blood types.)

Short Answer

Expert verified
The distribution results from Mendelian crosses between interrelated genotypic traits following independent assortment.

Step by step solution

01

Understanding the Genetic Background

The nail-patella syndrome is caused by a dominant allele (N) and it is inherited independently of the ABO blood group. Blood type A can be genotypically either homozygous (IAIA) or heterozygous (IAi). The blood type O is caused by the ii genotype. The N allele is dominant, so any individual having at least one N allele will exhibit the nail-patella syndrome, while nn individuals will have normal phenotypes.
02

Initial Cross and Genotype Analysis

We start with a cross between a heterozygous parent for nail-patella syndrome (Nn) and blood type A (IAi) with a homozygous normal nails and patellae (nn) and blood type O (ii). The offspring can inherit N or n from the parent with the syndrome and IA or i for blood type. This produces genotypes: NIAi (nail-patella, A), Ni (nail-patella, O), nIAi (normal nails, A), and nii (normal nails, O).
03

Second Generation Genotypic Analysis

Children from the first generation with the phenotype NIA and blood type A intermarry. Possible genotype for these children is NIAi. Crossing NIAi with another NIAi, we perform the Punnett square for both nail-patella and blood type independently. The ratio of phenotypes includes: NIA/NIA (nail-patella, A), NIA/i (nail-patella, A), NiA/NIA (nail-patella, A), NIA/i (nail-patella, A), and nii (normal nails, O) among others.
04

Calculating Phenotypic Ratios

Evaluating the Punnett squares, we see that: 66% of the offspring have nail-patella syndrome and blood type A, 16% have no syndrome and blood type O (nii), 9% show normal nails with blood type A, and 9% show napa with blood type O. These proportions arise from independent assortments of the genotypes and demonstrate expected phenotypic ratios from a dihybrid cross.
05

Validation with Observed Data

Observations from the problem statement include 66% NIA, 16% nii, 9% nIA, and 9% NiO. These match the expected theoretical values based on genotype independent assortment and Mendelian principles. Thus, the proportions correctly demonstrate how genetic traits segregate and show the compiled phenotypes in the second generation.

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Key Concepts

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

Autosomal Dominant Inheritance
Autosomal dominant inheritance refers to a pattern in which a single copy of a dominant allele on a non-sex chromosome (autosome) can cause a trait or disorder to be expressed. In disorders like Nail-Patella Syndrome, the presence of just one dominant allele (N) is sufficient for the trait to manifest. This occurs regardless if the second allele is normal (n).
Key characteristics of autosomal dominant inheritance include:
  • Vertical transmission: The trait is often present in every generation because only one copy of the gene is needed to express the trait.
  • Equal prevalence among genders: Since it is autosomal, both males and females are equally likely to be affected.
  • Children have a 50% chance of inheriting the trait if one parent is affected.
Understanding this concept is essential when analyzing genetic phenomena like Nail-Patella Syndrome, as it helps predict patterns and potentials for future generations.
ABO Blood Group System
The ABO blood group system is a classic example of co-dominance in human genetics. It consists of four possible blood types: A, B, AB, and O, determined by the combination of two alleles (IA, IB, i). Blood type A could be genotypically IAIA or IAi, where IA is dominant over i. Thus, individuals must have at least one IA allele to express type A. Type O results from a homozygous recessive genotype (ii).
Important features include:
  • Codominance: Both IA and IB alleles are codominant, meaning that if both are present (IAIB), the blood type will be AB, expressing both antigen types.
  • Genetic diversity: Because of multiple allele forms and combinations, the ABO system exemplifies how genes can contribute to population variation.
  • Inheritance pattern: The ABO blood type is inherited independently of other traits like the Nail-Patella Syndrome.
Grasping how these blood groups work is key for understanding complex genetic problems that involve multiple genetic traits.
Mendelian Genetics
Mendelian genetics forms the basis of understanding inheritance patterns based on Gregor Mendel's principles. His laws describe how traits pass from parents to offspring through genes. For the Nail-Patella Syndrome and blood group study, the three core Mendelian laws are pivotal:
  • Law of Segregation: Each organism carries two alleles for a trait, segregating independently into gametes, meaning offspring have a chance to inherit either.
  • Law of Independent Assortment: Different genes are inherited independently, as shown by the separate inheritance of the nail-patella trait and ABO blood type.
  • Law of Dominance: The dominant allele's trait will be expressed in a heterozygote, explaining why the Nail-Patella Syndrome appears with only one N allele.
These principles are instrumental in predicting outcomes in genetic crosses and understanding how traits like nail-patella syndrome and blood types are passed down.
Dihybrid Cross Analysis
Dihybrid cross analysis involves predicting the outcome of breeding experiments between organisms that differ in two traits. In the context of the given problem, the traits are Nail-Patella Syndrome and ABO blood types. A classic dihybrid cross can be explored using a 4x4 Punnett square to determine the genetic makeup and resulting phenotypes of offspring.
Key aspects include:
  • Setting up the Punnett square: List all combinations of alleles from each parent related to both traits. For example, NIAi crossed with NIAi focuses on dominant and recessive combinations.
  • Phenotypic ratios: The square helps visualize resulting phenotype ratios, such as the 66% Nail-Patella, blood type A outcome.
  • Independent assortment: Each trait pair segregates independently, demonstrating Mendel’s principle of independent assortment.
Understanding dihybrid crosses reveals how multiple genetic factors interplay in offspring phenotypes and facilitates more complex genetic prediction analyses.

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Most popular questions from this chapter

A corn geneticist wants to obtain a corn plant that has the three dominant phenotypes: anthocyanin (A), long tassels (L), and dwarf plant (D). In her collection of pure lines, the only lines that bear these alleles are \(A A L L d d\) and aa \(U D D\). She also has the fully recessive line aa \(u\) dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny a plant of the desired phenotype (which would have to be Aa Ll \(D d\) in this case). She knows that the three genes are linked in the order written and that the distance between the \(A / a\) and the \(L / l\) loci is 16 map units and that the distance between the \(L / l\) and the \(D / d\) loci is 24 map units. a. Draw a diagram of the chromosomes of the parents, the hybrid, and the tester. b. Draw a diagram of the crossover(s) necessary to produce the desired genotype. c. What percentage of the testcross progeny will be of the phenotype that she needs? d. What assumptions did you make (if any)?

In the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele \(t\) ) was crossed with a strain that is auxotrophic for methionine (mutant allele \(m\) ). Linear asci were isolated and classified into the following groups. $$\begin{array}{lcccccc}\text { Spore pair } & {}{} {\text { Ascus types }} \\\ \hline 1 \text { and } 2 & t+ & t+ & t+ & t+ & t m & t m \\ 3 \text { and } 4 & t+ & t m & +m & ++ & t m & ++ \\ 5 \text { and } 6 & +m & ++ & t+ & t m & ++ & t+ \\ 7 \text { and } 8 & +m & +m & +m & +m & ++ & +m \\ \hline \text { Number } & 260 & 76 & 4 & 54 & 1 & 5 \\ \hline\end{array}$$ a. Determine the linkage relations of these two genes to their centromere(s) and to each other. Specify distances in map units. b. Draw a diagram to show the origin of the ascus type with only one single representative (second from right

For a certain chromosomal region, the mean number of crossovers at meiosis is calculated to be two per meiosis. In that region, what proportion of meioses are predicted to have (a) no crossovers? (b) one crossover? (c) two crossovers?

Consider the Neurospora cross \(+;+\times f ; p\) It is known that the \(+/ f\) locus is very close to the centromere on chromosome \(7-\) in fact, so close that there are never any second-division segregations. It is also known that the \(+/ p\) locus is on chromosome \(5,\) at such a distance that there is usually an average of 12 percent second- division segregations. With this information, what will be the proportion of octads that are a. parental ditypes showing \(M_{1}\) patterns for both loci? b. nonparental ditypes showing \(\mathrm{M}_{1}\) patterns for both loci? c. tetratypes showing an \(\mathrm{M}_{1}\) pattern for \(+/ f\) and an \(\mathrm{M}_{\mathrm{ll}}\) pattern for \(+/ p ?\) d. tetratypes showing an \(\mathrm{M}_{\mathrm{II}}\) pattern for \(+/ f\) and an \(\mathrm{M}_{\mathrm{I}}\) pattern for \(+/ p ?\)

A female animal with genotype \(A / a \cdot B / b\) is crossed with a double- recessive male \((a / a \cdot b / b) .\) Their progeny include 442 A/a \(\cdot B / b, 458 a / a \cdot b / b, 46 A / a \cdot b / b,\) and 54 \(a / a \cdot B / b .\) Explain these results.

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