Understand Inheritance - IB Biology

1. Two linked genes, P and Q, have a recombination frequency of 18%. An individual is heterozygous for both genes, with a chromosome arrangement of PQ/pq. What percentage of gametes produced by this individual is expected to have the genotype Pq?
   Answer: 9% The parental gametes are PQ and pq. The recombinant gametes are Pq and pQ. The total recombination frequency is 18%, which represents the sum of all recombinant gametes. This frequency is split equally between the two types of recombinant gametes. Therefore, the percentage of Pq gametes is 18% / 2 = 9%. The remaining 82% of gametes would be parental (41% PQ and 41% pq).
2. A farmer has a bull with a dominant trait for black coat colour (B). To determine if the bull is homozygous (BB) or heterozygous (Bb), the farmer should cross it with a cow of which genotype and look for which outcome?
   Answer: A bb cow; any non-black offspring would prove the bull is heterozygous. This procedure is a test cross. To determine an unknown dominant genotype, the individual is crossed with a homozygous recessive individual (bb). If the bull is heterozygous (Bb), the cross Bb x bb will produce approximately 50% Bb (black) and 50% bb (non-black) offspring. The appearance of even one non-black (bb) calf proves the bull must carry the recessive 'b' allele and is therefore heterozygous.
3. Which aspect of Gregor Mendel's experimental design was most critical for his ability to deduce the principles of inheritance, which had eluded previous investigators?
   Answer: His application of mathematics to large sample sizes to identify statistical patterns. While all options were important parts of his experimental design, the most revolutionary aspect that allowed Mendel to deduce his laws was his quantitative approach. By counting large numbers of offspring and analyzing the results statistically, he was able to see the underlying probabilistic ratios (e.g., 3:1, 9:3:3:1) that revealed the principles of segregation and independent assortment. Previous studies were often purely observational and qualitative.
4. Human skin colour shows continuous variation, resulting from the combined effects of several genes. Which term best describes the genetic basis for such traits?
   Answer: Polygenic inheritance Continuous variation, where phenotypes show a range of values along a continuum (like height or skin colour), is characteristic of polygenic inheritance. This is when a single phenotypic trait is controlled by the additive effects of two or more genes. Pleiotropy is one gene affecting multiple traits. Codominance and multiple alleles refer to patterns of inheritance for a single gene.
5. In cats, the gene for orange fur colour (O) is on the X chromosome. The allele O results in orange fur, while the allele o results in black fur. Heterozygous females ($X^O$ $X^o$) have patches of both orange and black fur, a pattern called tortoiseshell. A tortoiseshell female mates with a black male. What is the expected phenotypic ratio of their male offspring?
   Answer: 1 orange : 1 black The tortoiseshell female has the genotype $X^O$ $X^o$. The black male has the genotype $X^o$ Y. Male offspring inherit their Y chromosome from the father and one of the two X chromosomes from the mother. There is a 50% chance a male kitten will inherit the $X^O$ chromosome from the mother, making him $X^O$ Y (orange). There is a 50% chance he will inherit the $X^o$ chromosome, making him $X^o$ Y (black). Therefore, the expected ratio for male offspring is 1 orange : 1 black. Tortoiseshell is not possible in males as it requires two X chromosomes.
6. A woman has blood type B, and her father had blood type O. She has a child with a man who has blood type A, but whose mother had blood type O. What is the probability that their first child will have blood type O?
   Answer: 0.25 The woman's blood type is B, but since her father was type O (genotype ii), she must have inherited an i allele from him. Therefore, her genotype is $I^B$ i. The man's blood type is A, but since his mother was type O (genotype ii), he must have inherited an i allele from her. His genotype is $I^A$ i. The cross is $I^B$ i x $I^A$ i. The possible offspring genotypes are $I^A$ $I^B$, $I^A$ i, $I^B$ i, and ii. Each has a probability of 1/4. The genotype for blood type O is ii, so the probability is 1/4 or 0.25.
7. A plant breeder crosses a homozygous red-flowered plant with a homozygous blue-flowered plant. The resulting F1 generation has flowers with distinct patches of red and blue. This is an example of which type of inheritance?
   Answer: Codominance Codominance occurs when both alleles in a heterozygous individual are fully and independently expressed. The presence of distinct patches of both red and blue indicates that both the red and blue alleles are being expressed simultaneously, which is the definition of codominance. Incomplete dominance would result in a blended, intermediate phenotype (e.g., purple flowers).
8. A single mutation in the human FBN1 gene can lead to Marfan syndrome, which affects the skeleton, eyes, and cardiovascular system. In fruit flies, the genes for body colour and wing length are located close together on the same chromosome and are often inherited together. What are the genetic principles that best explain these two observations, respectively?
   Answer: Pleiotropy and gene linkage Pleiotropy is the phenomenon where a single gene influences multiple, seemingly unrelated phenotypic traits. Marfan syndrome, where one mutated gene causes a wide range of symptoms, is a classic example. Gene linkage is the tendency of genes that are located physically close to each other on the same chromosome to be inherited together. The co-inheritance of body colour and wing length in fruit flies is a classic example of linkage.
9. In fruit flies, grey body (G) is dominant to ebony body (g), and normal wings (W) are dominant to vestigial wings (w). A fly with a grey body and normal wings is test-crossed. The cross produces offspring in a phenotypic ratio of 1 grey-normal : 1 ebony-normal. What was the genotype of the grey-bodied, normal-winged parent?
   Answer: GgWW A test cross involves crossing with a homozygous recessive individual (ggww). The offspring phenotypes reveal the gametes produced by the parent being tested. The offspring are grey-normal (G_W_) and ebony-normal (ggW_). Since ebony (gg) offspring are produced, the parent must carry a 'g' allele. Since only normal wings (W_) are produced, the parent must be homozygous dominant for the wing gene (WW), as crossing with 'ww' would otherwise produce some vestigial offspring. Therefore, the parent's genotype is GgWW.
10. A pedigree analysis reveals that an affected father has an unaffected daughter. This single observation definitively rules out which mode of inheritance?
    Answer: X-linked dominant In X-linked dominant inheritance, an affected father ($X^AY$) passes his single X chromosome, which carries the dominant allele, to all of his daughters. Therefore, all of his daughters must be affected. The observation of an unaffected daughter ($X^aX^a$) means she did not receive the dominant allele from her father, which is impossible. Thus, X-linked dominant inheritance is ruled out.
11. A dihybrid cross is performed between two plants heterozygous for both flower colour (Pp) and seed shape (Rr). The resulting offspring show a large excess of parental phenotypes (purple flowers/round seeds and white flowers/wrinkled seeds) and a significant deficit of recombinant phenotypes. Which statement is the most likely conclusion?
    Answer: The genes for flower colour and seed shape are linked. The expected phenotypic ratio for an unlinked dihybrid cross is 9:3:3:1. A significant deviation from this ratio, characterized by an overrepresentation of the parental combinations and underrepresentation of new combinations (recombinants), is the classic sign of gene linkage. This means the two genes are located on the same chromosome and tend to be inherited together.
12. In Labrador retrievers, coat colour is determined by two genes. One gene determines pigment production (E/e), where E_ allows pigment and ee results in a yellow coat regardless of the other gene. The second gene determines the pigment colour (B/b), where B_ is black and bb is brown. A cross is made between two dogs with the genotypes BbEe x Bbee. What proportion of the puppies is expected to be yellow?
    Answer: 1/2 This is an example of epistasis. A yellow coat occurs whenever the genotype is ee. We only need to analyze the cross for the E/e gene: Ee x ee. The possible offspring genotypes are 1/2 Ee and 1/2 ee. Therefore, 1/2 or 50% of the offspring will have the 'ee' genotype and will be yellow, regardless of the B/b gene alleles they inherit.
13. Coat colour in rabbits is controlled by a single gene with four alleles showing a dominance hierarchy: agouti (C) > chinchilla ($c^ch$) > Himalayan ($c^h$) > albino (c). What phenotypic ratio would be expected from a cross between a rabbit of genotype $Cc^h$ and a rabbit of genotype c^chc?
    Answer: 2 agouti : 1 chinchilla : 1 Himalayan The cross is $Cc^h$ × c^chc. The gametes from the first parent are C and $c^h$. The gametes from the second parent are c^ch and c. A Punnett square gives four offspring genotypes: Cc^ch, Cc, c^chc^h, and c^hc. According to the dominance hierarchy: Cc^ch is agouti (C > $c^ch$). Cc is agouti (C > c). c^chc^h is chinchilla ($c^ch$ > $c^h$). c^hc is Himalayan ($c^h$ > c). This results in a phenotypic ratio of 2 agouti : 1 chinchilla : 1 Himalayan.
14. Phenylketonuria (PKU) is an autosomal recessive disorder. A couple, John and Mary, are both phenotypically normal. John has a brother with PKU, and Mary's mother has PKU. What is the probability that their first child will be a carrier of the PKU allele?
    Answer: 1/2 Since Mary's mother has PKU (aa), Mary must be a carrier (Aa); P(Mary is Aa) = 1. Since John's brother has PKU (aa), his parents must both be carriers (Aa). John is phenotypically normal, so his genotype is either AA or Aa. The probability that a normal child from an Aa x Aa cross is a carrier (Aa) is 2/3. So, P(John is Aa) = 2/3. To find the probability of a carrier child (Aa), we consider two cases for John's genotype: Case 1: John is AA (P=1/3). The cross is AA x Aa. The probability of an Aa child is 1/2. Contribution: (1/3) (1/2) = 1/6. Case 2: John is Aa (P=2/3). The cross is Aa x Aa. The probability of an Aa child is 1/2. Contribution: (2/3) (1/2) = 2/6. The total probability is the sum of contributions: 1/6 + 2/6 = 3/6 = 1/2.
15. Monozygotic twins share an identical genotype. A study of adult identical twins who were separated at birth and raised in different socioeconomic conditions finds significant, consistent differences in their IQ scores and susceptibility to certain diseases. What is the most valid scientific conclusion?
    Answer: Phenotype is the result of the interaction between genotype and environmental factors. Since the twins are genetically identical, any consistent differences between them must be attributed to environmental influences. However, the fact that they are still more similar to each other than to the general population (a common finding in such studies) shows a genetic component. Therefore, the best conclusion is that phenotype arises from a complex interaction between a fixed genotype and variable environmental factors. The other options are too extreme or less likely.
16. A researcher performs a dihybrid cross assuming independent assortment and observes 480 offspring. To perform a chi-squared test on the results, the researcher needs to calculate the expected values. What is the expected number of offspring predicted to show both dominant phenotypes?
    Answer: 270 For a dihybrid cross with independent assortment, the expected phenotypic ratio is 9:3:3:1. The proportion of offspring expected to show both dominant phenotypes corresponds to the '9' part of the ratio. The total proportion is 9 + 3 + 3 + 1 = 16. So, the expected fraction is 9/16. The expected number is the total number of offspring multiplied by this fraction: 480 × (9/16) = 30 × 9 = 270.
17. Red-green colour blindness is an X-linked recessive trait. A woman with normal vision, whose father was colour-blind, marries a man with normal vision. What is the probability that their first son will be colour-blind?
    Answer: 0.50 The woman's father was colour-blind, so his genotype was X^bY. She inherited his X chromosome, so she must be a carrier ($X^BX^b$), as she has normal vision. The man has normal vision, so his genotype is X^BY. For a son to be born, he must receive a Y chromosome from the father. He will receive one of the mother's two X chromosomes. There is a 1/2 chance he receives her $X^B$ (normal vision) and a 1/2 chance he receives her $X^b$ (colour-blind). Therefore, the probability that a son born to this couple will be colour-blind is 1/2 or 0.50.

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