Biology Solver

Analysis

This question pertains to the reproductive biology and life cycle of gymnosperms (seed plants like pines and cycads). To answer it, we must understand the following concepts:

• Alternation of Generations: Gymnosperms have a dominant sporophyte (2n) phase and a highly reduced gametophyte (n) phase.
• Heterospory: They produce two types of spores: microspores (male) and megaspores (female).
• Gametophyte Retention: A key evolutionary step in seed plants is that the gametophytes are no longer free-living (unlike in mosses or ferns) but are instead retained within the protective tissues of the parent sporophyte.
• Terminology: Distinguishing between the sporangium (the spore-producing "container"), the spore, and the gametophyte (the multicellular structure that produces gametes).

Solution

1. The Female Reproductive Structure: In gymnosperms, the female reproductive unit is the ovule. The ovule consists of a central tissue called the megasporangium (also known as the nucellus), which is protected by an outer layer called the integument.
2. Formation of the Megaspore: Within the megasporangium, a specialized cell called the megaspore mother cell undergoes meiosis to produce four haploid megaspores. In most gymnosperms, three of these degenerate, leaving one functional megaspore.
3. Development of the Gametophyte: The functional megaspore does not leave the megasporangium. Instead, it divides mitotically while still inside the nucellus to form the multicellular female gametophyte. This gametophyte eventually develops archegonia (female sex organs) which contain the egg cells.
4. Retention and Dependence: Unlike bryophytes (where the gametophyte is the main plant body) or pteridophytes (where it is a small but independent "prothallus"), the female gametophyte of a gymnosperm is completely dependent on the parent sporophyte for nutrition and protection. It remains retained within the megasporangium throughout its entire existence.

Evaluation of Options:

• (A) Microsporangium: This is the "pollen sac" where male microspores are produced.
• (B) Pollen grain: This is the male gametophyte itself, which travels to the ovule.
• (C) Megasporangium: As explained above, this is the sporophytic tissue that encloses and protects the developing female gametophyte. This is the correct answer.
• (D) Archegonium: These are structures found inside the female gametophyte. The gametophyte contains the archegonium; it is not "retained within" it.

Answer

The correct option is (C) Megasporangium.

Analysis

This is a pedigree analysis problem involving Mendelian genetics. To solve it, we must first determine the mode of inheritance for albinism and then deduce the genotypes of the specific individuals mentioned to calculate the probability of their future offspring having the trait.

Key Concepts:

• Autosomal Recessive Inheritance: A trait that appears only when an individual has two copies of the recessive allele (homozygous recessive). Unaffected parents can have affected children if both parents are carriers (heterozygous).
• Genotype: The genetic makeup of an individual (e.g., $AA$, $Aa$, or $aa$).
• Phenotype: The observable trait (e.g., unaffected or albinism).
• Punnett Square: A tool used to predict the probability of genotypes in offspring.

Solution

Step 1: Determine the Mode of Inheritance

Looking at the pedigree:

1. Recessive vs. Dominant: Individuals I-I and I-II are unaffected, but they have an affected child (II-IV). This "skipping" of a generation is a hallmark of a recessive trait. If it were dominant, at least one parent of an affected individual would have to be affected.
2. Autosomal vs. Sex-linked: The trait affects both males (II-IV) and females (III-IV). While sex-linked traits can show similar patterns, albinism is a well-known autosomal recessive condition. Based on the pedigree alone, unaffected parents (I-I and I-II) having an affected son (II-IV) and an affected father (II-IV) having an affected daughter (III-IV) is consistent with autosomal recessive inheritance.

Let:

• $A$ = Dominant allele (Normal pigmentation)
• $a$ = Recessive allele (Albinism)

Step 2: Assign Genotypes to Individuals II-IV and II-V

• Individual II-IV (Father): He is shaded, meaning he has albinism. Since the trait is recessive, his genotype must be $aa$.
• Individual II-V (Mother): She is unshaded, so she must have at least one dominant allele ($A_$).
• Observation of Offspring: Individuals II-IV and II-V have a daughter, III-IV, who has albinism ($aa$). For a child to be $aa$, they must receive one $a$ allele from each parent.
• Conclusion: Since the mother (II-V) is unaffected but passed an $a$ allele to her daughter, her genotype must be $Aa$ (a carrier).

Step 3: Calculate the Likelihood for the Next Child

We now perform a cross between the father ($aa$) and the mother ($Aa$) using a Punnett Square:

Results of the cross:

• $Aa$ (Unaffected): 2 out of 4 (50%)
• $aa$ (Albinism): 2 out of 4 (50%)

Each pregnancy is an independent event, so the probability remains the same for every subsequent child.

Answer

The likelihood that individuals IV and V from generation II will have another child with albinism is 50% (or 1/2).

This biology problem involves pedigree analysis and Mendelian genetics. Below is the step-by-step breakdown to determine the likelihood of the child having albinism.

[Analysis]

• Type of Question: Pedigree analysis and probability calculation.
• Concepts Used: Autosomal recessive inheritance, genotypes, phenotypes, and Punnett squares.
• Inheritance Pattern: Albinism is an autosomal recessive trait. We can confirm this from the pedigree because unaffected parents (I-I and I-II) have an affected child (II-IV). For a trait to "skip" a generation in this manner, it must be recessive.
• Allele Key:
  • $A$: Dominant allele (Normal pigmentation)
  • $a$: Recessive allele (Albinism)

[Solution]

Step 1: Determine the genotype of Individual III-III

• Individual III-III is the son of Individual II-IV.
• Individual II-IV is shaded, meaning he has albinism. His genotype must be $aa$.
• Because Individual II-IV can only pass on the recessive allele ($a$), all of his biological children must receive one "$a$" from him.
• Individual III-III is unshaded (unaffected), meaning he must have at least one dominant allele ($A$).
• Therefore, the genotype of Individual III-III is $Aa$ (a carrier).

Step 2: Determine the genotype of Individual III-II

• The problem explicitly states that Individual III-II is homozygous dominant.
• Therefore, her genotype is $AA$.

Step 3: Perform the genetic cross

• We are crossing Individual III-II ($AA$) with Individual III-III ($Aa$).
• Punnett Square:

Step 4: Analyze the results

• The possible genotypes for the child are $AA$ (50%) and $Aa$ (50%).
• To have albinism, a child must have the genotype $aa$.
• In this specific cross, there is a 0% chance of the child inheriting two recessive alleles because Individual III-II only possesses dominant alleles to pass on.

[Answer]

The likelihood that the child will have albinism is 0%.