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inheritance

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卡片总数: 24内容版本: v4公开卡包更新时间: 8/1/2026

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#1
正面 (问题)

State three causes of genetic variation

背面 (解答)

Mutation Crossing over Independent segregation / assortment (of homologous chromosomes) Random fusion of gametes / fertilisation / mating

#2
正面 (问题)

What is meant by a genome?

背面 (解答)

(All) the DNA in a cell/organism; ‘(all) the ‘genes’/alleles’ ‘genetic material/code’ in a cell/organism/ person’ ‘the total number of DNA bases in a cell/organism

#3
正面 (问题)

In genetic crosses, the observed phenotypic ratios obtained in the offspring are often not the same as the expected ratios. Suggest two reasons why.

背面 (解答)

Small sample size; Fusion/fertilisation of gametes is random; Linked Genes; Sex-linkage / crossing over; Epistasis; Lethal genotypes;

#4
正面 (问题)

How do multiple alleles of a gene arise?

背面 (解答)

mutations; which are different / at different positions in the gene;

#5
正面 (问题)

What is a gene pool?

背面 (解答)

All the alleles in a population;

#6
正面 (问题)

What does Hardy Weinberg’s equation predict and assumption

背面 (解答)

The frequency/proportion of alleles (of a particular gene); Will stay constant from one generation to the next/over generations / no genetic change over time; Providing no mutation/no selection/population large/population genetically isolated/mating at random/no migration;

#7
正面 (问题)

What is meant by a recessive allele?

背面 (解答)

Only expressed in the homozygote / not expressed in the heterozygote / not expressed if dominant present;1

#8
正面 (问题)

Define gene linkage

背面 (解答)

(Genes/loci) on same chromosome;

#9
正面 (问题)

Define codominance

背面 (解答)

Both alleles expressed in the phenotype;

#10
正面 (问题)

Describe why observed phenotypes don’t match expected values

背面 (解答)

Fertilisation is random OR Fusion of gametes is random; Small/not-large population/sample; Selection advantage/disadvantage/lethal alleles;

#11
正面 (问题)

Define epistasis

背面 (解答)

The allele of one gene affects or masks the expression of another in the phenotype;

#12
正面 (问题)

Expected offspring phenotype ratios from heterozygous parents: Monohybrid Dihybrid Epistasis Autosomal linkage

背面 (解答)

Dominant : recessive Monohybrid 3:1 Dihybrid 9:3:3:1 Epistasis 9:4:3 or 15:1 or 9:7 Autosomal linkage 3:1 (no x over) (no other pattern other than 4 phenotypes with recombination of alleles)

#13
正面 (问题)

Male offspring are more likely than females to show recessive sex-linked characteristics. Explain why.

背面 (解答)

(Recessive) allele is always expressed in males / males have one (recessive) allele; Females need two recessive alleles / females need to be homozygous recessive / females could have dominant and recessive alleles / be heterozygous;

#14
正面 (问题)

Rules for recessive alleles

背面 (解答)

Unaffected parents can have an affected offspring (if they are Heterozygous)

#15
正面 (问题)

Rules for Dominant alleles

背面 (解答)

Affected offspring MUST have at least one affected parent. Unaffected parents ONLY have unaffected offspring. If both parents are affected and have an unaffected offspring, both parents must be Heterozygous

#16
正面 (问题)

Genotype:

背面 (解答)

The genetic constitution of an organism (All the alleles that an organism

#17
正面 (问题)

Explain how a single base substitution causes a change in the structure of a polypeptide

背面 (解答)

Change in (sequence of) amino acid(s)/primary structure; Change in hydrogen/ionic/disulfide bonds; Alters tertiary/30 structure;

#18
正面 (问题)

What is meant by the term phenotype (

背面 (解答)

Expression / appearance / characteristic due to) genetic constitution / genotype / allele(s); (Expression / appearance / characteristic due to) interaction with environment;

#19
正面 (问题)

2 equations in hardy weinburg

背面 (解答)

(alleles);P2 (homozygous dominant ) + 2Pq + q2 (homozygous recessive)= 1.0 P + q = 1.0 (genes)

#20
正面 (问题)

In human populations, cystic fibrosis is caused by a recessive allele. People who are unaffected maybe homozygous dominant (FF) or heterozygous (Ff) but anyone with the disease must be homozygous recessive (ff) In a population of 10 000 people, 5 people had the disease and the rest did not. How many of these people were heterozygotes?

背面 (解答)

The frequency of the disease is 5 in 10 000 = = 0.0005 i.e. the frequency of homozygous recessive individuals = q2 = 0.0005 Therefore the frequency of the recessive allele = q = Therefore the frequency of the dominant allele = P = 1 – q = 1 – 0.22 = 0.978 Therefore the frequency of heterozygous individuals (2Pq) = 2 x 0.978 x 0.022 2Pq= 0.044 Therefore the number of heterozygotes = 10 000 x 0.044 = 440

#21
正面 (问题)

assumptions made in hardy weinburg

背面 (解答)

⦁ Migration, i.e. the gene pool is isolated and there is no flow of other alleles in or out ⦁ Gene mutations ⦁ Selection for or against a particular allele IN ADDITION: ⦁ There should be a large population ⦁ Mating within the population should be random (all individuals have an equal chance of mating)

#22
正面 (问题)

A student investigated the monohybrid inheritance of eye shape in fruit flies. Two fruit flies with bar (narrow) eyes were crossed. Of the offspring, 1538 had bar eyes and 462 had round (normal) eyes. (a) Using suitable symbols, give the genotypes of the parents. (2) b)The ratio of bar-eyed flies and round-eyed flies in the student’s results were not the same as the ratio she had expected.(1) c)What ratio of bar-eyed to round-eyed flies was the student expecting? (2) d)The student wished to test her results with the ones she had expected. Which statistical test should she use

背面 (解答)

a)Bb / suitable equivalent; Both parents have bar eyes, but have some offspring with round eyes, so parents must be carriers of recessive allele for round eyes; b )3:1; (c) Fertilisation is random OR Fusion of gametes is random; Small / not large population / sample; Selection advantage / disadvantage / lethal alleles; d)χ2 / chi squared;

#23
正面 (问题)

This fruit fly has another characteristic controlled by a pair of codominant alleles, WN and WV. What is meant by codominant alleles?

背面 (解答)

Both alleles expressed in the phenotype (if both are present);

#24
正面 (问题)

In another population of 950 fruit flies, the frequency of the WV allele was 0.2. Use the Hardy-Weinberg equation to calculate the number of insects that would be expected to have the genotype WNWV.

背面 (解答)

304