孟德尔遗传与系谱
引入| English | 释义 | 拼音 |
|---|---|---|
| pedigree | 系谱图 | xì pǔ tú |
| consanguinity | 近亲婚配 | jìn qīn hūn pèi |
| proband | 先证者 | xiān zhèng zhě |
| autosomal dominant | 常染色体显性 | cháng rǎn sè tǐ xiǎn xìng |
| autosomal recessive | 常染色体隐性 | cháng rǎn sè tǐ yǐn xìng |
| X-linked recessive | X连锁隐性 | X lián suǒ yǐn xìng |
| X-linked dominant | X连锁显性 | X lián suǒ xiǎn xìng |
| de novo mutation | 新发突变 | xīn fā tū biàn |
What are the chances for the next child?
- A couple with one affected child asks the question every genetics clinic hears. The answer is a probability, and it comes from two tools: the family drawing and the grid of gametes.
- Four inheritance patterns cover most single-gene conditions, and each leaves its own signature on the drawing.
Reading a pedigree
- A pedigree 系谱图 uses squares for males and circles for females; a filled symbol means affected. A horizontal line joins a couple, and a double line marks consanguinity 近亲婚配 (the couple are related).
- An arrow marks the proband 先证者, the person through whom the family came to the clinic.
Dominant and recessive autosomes
- Autosomal dominant 常染色体显性: affected people in every generation, both sexes, and father-to-son transmission is possible. An affected parent (Aa) and an unaffected one (aa) have a 1/2 risk for each child.
- Autosomal recessive 常染色体隐性: affected brothers and sisters in one generation, healthy carrier parents, often consanguinity. Two carriers have a 1/4 risk for each child.
Which pedigree most suggests autosomal recessive inheritance?
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Recessive conditions cluster in one generation, with carrier parents, and consanguinity makes two carriers more likely.
The X-linked patterns
- X-linked recessive X连锁隐性: mostly males; a carrier mother gives each son a 1/2 risk and each daughter a 1/2 chance of being a carrier. A father passes his Y to every son, so male-to-male transmission rules this pattern out.
- X-linked dominant X连锁显性: both sexes affected; an affected father passes it to ALL his daughters and to NONE of his sons.
Match each pedigree signature to its pattern.
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Each pattern leaves its own signature on the drawing.
Which finding rules out X-linked recessive inheritance?
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A father passes his Y, not his X, to a son, so the son's condition cannot come from the father's X.
Ratios from gametes
- Aa × Aa gives AA, Aa, Aa, aa. For a recessive condition 1/4 are affected and 2/4 are carriers.
- The healthy brother or sister of an affected child is not aa, so only three boxes remain: their chance of being a carrier is 2/3, not 1/2.
Two carriers have a child with a recessive condition. Their healthy daughter wants to know her chance of being a carrier. Give it as a percentage.
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Of AA, Aa, Aa and aa, she cannot be aa. Two of the three remaining boxes are carriers, so 2/3 ≈ 67%.
Probabilities, not guarantees
- Each child is a new draw. Two carriers have three unaffected children $(3/4)^3 \approx 0.42$, or 42% of the time.
- So a recessive condition can hide in a small family, and "nobody else has it" is weak evidence.
Two carriers of a recessive condition have three children. What is the chance, as a percentage, that none of them is affected?
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Each child is unaffected with probability 3/4, so (3/4)³ = 27/64 ≈ 42.2%.
New mutations
- About 80% of children with achondroplasia (a dominant form of short stature) have average-height parents: the change is a de novo mutation 新发突变, more likely when the father is older.
- The brothers and sisters have a low risk, but the affected child will pass it on to each of their own children with a risk of 1/2.
A child with achondroplasia born to average-height parents has a 1/2 chance of passing it on to each of their own children.
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The parents' gametes carried a new mutation, but the child now has one copy of a dominant variant, so each of their children has a 1/2 risk.
Worked example: Bayes and age
- (Fictional) In a dominant late-onset condition, suppose 20% of carriers are still healthy at 60. A healthy 60-year-old child of an affected parent started at a risk of 1/2.
- $P(\text{carrier} \mid \text{healthy at 60}) = \dfrac{\tfrac12 \times 0.2}{\tfrac12 \times 0.2 + \tfrac12} = \dfrac{1}{6}$: being healthy so far is evidence.
Worked example: the healthy sister
- (Fictional) A woman's brother has cystic fibrosis, a recessive condition; she is healthy, so her carrier chance is 2/3. Her partner, with no family history, has the population carrier chance of 1/25.
- Risk for their child: $\tfrac23 \times \tfrac{1}{25} \times \tfrac14 = \tfrac{1}{150}$. Testing either partner changes the numbers at once.
A healthy woman whose brother has a recessive condition has a carrier chance of 2/3; her partner's is 1/25. Their child's risk is 1 in how many?
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2/3 × 1/25 × 1/4 = 2/300 = 1/150.
Three common slips
- "Dominant means common." Many dominant conditions are rare, and many cases are new mutations.
- "A healthy sibling has a 1/2 carrier risk." Once aa is excluded, it is 2/3.
- "Male-to-male transmission fits X-linked recessive." A father gives his son a Y, never his X.
Recap
- AD: every generation, 1/2 per child. AR: one generation, 1/4 per child, healthy sibs 2/3 carriers.
- X-linked recessive: mostly males, no male-to-male. X-linked dominant: all daughters, no sons of an affected father.
- Small families rarely show the ratios; age of onset and new mutations change the raw numbers, and Bayes combines them.