Analyze the provided pedigree data to determine the inheritance pattern and genotypes of the individuals…

Question

Analyze the provided pedigree data to determine the inheritance pattern and genotypes of the individuals shown.

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Here are the solutions for the pedigree diagram: 2.1 Step 1: Observe the phenotypes of the parents. Both the Father (unshaded square) and the Mother (unshaded circle) are normal. Step 2: Observe the phenotypes of their children. They have an affected daughter (shaded circle) and an affected son (shaded square). Step 3: Since the genetic disorder is controlled by a recessive allele (n), affected individuals must have the genotype nn. Step 4: For the affected children (nn) to inherit two recessive alleles, they must receive one n allele from the Father and one n allele from the Mother. Step 5: Since both parents are phenotypically normal, they must possess at least one dominant allele (N). Step 6: Therefore, both parents must carry the recessive allele (n) while also having the dominant allele (N), making their genotype heterozygous (Nn). Both parents are normal but have affected children, meaning they must both carry the recessive allele. 2.2 Step 1: From 2.1, we know both parents are heterozygous (Nn). Step 2: Perform a genetic cross between two heterozygous parents: Nn × Nn. Step 3: The possible genotypes for the offspring are NN, Nn, and nn. Step 4: Normal children are those who do not express the recessive disorder. Their genotypes can be homozygous dominant (NN) or heterozygous (Nn). The possible genotype(s) of the normal children are: NN and Nn 2.3 Step 1: Identify the parents and their children in the pedigree. The Father and Mother are normal. Step 2: Observe their first child, who is an affected female (shaded circle). Step 3: If the characteristic were sex-linked (specifically X-linked recessive, as most sex-linked disorders are), an affected female would have the genotype Xⁿ Xⁿ. Step 4: For a female to have the genotype Xⁿ Xⁿ, she must inherit one Xⁿ allele from her mother and one Xⁿ allele from her father. Step 5: However, the Father in the pedigree is normal (unshaded square). If the characteristic were X-linked recessive, a normal father would have the genotype X^N Y. He would pass his X^N allele to all his daughters. Step 6: Since a normal father (X^N Y) cannot have an affected daughter (Xⁿ Xⁿ), the characteristic cannot be sex-linked. A normal father has an affected daughter. If the characteristic were sex-linked recessive, a normal father could not have an affected daughter.

Analyze the provided pedigree data to determine the inheritance pattern and genotypes of the individuals shown.

Step-by-step solution

Part 2.1· 6 steps

Observe the phenotypes of the parents.

Observe the phenotypes of their children.

Since the genetic disorder is controlled by a recessive allele ( $n$ ), affected individuals must have the genotype $nn$ .

For the affected children ( $nn$ ) to inherit two recessive alleles, they must receive one $n$ allele from the Father and one $n$ allele from the Mother.

Since both parents are phenotypically normal, they must possess at least one dominant allele ( $N$ ).

Therefore, both parents must carry the recessive allele ( $n$ ) while also having the dominant allele ( $N$ ), making their genotype heterozygous ( $Nn$ ).

Part 2.2· 4 steps

From 2.1, we know both parents are heterozygous ( $Nn$ ).

Perform a genetic cross between two heterozygous parents

The possible genotypes for the offspring are $NN$ , $Nn$ , and $nn$ .

Normal children are those who do not express the recessive disorder.

Part 2.3· 6 steps

Identify the parents and their children in the pedigree.

Observe their first child, who is an affected female (shaded circle).

If the characteristic were sex-linked (specifically X-linked recessive, as most sex-linked disorders are), an affected female would have the genotype $X^n X^n$ .

For a female to have the genotype $X^n X^n$ , she must inherit one $X^n$ allele from her mother and one $X^n$ allele from her father.

However, the Father in the pedigree is normal (unshaded square).

Since a normal father ( $X^N Y$ ) cannot have an affected daughter ( $X^n X^n$ ), the characteristic cannot be sex-linked.

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Analyze the provided pedigree data to determine the inheritance pattern and genotypes of the individuals shown.

Step-by-step solution

Part 2.1· 6 steps

Observe the phenotypes of the parents.

Observe the phenotypes of their children.

Since the genetic disorder is controlled by a recessive allele (nnn), affected individuals must have the genotype nnnnnn.

For the affected children (nnnnnn) to inherit two recessive alleles, they must receive one nnn allele from the Father and one nnn allele from the Mother.

Since both parents are phenotypically normal, they must possess at least one dominant allele (NNN).

Therefore, both parents must carry the recessive allele (nnn) while also having the dominant allele (NNN), making their genotype heterozygous (NnNnNn).

Part 2.2· 4 steps

From 2.1, we know both parents are heterozygous (NnNnNn).

Perform a genetic cross between two heterozygous parents

The possible genotypes for the offspring are NNNNNN, NnNnNn, and nnnnnn.

Normal children are those who do not express the recessive disorder.

Part 2.3· 6 steps

Identify the parents and their children in the pedigree.

Observe their first child, who is an affected female (shaded circle).

If the characteristic were sex-linked (specifically X-linked recessive, as most sex-linked disorders are), an affected female would have the genotype XnXnX^n X^nXnXn.

For a female to have the genotype XnXnX^n X^nXnXn, she must inherit one XnX^nXn allele from her mother and one XnX^nXn allele from her father.

However, the Father in the pedigree is normal (unshaded square).

Since a normal father (XNYX^N YXNY) cannot have an affected daughter (XnXnX^n X^nXnXn), the characteristic cannot be sex-linked.

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Since the genetic disorder is controlled by a recessive allele ( $n$ ), affected individuals must have the genotype $nn$ .

For the affected children ( $nn$ ) to inherit two recessive alleles, they must receive one $n$ allele from the Father and one $n$ allele from the Mother.

Since both parents are phenotypically normal, they must possess at least one dominant allele ( $N$ ).

Therefore, both parents must carry the recessive allele ( $n$ ) while also having the dominant allele ( $N$ ), making their genotype heterozygous ( $Nn$ ).

From 2.1, we know both parents are heterozygous ( $Nn$ ).

The possible genotypes for the offspring are $NN$ , $Nn$ , and $nn$ .

If the characteristic were sex-linked (specifically X-linked recessive, as most sex-linked disorders are), an affected female would have the genotype $X^n X^n$ .

For a female to have the genotype $X^n X^n$ , she must inherit one $X^n$ allele from her mother and one $X^n$ allele from her father.

Since a normal father ( $X^N Y$ ) cannot have an affected daughter ( $X^n X^n$ ), the characteristic cannot be sex-linked.