Heredity (A): genes, alleles and Mendel’s first law
الأسبوع 10 · الوحدة الثالثة «تنوع الصفات في الكائنات الحية» (افتتاحية) والدرس الأول «الوراثة وتنوع الصفات» (١): الحمض النووي والجينات والكروموسومات والأليلات والسائد والمتنحي والطراز الجيني والمظهري وقانون مندل الأول ومربع بانيت · 3 دروس
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Mendel's first law and the Punnett square
Science Prep 3 Term 1, Unit 3: pp. 64–70
GoalState Mendel’s first law and use a Punnett square to predict a 3 : 1 ratio.
New words
Mendel’s first law · the two alleles of a gene separate when gametes form, so each gamete gets only onePunnett square · a table that lists all the gamete combinations of a crossfirst and second generation (F₁, F₂) · the offspring of the parents (P), and the offspring of two F₁ plants
Explanation
Gregor Mendel (1822–1884) crossed pea plants in the nineteenth century, because they have clear contrasting traits and can be pollinated under control. He crossed a pure tall plant with a pure short plant, and all the first generation, F₁, were tall: 100 % dominant. When he crossed two F₁ plants, the second generation, F₂, had tall and short plants in a 3 : 1 ratio. This led to his first law: the two alleles of a gene separate when gametes form, and each gamete gets one. We write P for the parents, G for the gametes, F₁ and F₂ for the generations, and × for a cross; in a Punnett square the female gametes go down the side and the male gametes across the top. A Punnett square lists all the combinations: Yy × Yy gives 1 YY : 2 Yy : 1 yy, so 3 yellow : 1 green. Careful: 3 : 1 is a probability, not a promise: a family of four plants need not have exactly one recessive plant.
P: ♀ Pp × ♂ Pp
♀ \ ♂
P
p
P
PP
Pp
p
Pp
pp
Genotype: 1 PP : 2 Pp : 1 pp Phenotype: 3 : 1
Worked examples
P: ♀ PP × ♂ pp
♀ \ ♂
p
P
Pp
Genotype: 1 Pp Phenotype: 1
A pure purple plant (PP) is crossed with a white plant (pp). P: PP × pp. G: P and p. F₁: all Pp, all purple. This is 100 % dominant, exactly as Mendel found.
Two F₁ plants Pp × Pp: each makes gametes P and p with equal chance. The Punnett square gives 1 PP : 2 Pp : 1 pp. Phenotypes: PP, Pp, Pp are purple (3), pp is white (1): 3 purple : 1 white. Among 48 F₂ plants we expect about 36 purple and 12 white (48 ÷ 4 = 12 white).
P: ♀ Aa × ♂ aa
♀ \ ♂
a
A
Aa
a
aa
Genotype: 1 Aa : 1 aa Phenotype: 1 : 1
A cross Aa × aa. Gametes: A, a and a, a. The square gives Aa, Aa, aa, aa, so 1 Aa : 1 aa, a ratio of 1 : 1. Among 20 offspring we expect about 10 of each.
Class activity
Two-coin Punnett simulation. Each parent is a pair of coins: heads (H) is the dominant allele, tails (t) the recessive one. Toss both pairs 40 times and tally the combinations: HH, Ht and tt. About 10 : 20 : 10 is expected, that is about 1 : 2 : 1, but your tally will differ, since it is chance (for example 12, 19 and 9). Explain why the result is only “about” the ratio.
Practice
1
State Mendel’s first law in your own words.
The two alleles of a gene separate when gametes form, so each gamete gets only one allele.
2
What do you get from PP × pp (F₁)?
All Pp, all showing the dominant trait.
3
Pp × Pp: give the genotype ratio and the phenotype ratio.
P: ♀ Pp × ♂ Pp
♀ \ ♂
P
p
P
PP
Pp
p
Pp
pp
Genotype: 1 PP : 2 Pp : 1 pp Phenotype: 3 : 1
1 PP : 2 Pp : 1 pp; 3 dominant : 1 recessive.
4
An F₂ of 60 plants has a 3 : 1 ratio. How many are expected to show the recessive trait?
60 ÷ 4 = 15 recessive (and 45 dominant).
5
A family of four plants has no recessive plant. Does this disprove the 3 : 1 ratio?
No: 3 : 1 is a probability for many plants; small numbers can differ by chance.