Heredity and Variation

159 soru

Soru 121Soru

In fruit flies (*Drosophila melanogaster*), the allele for normal wings (VV) is completely dominant over the allele for vestigial wings (vv). A monohybrid cross is carried out between two heterozygous normal-winged flies, producing a total of 320320 offspring in the F1F_1 generation. How many of these offspring are expected to be heterozygous normal-winged?

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Cevap: 160160

Cevap

The expected number of heterozygous normal-winged offspring is 160160.
Crossing two heterozygous individuals (Vv×VvVv \times Vv) yields a genotypic distribution of 1/4VV1/4\,VV, 1/2Vv1/2\,Vv, and 1/4vv1/4\,vv. Taking half of the total population (320320) gives 160160 expected heterozygous (VvVv) offspring.

Adım Adım Çözüm

1
Determine parental genotypes and set up the Punnett square
Both parents are heterozygous (Vv×VvVv \times Vv). Gametes produced by each parent are VV and vv.
Mendel's First Law (Law of Segregation) states that alleles segregate during gamete formation so that each gamete carries only one allele for each gene.
2
Determine the expected genotypic ratio of the F1F_1 generation
Genotypes produced: 1/4VV1/4\,VV (homozygous dominant), 2/4Vv2/4\,Vv (heterozygous dominant), 1/4vv1/4\,vv (homozygous recessive). Ratio is 1:2:11 : 2 : 1.
Combining gametes randomly yields 25%VV25\%\,VV, 50%Vv50\%\,Vv, and 25%vv25\%\,vv.
3
Calculate the expected number of heterozygous offspring
24×320=160\frac{2}{4} \times 320 = 160 offspring.
Multiply the fraction representing the heterozygous genotype (1/21/2 or 50%50\%) by the total offspring count (320320).

Anahtar Kavram

Mendel's Law of Segregation and Monohybrid Genotypic Ratios
Tahmini Süre:1m 0s
Soru 122Soru

Match each human phenotypic trait on the left with its corresponding pattern of variation and underlying genetic mechanism on the right.

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Öğeler

ABO blood group system
Adult body height
Rhesus factor status
Skin pigmentation gradient

Eşleşmeler

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Cevap

The correct pairings match ABO blood group system with discontinuous variation controlled by multiple alleles; Adult body height with continuous variation governed by polygenes; Rhesus factor status with discontinuous variation determined by monogenic inheritance; and Skin pigmentation gradient with continuous variation mediated by polygenes and environmental exposure.
Each phenotypic trait is correctly matched according to whether its phenotypic distribution is continuous (quantitative spectrum driven by polygenes) or discontinuous (qualitative discrete classes driven by single-gene or multiple-allele systems). ABO blood grouping involves multiple alleles yielding distinct blood groups; height is a classic polygenic quantitative continuous trait; Rhesus factor is a simple monogenic positive/negative discontinuous trait; and skin color is a polygenic continuous trait influenced by environmental factors.

Adım Adım Çözüm

1
Classify ABO blood group system according to its phenotypic pattern and genetic basis.
ABO blood grouping exhibits discrete categories (A, B, AB, O) controlled by multiple alleles at a single locus, matching discontinuous variation by multiple alleles.
Monogenic traits with multiple alleles do not produce intermediate spectrum values between blood types.
2
Analyze adult body height phenotypic distribution and inheritance.
Height displays continuous quantitative variation controlled by many genes working together, matching polygenic continuous variation.
Polygenic inheritance creates a continuous distribution curve without distinct phenotypic breaks.
3
Examine Rhesus factor inheritance and phenotype categories.
Rhesus factor displays clear-cut presence (Rh+) or absence (Rh-) governed by a single gene locus, matching monogenic discontinuous variation.
Single-gene traits with complete dominance form separate, non-overlapping phenotype classes.
4
Identify the variation profile of skin pigmentation gradient.
Skin tone varies continuously due to additive polygenes and is influenced by environmental UV exposure, matching environmentally modulated polygenic continuous variation.
Continuous traits often reflect environmental influences layered over polygenic inheritance.

Anahtar Kavram

Continuous and Discontinuous Variation
Soru 123Soru

Glucose-6-phosphate dehydrogenase (G6PD) deficiency is an X-linked recessive metabolic condition in humans. If a man with normal enzyme activity (XGYX^G Y) marries a heterozygous carrier woman (XGXgX^G X^g), what is the probability that any child born to them will be an affected male?

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Cevap: 25%

Cevap

The probability that any given child from this marriage will be an affected male is 25%.
The mother passes either XGX^G or XgX^g with equal probability (0.50.5). The father passes either XGX^G or YY with equal probability (0.50.5). An affected male must inherit the defective XgX^g allele from the mother (p=0.5p = 0.5) and the YY chromosome from the father (p=0.5p = 0.5). Multiplying these independent probabilities yields 0.5×0.5=0.250.5 \times 0.5 = 0.25, or 25%.

Adım Adım Çözüm

1
Identify parental genotypes and gametes
Father (XGYX^G Y) produces sperm XGX^G and YY. Mother (XGXgX^G X^g) produces eggs XGX^G and XgX^g.
Determining gamete types is required to construct the genetic cross.
2
Construct the Punnett square to find offspring genotypes
Possible genotypes: XGXGX^G X^G (25% normal female), XGXgX^G X^g (25% carrier female), XGYX^G Y (25% normal male), XgYX^g Y (25% affected male).
Combining parental gametes shows all possible genetic combinations for their children.
3
Calculate the specific probability for an affected male among all children
The target genotype XgYX^g Y occupies 1 out of 4 total squares = 1/4=25%1/4 = 25\%.
The question asks for the probability relative to any child born, not restricted to sons only.

Anahtar Kavram

X-linked Recessive Inheritance Probability
Tahmini Süre:1m 30s
Soru 124Soru

Human phenotypic traits exhibiting discontinuous variation, such as the ABO blood group system and tongue rolling ability, are governed by polygenic inheritance and can be significantly modified by environmental factors during an individual's lifetime.

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Cevap: False

Cevap

The statement is False. Discontinuous variation is characterized by monogenic inheritance producing non-overlapping phenotypic categories that are not modified by environmental factors.
The statement is false because discontinuous variation results from monogenic inheritance, producing clear-cut phenotypic classes that are completely independent of environmental factors.

Adım Adım Çözüm

1
Analyze the phenotypic expression of the traits listed (ABO blood group and tongue rolling).
These traits fall into distinct, non-overlapping categories without intermediate forms.
Clear-cut distinct classes are the hallmark of discontinuous variation.
2
Evaluate the genetic control and environmental influence described in the statement.
Discontinuous traits are governed by monogenic inheritance (single gene or allele pairs) and are unaffected by environment, whereas polygenic inheritance and environmental sensitivity characterize continuous variation.
Distinguishing the genetic basis (monogenic vs. polygenic) and environmental interaction is essential to classifying variation types accurately.

Anahtar Kavram

Genetic basis and environmental stability of discontinuous variation
Soru 125Soru

In pea plants (*Pisum sativum*), the allele for green pod color (GG) is completely dominant over the allele for yellow pod color (gg). Which expected offspring phenotypic ratio corresponds to each parental cross?

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Öğeler

GG×ggGG \times gg
Gg×GgGg \times Gg
Gg×ggGg \times gg
gg×gggg \times gg

Eşleşmeler

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Cevap

Crossing homozygous dominant with homozygous recessive (GG×ggGG \times gg) yields 100% green pods; crossing two heterozygotes (Gg×GgGg \times Gg) produces a 3 green to 1 yellow phenotypic ratio; crossing a heterozygote with a homozygous recessive (Gg×ggGg \times gg) results in a 1 green to 1 yellow phenotypic ratio; and crossing two homozygous recessives (gg×gggg \times gg) yields 100% yellow pods.
According to Mendel's Law of Segregation, parental alleles segregate during gamete formation and combine at fertilization. A homozygous dominant crossed with homozygous recessive (GG×ggGG \times gg) produces only green offspring (GgGg). A monohybrid cross between two heterozygotes (Gg×GgGg \times Gg) produces 3 green (1GG+2Gg1 GG + 2 Gg) to 1 yellow (1gg1 gg). A testcross of a heterozygote (Gg×ggGg \times gg) produces a 1:1 ratio of green to yellow. Crossing two recessive parents (gg×gggg \times gg) results exclusively in yellow offspring.

Adım Adım Çözüm

1
Determine allele interaction and parent gametes for GG×ggGG \times gg.
The GGGG parent contributes only GG gametes, and the gggg parent contributes only gg gametes. All offspring have genotype GgGg.
Mendel's Law of Segregation states that allele pairs separate during gamete formation.
2
Determine phenotypes for GG×ggGG \times gg.
Since GG (green) is dominant to gg (yellow), all GgGg offspring are green (100% Green pods).
Heterozygotes express the dominant trait.
3
Construct Punnett square for Gg×GgGg \times Gg.
Genotypes: 1GG,2Gg,1gg1 GG, 2 Gg, 1 gg. Phenotypes: GGGG and GgGg are green, gggg is yellow.
Combining 1GG+2Gg=31 GG + 2 Gg = 3 green vs 1gg=11 gg = 1 yellow gives a 3:1 phenotypic ratio.
4
Analyze testcross Gg×ggGg \times gg.
Genotypes: 50%Gg50\% Gg and 50%gg50\% gg. Phenotypes: 1 Green : 1 Yellow.
The heterozygous parent contributes GG and gg in equal proportions (1:1).
5
Evaluate gg×gggg \times gg.
All offspring are gggg (100% Yellow pods).
No dominant GG allele is present in either parent.

Anahtar Kavram

Mendel's First Law and Monohybrid Cross Phenotypic Ratios
Soru 126Soru

Continuous variation in human morphological traits, such as adult height and body mass index, produces a smooth spectrum of intermediate phenotypes within a population because these traits are governed by polygenic inheritance and modified by environmental factors.

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Cevap: True

Cevap

The statement is True.
Continuous variations in humans involve quantitative traits where phenotypic traits range smoothly from one extreme to another. This occurs because multiple genes (polygenes) contribute additively to the trait, and environmental influences further smooth out differences between genotypes.

Adım Adım Çözüm

1
Identify the type of variation described for human height and body mass index.
Height and body weight display a spectrum of overlapping phenotypes across a population, which defines continuous variation.
Continuous traits show quantitative variation across a range rather than discrete qualitative classes.
2
Analyze the genetic and environmental basis of continuous human morphological variation.
Continuous traits are polygenic (controlled by many gene loci) and their expression is noticeably influenced by environmental factors such as diet and exercise.
Multiple genes provide an additive genetic foundation, while environmental conditions create fine gradations between genetic potential.
3
Evaluate the accuracy of the statement based on biological principles.
The statement correctly links continuous morphological traits to polygenic control and environmental modification.
The explanation aligns with the established definition and mechanisms of human continuous variation.

Anahtar Kavram

Polygenic inheritance and environmental impact on continuous human variations
Soru 127Soru

A Rhesus-negative (RhRh^-) woman marries a Rhesus-positive (Rh+Rh^+) man whose father was Rhesus-negative (RhRh^-). What is the probability that their second child will inherit the Rhesus-positive trait and be at risk of developing erythroblastosis fetalis?

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Cevap: 50%

Cevap

The probability that the child will be Rhesus-positive and at risk is 50%.
The RhRh^- mother has genotype rrrr, while the Rh+Rh^+ father has genotype RrRr because he inherited a recessive rr allele from his RhRh^- father. A monohybrid cross between RrRr and rrrr yields 50% RrRr (Rh+Rh^+) and 50% rrrr (RhRh^-). Because erythroblastosis fetalis occurs when an RhRh^- mother bears an Rh+Rh^+ fetus, there is a 50% chance that any child will inherit the Rh+Rh^+ phenotype and be at risk.

Adım Adım Çözüm

1
Determine the genotype of the mother
Since Rhesus negativity is an autosomal recessive trait, the RhRh^- mother must have the genotype rrrr.
Recessive phenotypes only express when homozygous.
2
Determine the genotype of the father
The father is Rh+Rh^+, so he carries at least one dominant allele (RR). Because his father was RhRh^- (rrrr), he must have inherited a recessive allele (rr). Thus, the father's genotype is RrRr.
An individual receives one allele from each parent.
3
Perform a test cross between the mother (rrrr) and father (RrRr)
The cross Rr×rrRr \times rr produces genotypes RrRr (50%) and rrrr (50%).
Punnett square analysis reveals half the offspring will inherit the RR allele from the father and rr from the mother.
4
Calculate the risk of erythroblastosis fetalis
Erythroblastosis fetalis occurs when an RhRh^- mother carries an Rh+Rh^+ fetus. The probability of the fetus being Rh+Rh^+ (RrRr) is 50%.
The risk applies only to Rh+Rh^+ offspring born to sensitized RhRh^- mothers.

Anahtar Kavram

Rhesus factor inheritance and hemolytic disease of the newborn (erythroblastosis fetalis)
Soru 128Soru

Match each non-Mendelian genetic concept on the left with its correct phenotypic expression pattern or population characteristic on the right.

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Öğeler

Incomplete Dominance
Codominance
Multiple Alleles

Eşleşmeler

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Cevap

Incomplete Dominance matches with the statement describing an intermediate phenotype distinct from both parents; Codominance matches with the statement describing simultaneous and full expression of both parental alleles without blending; Multiple Alleles matches with the statement describing more than two alternative gene forms existing at a single locus within a population.
Incomplete dominance produces a blended intermediate phenotype in heterozygotes. Codominance leads to the simultaneous, distinct expression of both alleles without blending. Multiple alleles refer to the existence of more than two gene variants at a given locus within a population.

Adım Adım Çözüm

1
Analyze Incomplete Dominance
In incomplete dominance, the heterozygous phenotype is a blend or quantitative intermediate between the two homozygous traits.
Neither allele completely masks the expression of the other, resulting in a distinct third phenotype.
2
Analyze Codominance
In codominance, both alleles are fully expressed alongside each other without producing an intermediate blend.
Both gene products are fully functional and observable simultaneously in the heterozygote.
3
Analyze Multiple Alleles
Multiple alleles describe genetic systems where three or more alleles exist at the same locus across a population.
While an individual carries at most two alleles, population gene pools can contain several alternative allele variants (e.g., IAI^A, IBI^B, and ii in ABO blood grouping).

Anahtar Kavram

Non-Mendelian Inheritance Modes
Soru 129Soru

A cytogenetic analysis of a patient reveals a somatic cell chromosomal complement of 4747 chromosomes with an additional sex chromosome (47,XXY47, XXY). Which mechanism is directly responsible for this structural or numerical chromosomal aberration, and what clinical condition does it produce?

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Cevap: Nondisjunction of sex chromosomes during meiotic division, resulting in Klinefelter syndrome

Cevap

Nondisjunction of sex chromosomes during meiotic division, resulting in Klinefelter syndrome
The correct option correctly attributes the karyotype 47,XXY47, XXY to nondisjunction during gametogenesis. When pair 23 fails to disjoin properly during meiosis, an egg carrying XXXX fertilized by a YY sperm (or an XX egg fertilized by an XYXY sperm) produces an individual with 4747 chromosomes (47,XXY47, XXY), which manifests as Klinefelter syndrome.

Adım Adım Çözüm

1
Analyze the given karyotype
The individual has 4747 chromosomes including an extra X chromosome (47,XXY47, XXY).
Normal human somatic cells have 4646 chromosomes (4444 autosomes + 22 sex chromosomes).
2
Identify the genetic mechanism leading to gain of a whole chromosome
The mechanism is meiotic nondisjunction, where chromosomes fail to separate properly during Anaphase I or II.
Gene mutations (substitutions, insertions, deletions) alter nucleotide sequences within a single gene but do not change the total number of chromosomes.
3
Match karyotype 47,XXY47, XXY to its clinical condition
An extra X chromosome in a male (47,XXY47, XXY) causes Klinefelter syndrome.
Trisomy 21 causes Down syndrome, whereas sex-chromosome trisomy 47,XXY47, XXY defines Klinefelter syndrome.

Anahtar Kavram

Chromosomal Aberrations and Nondisjunction
Tahmini Süre:1m 0s
Soru 130Soru

A biological study recorded two distinct human characteristics across a high school population: Trait X (skin pigmentation shade measured by spectrophotometry, exhibiting a continuous bell-shaped distribution curve) and Trait Y (phenylthiocarbamide tasting ability, where individuals strictly belong to either the taster or non-taster phenotype). Based on these phenotypic distribution patterns, which of the following statements correctly contrasts the underlying genetic mechanisms and environmental influences governing both traits?

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Cevap: Trait X is a continuous morphological variation governed by polygenic inheritance and modified by environmental factors, whereas Trait Y is a discontinuous physiological variation controlled by a single gene pair with negligible environmental modification.

Cevap

Trait X (skin pigmentation) is a continuous morphological variation under polygenic control and environmental influence, while Trait Y (PTC tasting) is a discontinuous physiological variation under single-gene control.
The correct answer accurately identifies Trait X (skin color spectrum) as a continuous morphological trait governed by multiple additive genes (polygenes) and modified by environment (sunlight), while correctly identifying Trait Y (PTC tasting) as a discontinuous physiological trait governed by a single pair of Mendelian alleles unaffected by environmental conditions.

Adım Adım Çözüm

1
Classify Trait X based on its phenotypic distribution curve
The continuous bell-shaped spectrum indicates quantitative (continuous) variation, which is characteristic of morphological features like skin color controlled by multiple gene pairs (polygenic) and influenced by solar exposure.
Continuous variation produces a gradient of phenotypes without clear-cut divisions.
2
Classify Trait Y based on its categorical grouping
Clear-cut categories (tasters vs non-tasters) without intermediate forms define qualitative (discontinuous) physiological variation, driven by a single gene pair (e.g., TAS2R38 gene locus).
Discontinuous variation results in distinct phenotypic classes unaffected by environmental factors.
3
Synthesize and contrast the genetic and environmental controls of both traits
Trait X is continuous, morphological, polygenic, and environmentally modified. Trait Y is discontinuous, physiological, monogenic, and environmentally stable.
Matching phenotypic frequency patterns with genetic basis is key to differentiating human variation types.

Anahtar Kavram

Classification of continuous versus discontinuous morphological and physiological human variations
Soru 131Soru

In genetic analysis of diploid organisms, the specific combination of alleles inherited by an individual for a particular gene (such as TtTt) determines its observable physical characteristics. Which pair of terms correctly identifies the symbolic genetic makeup of the organism (TtTt) and its state of having two different alleles for that gene, respectively?

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Cevap: Genotype and heterozygous

Cevap

The correct pair of terms is genotype for the genetic constitution and heterozygous for the presence of two different alleles.
The term genotype refers specifically to the allele combination (TtTt) of an organism, while heterozygous describes an individual carrying two different alleles (TT and tt) for a specific locus.

Adım Adım Çözüm

1
Identify the term for the genetic constitution
The total genetic constitution of an individual, represented by allele symbols like TtTt, is called its genotype.
Genotype specifies the precise allele combination inherited from the parents.
2
Determine the state of the alleles in TtTt
Because TtTt consists of two non-identical alleles (TT and tt), the organism is in a heterozygous condition.
A homozygous condition requires two identical alleles (e.g., TTTT or tttt).

Anahtar Kavram

Genotype vs Phenotype and Homozygous vs Heterozygous terminology
Soru 132Soru

A man with normal blood clotting marries a phenotypically normal woman whose father had hemophilia A, an X-linked recessive disorder. What is the probability that any child born to this couple will be a carrier of the hemophilia allele?

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Cevap: 25%

Cevap

The probability that any child born to this couple will be a carrier of the hemophilia allele is 25%.
Since the woman's father had hemophilia (XhYX^h Y), she inherited his XhX^h allele and is a carrier (XHXhX^H X^h). When crossed with a normal male (XHYX^H Y), four total offspring genotypes are produced with equal probability: normal female (XHXHX^H X^H), carrier female (XHXhX^H X^h), normal male (XHYX^H Y), and affected male (XhYX^h Y). Only XHXhX^H X^h individuals are carriers, representing 1 out of 4 total possible outcomes, or 25%.

Adım Adım Çözüm

1
Determine the parental genotypes from the pedigree description.
Father = XHYX^H Y, Mother = XHXhX^H X^h.
Because the woman's father had hemophilia (XhYX^h Y), she must have inherited his affected XhX^h chromosome, making her a heterozygous carrier (XHXhX^H X^h).
2
Construct a genetic cross between XHYX^H Y and XHXhX^H X^h.
The possible offspring genotypes are XHXHX^H X^H (25%), XHXhX^H X^h (25%), XHYX^H Y (25%), and XhYX^h Y (25%).
A Punnett square combines the maternal gametes (XH,XhX^H, X^h) and paternal gametes (XH,YX^H, Y) in equal proportions.
3
Identify the carrier genotype among total offspring possibilities.
1 out of 4 total possibilities is XHXhX^H X^h, which equals 25%.
Carrier status requires one recessive allele on an X chromosome in a phenotypically normal female (XHXhX^H X^h).

Anahtar Kavram

X-linked recessive inheritance and offspring probability calculation
Soru 133Soru

Down syndrome is a genetic disorder caused by a single nucleotide point mutation in a nuclear gene.

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Cevap: False

Cevap

False
The statement is false because Down syndrome is caused by a chromosomal aberration (trisomy 21 due to non-disjunction), not by a single nucleotide gene mutation.

Adım Adım Çözüm

1
Identify the genetic cause of Down syndrome
Down syndrome is caused by non-disjunction of chromosome 21 during meiosis, leading to trisomy 21 (47 chromosomes in total).
Down syndrome involves an extra chromosome set rather than a change in a single gene sequence.
2
Classify the mutation type
An alteration in total chromosome count is classified as a numerical chromosomal aberration, not a gene (point) mutation.
Gene mutations involve point alterations, insertions, or deletions within a specific gene sequence, whereas chromosomal aberrations affect whole chromosomes or large segments.

Anahtar Kavram

Distinction between gene mutations and chromosomal aberrations
Soru 134Soru

A color-blind man marries a phenotypically normal woman whose mother was color-blind. What is the probability that any female child born to this couple will be color-blind?

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Cevap: 50%50\%

Cevap

The probability that a female child born to this couple will be color-blind is 50%50\%.
The father has the genotype XcYX^c Y and passes his XcX^c chromosome to all female children. The mother is phenotypically normal but carries the allele from her color-blind mother, making her genotype XCXcX^C X^c. Half of her egg cells carry XCX^C and half carry XcX^c. Therefore, 50%50\% of female offspring receive XcX^c from both parents (XcXcX^c X^c) and are color-blind.

Adım Adım Çözüm

1
Determine the genotypes of the parents.
The father is color-blind (XcYX^c Y). The mother is phenotypically normal but her mother was color-blind (XcXcX^c X^c), meaning the mother must be a carrier (XCXcX^C X^c).
Sex-linked recessive traits on the X chromosome require identifying maternal and paternal allele contributions.
2
Perform a genetic cross for female offspring.
Female children inherit XcX^c from the father and either XCX^C or XcX^c from the mother, resulting in genotypes XCXcX^C X^c (carrier) and XcXcX^c X^c (color-blind) in a 1:11:1 ratio.
Determining the phenotypic ratio specifically among female offspring requires considering only the XX combinations.
3
Calculate the probability for female offspring.
11 out of 22 female children (50%50\%) will have the XcXcX^c X^c genotype and express color blindness.
The question asks specifically for the probability among female children.

Anahtar Kavram

Sex-Linked Inheritance and Female Phenotypic Probability
Soru 135Soru

An agricultural breeder crosses two pure-breeding lines of oil palm: Line A, which produces high oil yield but is highly susceptible to fungal wilt, and Line B, which produces low oil yield but is completely resistant to fungal wilt. The resulting F1F_1 hybrids exhibit exceptional vigor, producing higher oil yields and stronger disease resistance than either parent. However, when these F1F_1 hybrid palms are self-pollinated, the F2F_2 generation displays wide phenotypic variation, with many individuals showing reduced yield and high disease susceptibility. Which of the following biological principles best explains the performance decline observed in the F2F_2 generation?

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Cevap: The segregation and independent assortment of alleles during meiosis break down high heterozygosity, unmasking deleterious recessive genes in homozygous F2F_2 offspring.

Cevap

The decline in performance (loss of heterosis) in the F2F_2 generation is caused by the segregation and independent assortment of alleles during meiosis, which reduces heterozygosity and leads to the expression of harmful homozygous recessive genotypes.
The superior performance of F1F_1 hybrids is due to heterosis (hybrid vigor), which relies on a high degree of heterozygosity masking harmful recessive alleles. When F1F_1 plants self-pollinate, meiotic segregation segregates these alleles into homozygous combinations (aaaa or bbbb). This unmasks undesirable recessive traits, causing a breakdown of hybrid vigor and wide phenotypic variation in the F2F_2 generation.

Adım Adım Çözüm

1
Identify the genetic principle demonstrated by the F1F_1 generation
Crossing two genetically distinct inbred lines produces F1F_1 hybrids that exhibit superior traits (hybrid vigor or heterosis) due to optimal heterozygous allele combinations (AaBbAaBb).
Heterosis masks harmful recessive alleles present in the parental lines.
2
Analyze what happens genetically during F1×F1F_1 \times F_1 self-pollination
Meiosis in F1F_1 parents causes segregation and independent assortment of homologous chromosomes, producing gametes with different allele combinations.
Recombination leads to homozygous genotypes (AAAA, aaaa, BBBB, bbbb) in the F2F_2 generation.
3
Determine the phenotypic outcome in the F2F_2 generation
Homozygosity increases across the population, unmasking deleterious recessive alleles and leading to inbreeding depression and loss of hybrid vigor.
Commercial breeders must produce fresh F1F_1 seeds every season rather than saving F2F_2 seeds.

Anahtar Kavram

Hybrid Vigor (Heterosis) and Inbreeding Depression in Agricultural Crop Breeding
Soru 136Soru

In rabbits, the allele for black coat color (BB) is completely dominant over the allele for brown coat color (bb). A black-furred rabbit mated with a brown-furred rabbit and produced 28 black-furred offspring and 26 brown-furred offspring over several litters. Which of the following represents the most likely genotypes of the parent rabbits?

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Cevap: BbBb and bbbb

Cevap

The black-furred parent is heterozygous (BbBb) and the brown-furred parent is homozygous recessive (bbbb).
The observed offspring ratio of 28 black to 26 brown is approximately a 1:1 ratio. In monohybrid inheritance, a 1:1 phenotypic outcome occurs when a heterozygous dominant individual (BbBb) is crossed with a homozygous recessive individual (bbbb). The brown parent must contribute a recessive allele (bb) to every offspring, so the presence of brown offspring (bbbb) proves that the black parent must also carry a hidden recessive allele (BbBb).

Adım Adım Çözüm

1
Determine the phenotype and genotype of the recessive parent
Since brown coat color is recessive (bb), the brown parent must be homozygous recessive (bbbb).
Recessive traits are only expressed phenotypically when two copies of the recessive allele are present.
2
Analyze the ratio of the offspring phenotypes
The offspring ratio is 28 black : 26 brown, which approximates a 1:1 ratio (1/21/2 black : 1/21/2 brown).
A 1:1 offspring ratio in a monohybrid cross indicates a test cross between a heterozygous dominant individual and a homozygous recessive individual.
3
Verify parental genotypes using a Punnett square
Crossing Bb×bbBb \times bb produces gametes BB and bb from the black parent, and bb from the brown parent, yielding 50% Bb50\%\ Bb (black) and 50% bb50\%\ bb (brown).
According to Mendel's First Law of Segregation, alleles segregate during gamete formation so each gamete carries only one allele for each gene.

Anahtar Kavram

Mendel's First Law and Monohybrid Test Cross Ratios
Tahmini Süre:1m 0s
Soru 137Soru

A medical survey categorizes human traits according to their biological nature (morphological versus physiological) and their population distribution pattern (continuous versus discontinuous). Which of the following pairings correctly matches a physiological trait exhibiting discontinuous variation with a morphological trait exhibiting continuous variation?

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Cevap: Ability to taste phenylthiocarbamide (PTC) and adult body height

Cevap

Ability to taste phenylthiocarbamide (PTC) and adult body height
The ability to taste phenylthiocarbamide (PTC) is a physiological trait involving chemical receptor function that splits the population into clear qualitative classes (tasters and non-tasters, showing discontinuous variation). Adult body height is a structural feature (morphological trait) influenced by multiple gene pairs and environmental factors, producing a continuous spectrum of quantitative values.

Adım Adım Çözüm

1
Classify traits by biological type (morphological vs. physiological)
PTC tasting ability and blood pressure are physiological traits (functional/biochemical), whereas fingerprint patterns, earlobe attachment, and height are morphological traits (physical form/structure).
Physiological traits pertain to internal function and biochemistry, whereas morphological traits pertain to physical appearance and body form.
2
Analyze distribution patterns (continuous vs. discontinuous)
PTC tasting is discontinuous (clear-cut categories without intermediate forms). Adult height is continuous (controlled by polygenes and modified by environment, forming a smooth gradient).
Discontinuous traits are typically monogenic with distinct categories, while continuous traits are polygenic with a range of intermediate values.
3
Match the required pairing conditions
The pairing of PTC tasting ability (physiological, discontinuous) with adult body height (morphological, continuous) fulfills both conditions perfectly.
This is the only pair containing a physiological discontinuous trait alongside a morphological continuous trait.

Anahtar Kavram

Classification of Human Morphological and Physiological Variations by Pattern of Inheritance
Soru 138Soru

Which of the following mutagenic agents primarily induces gene mutations by causing adjacent thymine bases on a single DNA strand to covalently bond and form pyrimidine dimers?

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Cevap: Ultraviolet radiation

Cevap

Ultraviolet radiation is the mutagenic agent that causes adjacent thymine bases on a DNA strand to form pyrimidine dimers.
Ultraviolet radiation is absorbed strongly by pyrimidines, promoting covalent linkage between adjacent thymine bases on the same DNA strand. This forms thymine dimers that block normal DNA replication and transcription.

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1
Identify the primary mechanism of action described in the stem.
The mechanism is the formation of intrastrand thymine (pyrimidine) dimers due to absorption of radiant energy.
Different physical and chemical mutagens operate through distinct biochemical pathways on nucleic acids.
2
Evaluate physical versus chemical mutagens.
Non-ionizing ultraviolet (UV) light specifically excites pyrimidine rings (thymine and cytosine), leading to covalent dimer formation.
UV light lacks the energy to ionize atoms or break double strands, but has sufficient energy to alter nitrogenous base bonding.

Anahtar Kavram

Physical mutagens and the molecular mechanism of UV-induced pyrimidine dimerization
Soru 139Soru

Nitrous acid (HNO2\text{HNO}_2) acts as a chemical mutagen by inducing the oxidative deamination of cytosine, converting it into uracil (UU). If a double-stranded DNA molecule containing a newly formed uracil base undergoes two consecutive rounds of DNA replication without prior enzymatic repair, which specific base-pair substitution will ultimately be fixed in the mutant daughter DNA molecule?

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Cevap: A CGTAC\cdot G \rightarrow T\cdot A transition mutation

Cevap

A CGTAC\cdot G \rightarrow T\cdot A transition mutation will be fixed in the daughter DNA molecule.
Deamination of cytosine creates uracil. Uracil pairs with adenine during the first replication round. In the subsequent round, adenine pairs with thymine, substituting the original cytosine-guanine pair with a thymine-adenine pair (CGTAC\cdot G \rightarrow T\cdot A).

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1
Identify the primary chemical effect of nitrous acid on cytosine.
Nitrous acid removes an amino group (oxidative deamination) from cytosine, turning it into uracil (UU).
Uracil is a nitrogenous base with hydrogen-bonding properties similar to thymine.
2
Trace the base pairing during the first cycle of DNA replication.
The template strand containing uracil (UU) pairs with adenine (AA) during DNA synthesis, producing a UAU-A base pair intermediate.
DNA polymerase recognizes uracil as equivalent to thymine during template reading.
3
Trace the base pairing during the second cycle of DNA replication.
The strand with adenine (AA) serves as a template and pairs with thymine (TT), producing a stable ATA-T (or TAT-A) double helix.
Standard complementary base pairing pairs adenine with thymine in normal DNA.
4
Compare the original wild-type base pair with the final mutant base pair.
The original CGC\cdot G base pair has been replaced by a TAT\cdot A base pair. Because a pyrimidine-purine pair is replaced by another pyrimidine-purine pair, it is classified as a transition mutation.
Point mutations that exchange a pyrimidine for a pyrimidine (CTC \rightarrow T) are transitions.

Anahtar Kavram

Chemical Mutagenesis and Base-Pair Transition Mutations
Tahmini Süre:2m 0s
Soru 140Soru

Agricultural breeders often apply Mendelian genetics to select desirable traits in farm animals. A livestock breeder crosses two cattle that are both heterozygous for high milk production (MmMm) and disease resistance (RrRr). Assuming independent assortment of these dominant autosomal traits, what proportion of the offspring is expected to display both high milk production and disease resistance?

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Cevap: 916\frac{9}{16}

Cevap

916\frac{9}{16} of the offspring will display both dominant traits (high milk production and disease resistance).
Crossing two dihybrid individuals (MmRr×MmRrMmRr \times MmRr) produces offspring with a phenotypic ratio of 9 (both dominant) : 3 (dominant/recessive) : 3 (recessive/dominant) : 1 (both recessive). Therefore, 916\frac{9}{16} of the offspring inherit both high milk production and disease resistance.

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1
Determine the probability of expressing the dominant phenotype for the first trait (milk production).
From Mm×MmMm \times Mm, the probability of high milk yield (M_M\_) is 34\frac{3}{4}.
Genotypes MMMM, MmMm, and MmMm out of MM,Mm,Mm,mmMM, Mm, Mm, mm express the dominant phenotype.
2
Determine the probability of expressing the dominant phenotype for the second trait (disease resistance).
From Rr×RrRr \times Rr, the probability of disease resistance (R_R\_) is 34\frac{3}{4}.
Genotypes RRRR, RrRr, and RrRr out of RR,Rr,Rr,rrRR, Rr, Rr, rr express the dominant phenotype.
3
Apply the product rule of probability for independent assortment.
34×34=916\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}.
The inheritance of milk yield and disease resistance assort independently during gamete formation.

Anahtar Kavram

Selective breeding using Mendelian dihybrid inheritance principles to combine desirable agricultural traits.
Tahmini Süre:1m 0s
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