Binomial Theorem for Offspring - Genetics

1. In a specific plant species, an allele for red flowers (R) is incompletely dominant over the allele for white flowers (r), with heterozygotes (Rr) producing pink flowers. From a cross between two pink-flowered plants, 5 progeny are grown. What is the probability that a majority of the progeny have pink flowers?
   Answer: 1/2 The cross is Rr x Rr, which produces offspring in the ratio 1/4 red (RR), 1/2 pink (Rr), and 1/4 white (rr). The probability of a single progeny being pink is p = 1/2, and not pink is q = 1/2. For a family of n=5, a 'majority' means 3, 4, or 5 progeny are pink. We must sum the probabilities for these outcomes. P(x=3) = [5!/(3!2!)]*$(1/2)^5$ = 10/32. P(x=4) = [5!/(4!1!)]*$(1/2)^5$ = 5/32. P(x=5) = [5!/(5!0!)]*$(1/2)^5$ = 1/32. The total probability is 10/32 + 5/32 + 1/32 = 16/32 = 1/2.
2. A geneticist performs a dihybrid cross (AaBb x AaBb). If six progeny are analyzed, what is the probability that exactly two will have the genotype aabb?
   Answer: 15 * $(1/16)^2$ * $(15/16)^4$ In a dihybrid cross, the probability of obtaining an offspring with the genotype aabb is p = 1/16. The probability of any other genotype is q = 1 - 1/16 = 15/16. The number of progeny is n=6, and we want to find the probability of exactly x=2 having the aabb genotype. The binomial coefficient is C(n,x) = n!/(x!(n-x)!) = 6!/(2!4!) = (6*5)/2 = 15. The full formula is P(x) = C(n,x) * $p^x$ * $q^{n-x}$, which in this case is 15 * $(1/16)^2$ * $(15/16)^4$.
3. Two parents, both heterozygous for a recessive allele (Aa), have 4 children. Which of the following expressions correctly calculates the probability that they have an equal number of children with the dominant phenotype as with the recessive phenotype?
   Answer: [4!/(2!2!)] * $(3/4)^2$ * $(1/4)^2$ When you encounter genetics problems involving multiple offspring and specific outcome patterns, you're dealing with binomial probability. This requires understanding both the basic genetics (phenotype ratios from crosses) and the combinatorial mathematics of multiple independent events. First, establish the basic genetics: two heterozygous parents (Aa × Aa) produce offspring with a 3:1 phenotype ratio - 3/4 dominant phenotype and 1/4 recessive phenotype. For 4 children to have equal numbers of each phenotype, you need exactly 2 dominant and 2 recessive. The correct calculation uses the binomial probability formula: $\frac{n!}{k!(n-k)!} \times p^k \times (1-p)^{n-k}$, where n=4 total children, k=2 children with dominant phenotype, p=3/4 probability of dominant phenotype. This gives us $\frac{4!}{2!2!} \times (\frac{3}{4})^2 \times (\frac{1}{4})^2$, which is answer D. Answer A incorrectly uses 1/2 probabilities instead of the actual 3/4 and 1/4 phenotype probabilities from the Aa × Aa cross. Answer B omits the combinatorial coefficient entirely - it only calculates the probability of one specific sequence (like dominant-dominant-recessive-recessive) rather than all possible arrangements. Answer C switches the exponents, calculating the probability of 2 recessive children having dominant phenotypes, which is nonsensical. Remember: binomial genetics problems always need three components - the combinatorial coefficient for arrangements, the probability of success raised to the number of successes, and the probability of failure raised to the number of failures.
4. A couple are both heterozygous carriers for an autosomal recessive disorder. They plan to have four children. What is the probability that at least one of their children will be affected by the disorder?
   Answer: 175/256 The cross is Aa x Aa. The probability of an affected child (aa) is p = 1/4. The probability of an unaffected child (A_) is q = 3/4. For a family of n=4 children, the probability of 'at least one' affected child is most easily calculated as 1 minus the probability of zero affected children (P(x=0)). P(x=0) = (4!/(0!4!)) * $(1/4)^0$ * $(3/4)^4$ = 1 1 (81/256) = 81/256. The probability of at least one affected child is 1 - P(x=0) = 1 - 81/256 = 175/256.
5. A couple, both carriers for an autosomal recessive disease, have five children. What is the ratio of the probability of having exactly one affected child to the probability of having exactly two affected children?
   Answer: 3:2 For the cross Aa x Aa, p(affected) = 1/4 and q(unaffected) = 3/4. n=5. We need to calculate P(x=1) and P(x=2) and find their ratio. P(x=1) = [5!/(1!4!)] * $(1/4)^1$ * $(3/4)^4$ = 5 (1/4) (81/256) = 405/1024. P(x=2) = [5!/(2!3!)] * $(1/4)^2$ * $(3/4)^3$ = 10 (1/16) (27/64) = 270/1024. The ratio P(1)/P(2) is (405/1024) / (270/1024) = 405/270. Dividing both by 135 gives 3/2, or a ratio of 3:2.
6. In a certain species of fish, scale color is determined by a single gene with two alleles. When two heterozygous fish are crossed, 3/4 of the offspring are black and 1/4 are white. If a heterozygous fish is instead crossed with a white fish and they produce a school of 6 offspring, what is the probability that the number of black fish is not equal to the number of white fish?
   Answer: 11/16 This problem tests your understanding of Mendelian genetics and binomial probability. When you see genetics crosses with specific offspring ratios, start by determining the inheritance pattern and genotypes involved. From the first cross (heterozygous × heterozygous → 3/4 black : 1/4 white), you can identify that black is dominant over white. Let's call the alleles B (black) and b (white). The heterozygous parents are Bb × Bb, producing offspring in a 3:1 ratio typical of dominant inheritance. For the test cross (Bb × bb), each offspring has a 1/2 probability of being black (Bb) and 1/2 probability of being white (bb). With 6 offspring, you need to find the probability that black ≠ white, which means finding 1 minus the probability that black = white. For equal numbers in 6 offspring, you need exactly 3 black and 3 white. Using the binomial probability formula: $P(3 \text{ black}) = \binom{6}{3} \times (1/2)^3 \times (1/2)^3 = 20 \times (1/2)^6 = 20/64 = 5/16$ Therefore, the probability of unequal numbers is $1 - 5/16 = 11/16$. Answer choice A (5/16) represents the probability of getting exactly equal numbers, not unequal. Choice B (7/8) and C (3/4) likely come from incorrectly calculating probabilities for different offspring totals or misunderstanding the question. Study tip: In genetics probability problems, always identify the inheritance pattern first, then set up your binomial probability calculations carefully. Pay close attention to whether the question asks for a specific outcome or its complement.
7. A rare autosomal dominant disorder exhibits 75% penetrance. A man heterozygous for the causative allele marries a woman who is homozygous recessive. If they have four children, what is the probability that exactly two of them will be phenotypically affected?
   Answer: 0.2109 First, determine the probability (p) of a single child being phenotypically affected. The cross is Aa x aa, so the probability of inheriting the dominant allele (A) is 1/2. With 75% penetrance, p = P(inheriting A) P(penetrance) = 0.5 0.75 = 0.375. The probability of not being affected is q = 1 - 0.375 = 0.625. For n=4 children and x=2 affected, we use the binomial formula: P(x=2) = [4!/(2!2!)] * $p^2$ * $q^2$ = 6 * $(0.375)^2$ * $(0.625)^2$ = 6 0.140625 0.390625 ≈ 0.2109.
8. A researcher crosses a plant that is heterozygous for a dominant allele (Dd) with a homozygous recessive plant (dd). Out of 5 resulting seeds, what is the probability that more seeds are of the recessive phenotype than the dominant phenotype?
   Answer: 1/2 The cross Dd x dd yields 1/2 dominant phenotype (Dd) and 1/2 recessive phenotype (dd). So, p(recessive) = 1/2 and q(dominant) = 1/2. In a family of n=5, 'more recessive than dominant' means having 3, 4, or 5 recessive offspring. P(x=3) = [5!/(3!2!)]*$(1/2)^5$ = 10/32. P(x=4) = [5!/(4!1!)]*$(1/2)^5$ = 5/32. P(x=5) = [5!/(5!0!)]*$(1/2)^5$ = 1/32. The total probability is 10/32 + 5/32 + 1/32 = 16/32 = 1/2. Due to the symmetry of p=q=1/2, the probability of more recessive is the same as more dominant, and both must be 1/2.
9. A woman who is a carrier for an X-linked recessive disorder and a man who is unaffected plan to have four children. What is the probability that exactly two of their children will be carrier daughters?
   Answer: 27/128 The cross is $X^A$ $X^a$ x $X^A$ Y. The possible offspring are: $X^A$ $X^A$ (unaffected daughter), $X^A$ $X^a$ (carrier daughter), $X^A$ Y (unaffected son), and $X^a$ Y (affected son), each with a 1/4 probability. Let the desired outcome be a carrier daughter, so p = 1/4. Any other outcome is the alternative, so q = 3/4. For n=4 children, we want to find the probability that exactly x=2 are carrier daughters. P(x=2) = [4!/(2!2!)] * $(1/4)^2$ * $(3/4)^2$ = 6 (1/16) (9/16) = 54/256 = 27/128.
10. In a test cross involving a heterozygous individual (Gg), six progeny are produced. What is the probability that at least five of the progeny will exhibit the dominant phenotype?
    Answer: 7/64 A test cross is Gg x gg. The offspring genotypes are 1/2 Gg (dominant phenotype) and 1/2 gg (recessive phenotype). The probability of the dominant phenotype is p = 1/2, and recessive is q = 1/2. For n=6 progeny, 'at least five' means P(x=5) + P(x=6). P(x=5) = [6!/(5!1!)] * $(1/2)^5$ * $(1/2)^1$ = 6 (1/32) (1/2) = 6/64. P(x=6) = [6!/(6!0!)] * $(1/2)^6$ * $(1/2)^0$ = 1 (1/64) 1 = 1/64. The total probability is 6/64 + 1/64 = 7/64.
11. A man with blood type A (genotype $I^A$ i) and a woman with blood type B (genotype $I^B$ i) have five children. What is the probability that none of their children have blood type O?
    Answer: 243/1024 The cross $I^A$ i x $I^B$ i produces four equally likely offspring genotypes: $I^A$ $I^B$ (Type AB), $I^A$ i (Type A), $I^B$ i (Type B), and ii (Type O). The probability of a single child having type O blood (genotype ii) is p = 1/4. The probability of a single child NOT having type O blood is q = 1 - 1/4 = 3/4. We are looking for the probability that out of n=5 children, exactly x=0 have type O blood. P(x=0) = [5!/(0!5!)] * $(1/4)^0$ * $(3/4)^5$ = 1 1 (243/1024) = 243/1024.
12. A couple are both carriers for a recessive allele that is lethal during embryonic development when homozygous. If they have three living children, what is the probability that exactly one of their living children is a carrier of the lethal allele?
    Answer: 2/9 The cross is Aa x Aa, yielding genotypes 1/4 AA, 2/4 Aa, and 1/4 aa. Since 'aa' is lethal, it is excluded from the population of living children. The adjusted probabilities for living children are 1/3 AA (unaffected non-carrier) and 2/3 Aa (unaffected carrier). The probability that a living child is a carrier is p = 2/3, and the probability of being a non-carrier is q = 1/3. For n=3 living children, the probability of exactly x=1 carrier is: P(x=1) = [3!/(1!2!)] * $(2/3)^1$ * $(1/3)^2$ = 3 (2/3) (1/9) = 6/27 = 2/9.
13. A pea plant heterozygous for both height (Tt) and seed color (Gg) is self-pollinated. If five offspring are analyzed, what is the probability that exactly three are dwarf?
    Answer: 45/512 This question involves a dihybrid cross, but only one trait (height) is relevant to the question. The cross for height is Tt x Tt. The probability of an offspring being dwarf (tt) is p = 1/4. The probability of being tall (T_) is q = 3/4. For n=5 offspring, we want to find the probability that exactly x=3 are dwarf. P(x=3) = [5!/(3!2!)] * $(1/4)^3$ * $(3/4)^2$ = 10 (1/64) (9/16) = 90/1024 = 45/512.
14. A woman who is a carrier for red-green color blindness, an X-linked recessive trait, marries a man with normal vision. They have three sons. What is the probability that at least one of their sons is color-blind?
    Answer: 7/8 The cross is $X^C$ $X^c$ x $X^C$ Y. The probability of having a color-blind son ($X^c$ Y) is 1/4. The probability of having an unaffected son ($X^C$ Y) is 1/4. For any given son, the probability that he is color-blind is 1/2 (50% of male offspring). Let p = 1/2 for a son being color-blind. For n=3 sons, P(at least one color-blind) = 1 - P(zero color-blind). P(x=0) = $(1/2)^3$ = 1/8. Therefore, P(at least one) = 1 - 1/8 = 7/8.
15. Two heterozygous parents (Aa) have 4 children. What is the probability that their first child is unaffected and that, out of the total four children, exactly one is affected?
    Answer: 81/256 This is a conditional probability problem combining simple probability with the binomial theorem. The probability of the first child being unaffected (A_) is q = 3/4. After this event, we need to find the probability that among the remaining n=3 children, exactly one is affected (aa). For this part, p(affected) = 1/4 and q(unaffected) = 3/4. P(1 affected in 3) = [3!/(1!2!)] * $(1/4)^1$ * $(3/4)^2$ = 3 (1/4) (9/16) = 27/64. The total probability is the product of these two independent events: P(1st unaffected) P(1 affected in next 3) = (3/4) (27/64) = 81/256.
16. A couple learns they are both carriers for an autosomal recessive condition. They have a child who is affected. What is the probability that if they have two more children, both of these next two children will be unaffected?
    Answer: 9/16 When you encounter autosomal recessive inheritance problems involving multiple children, you're dealing with independent probability events. Each pregnancy has the same odds regardless of previous outcomes. Since both parents are carriers (Aa × Aa), each child has a $\frac{1}{4}$ chance of being affected (aa) and a $\frac{3}{4}$ chance of being unaffected (AA or Aa). The fact that their first child is affected doesn't change these probabilities for future children. For both of the next two children to be unaffected, you multiply the individual probabilities: $\frac{3}{4} × \frac{3}{4} = \frac{9}{16}$. This makes D correct. Choice A ($\frac{1}{2}$) incorrectly assumes each child has a 50% chance of being unaffected, which would apply to autosomal dominant conditions or X-linked inheritance in certain scenarios. Choice B ($\frac{1}{4}$) represents the probability that both children would be affected, not unaffected. Choice C ($\frac{3}{4}$) is the probability for just one child to be unaffected, not both children. The key strategy here is recognizing that genetic inheritance follows independent assortment—each pregnancy is a separate event with identical odds. Always start by determining the probability for a single event, then apply multiplication rules for multiple independent events. Remember that in autosomal recessive crosses between carriers, the ratio is always 1:2:1 (affected:carrier:normal), giving $\frac{3}{4}$ unaffected overall.
17. In Labrador retrievers, coat color is determined by epistatic interactions between two genes (B/b and E/e). A cross between two black labs of genotype BbEe produces a litter of 4 pups. What is the probability that there are no yellow pups in the litter?
    Answer: 81/256 The cross BbEe x BbEe results in a 9:3:4 phenotypic ratio of black:brown:yellow pups. The yellow phenotype (genotype __ee) occurs with a probability of 4/16 = 1/4. Therefore, the probability of a pup being yellow is p = 1/4, and the probability of it not being yellow (i.e., black or brown) is q = 3/4. The question asks for the probability of zero yellow pups (x=0) in a litter of n=4. P(x=0) = [4!/(0!4!)] * $(1/4)^0$ * $(3/4)^4$ = 1 1 (81/256) = 81/256.

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